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Sympy strange integral output

I'm trying to solve an integral with sympy. But it gives me a wrong solution. Why?
import sympy
from sympy import Integral, exp, oo
x, y = sympy.symbols("x y", real=True)
b, u, l, t = sympy.symbols("b u l t ", real=True, positive=True)
Fortet = Integral(exp(-l * t) * (sympy.sqrt(2 * sympy.pi * t)) ** (-1) * exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, oo))
Fortet.doit()
Result (wrong):
Piecewise((-(-b/2 + y)*sqrt(2*l +
u**2)*(-sqrt(pi)*sinh(sqrt(2)*sqrt(b)*sqrt(l +
u**2/2)*sqrt(polar_lift(1 + y**2/(b*polar_lift(b -
2*y))))*sqrt(polar_lift(b - 2*y))) +
sqrt(pi)*cosh(sqrt(2)*sqrt(b)*sqrt(l + u**2/2)*sqrt(polar_lift(1 +
y**2/(b*polar_lift(b - 2*y))))*sqrt(polar_lift(b - 2*y))))*exp(b*u -
u*y)/(sqrt(pi)*(b - 2*y)*(l + u**2/2)), Abs(arg(1 +
y**2/(b*polar_lift(b - 2*y))) + arg(b - 2*y)) <= pi/2),
(Integral(sqrt(2)*exp(-l*t)*exp(-(b - t*u -
y)**2/(2*t))/(2*sqrt(pi)*sqrt(t)), (t, 0, oo)), True))
Expected (correct) solution:
Solution = (exp((-u)*(b - y)) * exp(sympy.sqrt(u**2 + 2*l)*(b-y)))/(sympy.sqrt(2*l + u**2)) #RIGHT solution

Both results are in fact the same. The first one is probably slightly more correct. You tend see these polar_lift functions whenever SymPy tries to do something like square rooting something when it does not know the signs of the things inside (after integrating)
A polar_lift does not appear below, but this basic Gaussian example shows that SymPy tries to be as general as possible:
from sympy import *
x = Symbol("x", real=True)
y = Symbol("y", real=True)
s = Symbol("s", real=True) # , positive=True
gaussian = exp(-((x-y)**2)/(2*(s**2)))
nfactor = simplify(integrate(gaussian, (x,-oo,oo)))
print(nfactor)
You need s to be declared as positive: s = Symbol("s", real=True, positive=True). A similar thing happens with these kinds of polar_lift(b - 2*y) functions in your example. It also happens with the question I reference below.
I have no idea why, but N(polar_lift(x)) for any float or int x gives x again; yet, SymPy does not simplify nicely with symbolic x. Turns out if you keep on simplifying, you get nicer and nicer looking answers. I couldn't find anything about polar_lift related to pure math so I don't know what it actually does.
Remember for the simple example above how it gave a piece-wise? Same thing here. So we just take the first piece since the second piece is an un-evaluated integral.
In the code below, I use this question to remove the piece-wise function and then I simplify twice. And finally, I manually remove the polar_lift.
import sympy as sp
x, y = sp.symbols("x y", real=True)
b, u, l, t = sp.symbols("b u l t ", real=True, positive=True)
Fortet = sp.integrate(sp.exp(-l * t) * (sp.sqrt(2 * sp.pi * t)) ** (-1) *
sp.exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, sp.oo), conds='none')
incorrect = Fortet.simplify().simplify()
correct = eval(str(incorrect).replace("polar_lift", ""))
correct = correct.factor()
print(correct)
The result is:
exp(b*u)*exp(-u*y)*exp(-sqrt(2*l + u**2)*sqrt(b**2 - 2*b*y + y**2))/sqrt(2*l + u**2)
That is close enough to your expression. I couldn't make SymPy simplify the sqrt(b**2 - 2*b*y + y**2) to Abs(b-y) no matter how hard I tried.
Note that either SymPy is still wrong or you are wrong since the powers in the numerator are opposite. So I checked on the Desmos for a numeric answer (top one is yours):

How to evaluate this line integral (Python-Sympy)

The idea is to compute the line integral of the following vector field and curve:
This is the code I have tried:
import numpy as np
from sympy import *
from sympy import Curve, line_integrate
from sympy.abc import x, y, t
C = Curve([cos(t) + 1, sin(t) + 1, 1 - cos(t) - sin(t)], (t, 0, 2*np.pi))
line_integrate(y * exp(x) + x**2 + exp(x) + z**2 * exp(z), C, [x, y, z])
But the ValueError: Function argument should be (x(t), y(t)) but got [cos(t) + 1, sin(t) + 1, -sin(t) - cos(t) + 1] comes up.
How can I compute this line integral then?
I think that maybe this line integral contains integrals that don't have exact solution. It is also fine if you provide a numerical approximation method.
Thanks

In this case you can compute the integral using line_integrate because we can reduce the 3d integral to a 2d one. I'm sorry to say I don't know python well enough to write the code, but here's the drill:
If we write
C(t) = x(t),y(t),z(t)
then the thing to notice is that
z(t) = 3 - x(t) - y(t)
and so
dz = -dx - dy
So, we can write
F.dr = Fx*dx + Fy*dy + Fz*dz
= (Fx-Fz)*dx + (Fy-Fz)*dy
So we have reduced the problem to a 2d problem: we integrate
G = (Fx-Fz)*i + (Fx-Fz)*j
round
t -> x(t), y(t)
Note that in G we need to get rid of z by substituting
z = 3 - x - y

The value error you receive does not come from your call to the line_integrate function; it comes because according to the source code for the Curve class, only functions in 2D Euclidean space are supported. This integral can still be computed without using sympy according to this research blog that I found by simply searching for a workable method on Google.
The code you need looks like this:
import autograd.numpy as np
from autograd import jacobian
from scipy.integrate import quad
def F(X):
x, y, z = X
return [y * np.exp(x), x**2 + np.exp(x), z**2 * np.exp(z)]
def C(t):
return np.array([np.cos(t) + 1, np.sin(t) + 1, 1 - np.cos(t) - np.sin(t)])
dCdt = jacobian(C, 0)
def integrand(t):
return F(C(t)) # dCdt(t)
I, e = quad(integrand, 0, 2 * np.pi)
The variable I then stores the numerical solution to your question.

You can define a function:
import sympy as sp
from sympy import *
def linea3(f,C):
P = f[0].subs([(x,C[0]),(y,C[1]),(z,C[2])])
Q = f[1].subs([(x,C[0]),(y,C[1]),(z,C[2])])
R = f[2].subs([(x,C[0]),(y,C[1]),(z,C[2])])
dx = diff(C[0],t)
dy = diff(C[1],t)
dz = diff(C[2],t)
m = integrate(P*dx+Q*dy+R*dz,(t,C[3],C[4]))
return m
Then use the example:
f = [x**2*z**2,y**2*z**2,x*y*z]
C = [2*cos(t),2*sin(t),4,0,2*sp.pi]

Sympy Subs not replacing a symbol when its power is being replaced

I'm trying to make some basic substitutions but SymPy doesn't want to help me out
x, y, z, k = symbols("x y z k", positive=True, real=True)
exp = x**4 + x**3 + x**2 + x
what_im_expecting = simplify(y**(Rational(1/4)) + y**(Rational(3/4)) + sqrt(y) + y)
what_i_actually_get = exp.subs(x**4,y)
exp, what_i_actually_get, what_im_expecting
returns
x + y**(Rational(3, 4)) + sqrt(y) + y
Can anyone help me out?
a more complex example:

The method subs can be trusted to replace the terms that exactly match the given "old" expression, which x**4 here. The replacement of other things related to x**4 is not so certain. (There are many open issues with subs: some say it substitutes too much, some say too little.) There is some substitution logic specific to powers, but x by itself is not formally a power, so it escapes that logic. A workaround: temporarily replace x by x**1, preventing automatic evaluation of that power to x.
x1 = sp.Pow(x, 1, evaluate=False)
subbed = exp.subs(x, x1).subs(x**4, y).subs(x1, x)
Now subbed is y**(3/4) + y**(1/4) + sqrt(y) + y.
But, don't expect human-like ingenuity from subs. With the same workaround, trying to do subs(x**4 - 1, y) results in x**3 + x**2 + x + y + 1: nothing like sqrt(y+1), etc, appears. It's better to substitute in the most direct way possible:
subs(x, (y+1)**Rational(1, 4))
Then you don't need any workarounds.

Assuming positive real part of a complex number in Sympy

I cannot figure it out how to assume positive real part of a complex number in Sympy.
Example of an Mathematica code:
a = InverseFourierTransform[ R/(I omega - lambda) + Conjugate[R]/(I omega - Conjugate[lambda]), omega, t, FourierParameters -> {1, -1}]
Simplify[a, {Re[lambda] < 0, t > 0}]
Similar Sympy code:
import sympy as sym
sym.init_printing()
ω = sym.symbols('omega', real=True, positive=True)
R, λ = sym.symbols('R, lambda', complex=True)
t = sym.symbols('t', real=True, positive=True)
α = R/(sym.I*ω-λ)+sym.conjugate(R)/(sym.I*ω-sym.conjugate(λ))
sym.inverse_fourier_transform(α, ω, t)
How could I assume real part of lambda to be positive? If I assume lambda to have positive=True, then sympy assumes imaginary=False.
Any ideas?

Create two real symbols x, y, assume x positive, and let λ be x + I*y.
import sympy as sym
ω, x, t = sym.symbols('omega x t', positive=True)
y = sym.symbols('y', real=True)
R = sym.symbols('R')
λ = x + sym.I*y
α = R/(sym.I*ω-λ)+sym.conjugate(R)/(sym.I*ω-sym.conjugate(λ))
res = sym.inverse_fourier_transform(α, ω, t)
The result is
2*pi*R*exp(2*pi*t*(x + I*y)) + 2*pi*exp(2*pi*t*(x - I*y))*conjugate(R)
You can then return to single-symbol λ with substitution:
λ = sym.symbols('lambda')
res.subs(x + sym.I*y, λ).conjugate().subs(x + sym.I*y, λ).conjugate()
obtaining
2*pi*R*exp(2*pi*lambda*t) + 2*pi*exp(2*pi*t*conjugate(lambda))*conjugate(R)
(The trick with two conjugations is needed because subs isn't going to replace x - I*y with conjugate(lambda) otherwise.)
Remarks on assumptions
complex=True is redundant; real numbers are included in complex numbers (7 is a complex number), so this has no effect
real=True is redundant when positive=True is given

What does Sympys' doit() does on derivatives?

I'm trying to compute higher derivatives of f(y(x)) w.r.t to x using sympys diff.
from sympy import *
from IPython.display import display
init_printing(use_latex=True)
x = symbols('x')
f, y = symbols('f, y', cls=Function)
d2 = diff(f(y(x)),x,2)
print(d2)
print(d2.doit())
Sympy returns :
Derivative(y(x), x)**2*Derivative(f(y(x)), y(x), y(x)) + Derivative(y(x), x, x)*Subs(Derivative(f(_xi_1), _xi_1), (_xi_1,), (y(x),))
Derivative(f(y(x)), y(x))*Derivative(y(x), x, x) + 2*Derivative(y(x), x)**2*Derivative(f(y(x)), y(x), y(x))
Latex image: Sympy result.
While the first result seems to be correct, I do not understand the factor 2 in the second expression after the doit() operation.

It looks like you stumbled upon a bug, which was just fixed a few weeks ago.
You can test this by substituting f, y, and x with some functions or values respectively (append to your code):
f_ex = Lambda(x, x**2)
y_ex = Lambda(x, sin(x))
x_ex = 2
substitutions = [ (f,f_ex), (y,y_ex), (x,x_ex) ]
print( d2.subs(substitutions).doit().n() ) #-1.30728724172722
print( d2.doit().subs(substitutions).doit().n() ) #-0.960930862590836
The printed values should be the same.
The issue can be further isolated to:
print((Derivative(f(y(x)), x, x)))
print((Derivative(f(y(x)), y(x), y(x))).doit())
Here, a plain doit simply adds a factor of 2, which is obviously wrong.