3.74.54 \(\int \frac {2 x-x^2+e^x (-4 x+2 x^2-2 x^3)+(-1+e^x (2-2 x)) \log (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})}{x^4+e^x (-2 x^4+2 x^5)+(-2 x^2+e^x (4 x^2-4 x^3)) \log (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})+(1+e^x (-2+2 x)) \log ^2(\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} (8-16 x+8 x^2)})} \, dx\)

Optimal. Leaf size=29 \[ \frac {x}{x^2-\log \left (\frac {1}{2 \left (2-e^{-x}-2 x\right )^2}\right )} \]

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Rubi [F]  time = 3.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2 x-x^2+e^x \left (-4 x+2 x^2-2 x^3\right )+\left (-1+e^x (2-2 x)\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )}{x^4+e^x \left (-2 x^4+2 x^5\right )+\left (-2 x^2+e^x \left (4 x^2-4 x^3\right )\right ) \log \left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )+\left (1+e^x (-2+2 x)\right ) \log ^2\left (\frac {e^{2 x}}{2+e^x (-8+8 x)+e^{2 x} \left (8-16 x+8 x^2\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x - x^2 + E^x*(-4*x + 2*x^2 - 2*x^3) + (-1 + E^x*(2 - 2*x))*Log[E^(2*x)/(2 + E^x*(-8 + 8*x) + E^(2*x)*(
8 - 16*x + 8*x^2))])/(x^4 + E^x*(-2*x^4 + 2*x^5) + (-2*x^2 + E^x*(4*x^2 - 4*x^3))*Log[E^(2*x)/(2 + E^x*(-8 + 8
*x) + E^(2*x)*(8 - 16*x + 8*x^2))] + (1 + E^x*(-2 + 2*x))*Log[E^(2*x)/(2 + E^x*(-8 + 8*x) + E^(2*x)*(8 - 16*x
+ 8*x^2))]^2),x]

[Out]

-2*Defer[Int][(x^2 - Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^(-2), x] - 2*Defer[Int][1/((-1 + x)*(x^2 - Log[E
^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^2), x] - 2*Defer[Int][x^2/(x^2 - Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^
2, x] + 2*Defer[Int][1/((1 - 2*E^x + 2*E^x*x)*(x^2 - Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^2), x] + 2*Defer
[Int][1/((-1 + x)*(1 - 2*E^x + 2*E^x*x)*(x^2 - Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^2), x] + 2*Defer[Int][
x/((1 - 2*E^x + 2*E^x*x)*(x^2 - Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)])^2), x] + Defer[Int][(x^2 - Log[E^(2*x
)/(2*(1 + 2*E^x*(-1 + x))^2)])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x \left (-2+x+2 e^x \left (2-x+x^2\right )\right )+\left (-1-2 e^x (-1+x)\right ) \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{\left (1+2 e^x (-1+x)\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=\int \left (\frac {2 x^2}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {-2 x+x^2-x^3+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )-x \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx\\ &=2 \int \frac {x^2}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {-2 x+x^2-x^3+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )-x \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=2 \int \left (\frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx+\int \frac {-x \left (-2+x-x^2\right )+(-1+x) \log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}{(1-x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\\ &=2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \left (-\frac {2 x \left (1-x+x^2\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )}\right ) \, dx\\ &=2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx-2 \int \frac {x \left (1-x+x^2\right )}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ &=-\left (2 \int \left (\frac {1}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {1}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}+\frac {x^2}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2}\right ) \, dx\right )+2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ &=-\left (2 \int \frac {1}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx\right )-2 \int \frac {1}{(-1+x) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx-2 \int \frac {x^2}{\left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {1}{(-1+x) \left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+2 \int \frac {x}{\left (1-2 e^x+2 e^x x\right ) \left (x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )\right )^2} \, dx+\int \frac {1}{x^2-\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 33, normalized size = 1.14 \begin {gather*} -\frac {x}{-x^2+\log \left (\frac {e^{2 x}}{2 \left (1+2 e^x (-1+x)\right )^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - x^2 + E^x*(-4*x + 2*x^2 - 2*x^3) + (-1 + E^x*(2 - 2*x))*Log[E^(2*x)/(2 + E^x*(-8 + 8*x) + E^(
2*x)*(8 - 16*x + 8*x^2))])/(x^4 + E^x*(-2*x^4 + 2*x^5) + (-2*x^2 + E^x*(4*x^2 - 4*x^3))*Log[E^(2*x)/(2 + E^x*(
-8 + 8*x) + E^(2*x)*(8 - 16*x + 8*x^2))] + (1 + E^x*(-2 + 2*x))*Log[E^(2*x)/(2 + E^x*(-8 + 8*x) + E^(2*x)*(8 -
16*x + 8*x^2))]^2),x]

[Out]

-(x/(-x^2 + Log[E^(2*x)/(2*(1 + 2*E^x*(-1 + x))^2)]))

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fricas [A]  time = 0.83, size = 42, normalized size = 1.45 \begin {gather*} \frac {x}{x^{2} - \log \left (\frac {e^{\left (2 \, x\right )}}{2 \, {\left (4 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x - 1\right )} e^{x} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp
(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*ex
p(x)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)*exp(x)+x^4),x, algorithm="f
ricas")

[Out]

x/(x^2 - log(1/2*e^(2*x)/(4*(x^2 - 2*x + 1)*e^(2*x) + 4*(x - 1)*e^x + 1)))

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giac [A]  time = 10.39, size = 45, normalized size = 1.55 \begin {gather*} \frac {x}{x^{2} - 2 \, x + \log \left (8 \, x^{2} e^{\left (2 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + 8 \, x e^{x} + 8 \, e^{\left (2 \, x\right )} - 8 \, e^{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp
(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*ex
p(x)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)*exp(x)+x^4),x, algorithm="g
iac")

[Out]

x/(x^2 - 2*x + log(8*x^2*e^(2*x) - 16*x*e^(2*x) + 8*x*e^x + 8*e^(2*x) - 8*e^x + 2))

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maple [C]  time = 0.70, size = 332, normalized size = 11.45

method	result	size
risch	\(\frac {2 x}{i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{2}+2 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )^{3}-i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-i \pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )^{2}}\right )+i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+2 x^{2}+6 \ln \relax (2)+4 \ln \left ({\mathrm e}^{x} x -{\mathrm e}^{x}+\frac {1}{2}\right )-4 \ln \left ({\mathrm e}^{x}\right )}\)	\(332\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x+2)*exp(x)-1)*ln(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp(x)-x^2
+2*x)/(((2*x-2)*exp(x)+1)*ln(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*exp(x)-2*x
^2)*ln(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)*exp(x)+x^4),x,method=_RETURNVERBOSE)

[Out]

2*x/(I*Pi*csgn(I*exp(x))^2*csgn(I*exp(2*x))-I*Pi*csgn(I*exp(2*x))*csgn(I*exp(2*x)/(exp(x)*x-exp(x)+1/2)^2)^2-I
*Pi*csgn(I/(exp(x)*x-exp(x)+1/2)^2)*csgn(I*exp(2*x)/(exp(x)*x-exp(x)+1/2)^2)^2+2*I*Pi*csgn(I*(exp(x)*x-exp(x)+
1/2))*csgn(I*(exp(x)*x-exp(x)+1/2)^2)^2+I*Pi*csgn(I*exp(2*x)/(exp(x)*x-exp(x)+1/2)^2)^3-I*Pi*csgn(I*(exp(x)*x-
exp(x)+1/2))^2*csgn(I*(exp(x)*x-exp(x)+1/2)^2)-2*I*Pi*csgn(I*exp(x))*csgn(I*exp(2*x))^2-I*Pi*csgn(I*(exp(x)*x-
exp(x)+1/2)^2)^3+I*Pi*csgn(I/(exp(x)*x-exp(x)+1/2)^2)*csgn(I*exp(2*x))*csgn(I*exp(2*x)/(exp(x)*x-exp(x)+1/2)^2
)+I*Pi*csgn(I*exp(2*x))^3+2*x^2+6*ln(2)+4*ln(exp(x)*x-exp(x)+1/2)-4*ln(exp(x)))

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maxima [A]  time = 0.98, size = 25, normalized size = 0.86 \begin {gather*} \frac {x}{x^{2} - 2 \, x + \log \relax (2) + 2 \, \log \left (2 \, {\left (x - 1\right )} e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)-1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(-2*x^3+2*x^2-4*x)*exp
(x)-x^2+2*x)/(((2*x-2)*exp(x)+1)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))^2+((-4*x^3+4*x^2)*ex
p(x)-2*x^2)*log(exp(x)^2/((8*x^2-16*x+8)*exp(x)^2+(8*x-8)*exp(x)+2))+(2*x^5-2*x^4)*exp(x)+x^4),x, algorithm="m
axima")

[Out]

x/(x^2 - 2*x + log(2) + 2*log(2*(x - 1)*e^x + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )-2\,x+x^2+{\mathrm {e}}^x\,\left (2\,x^3-2\,x^2+4\,x\right )}{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (4\,x^2-4\,x^3\right )-2\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^4-2\,x^5\right )+x^4+{\ln \left (\frac {{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^{2\,x}\,\left (8\,x^2-16\,x+8\right )+{\mathrm {e}}^x\,\left (8\,x-8\right )+2}\right )}^2\,\left ({\mathrm {e}}^x\,\left (2\,x-2\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(2*x - 2) + 1) - 2*x + x^2 +
exp(x)*(4*x - 2*x^2 + 2*x^3))/(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(4*x
^2 - 4*x^3) - 2*x^2) - exp(x)*(2*x^4 - 2*x^5) + x^4 + log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x
- 8) + 2))^2*(exp(x)*(2*x - 2) + 1)),x)

[Out]

int(-(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(2*x - 2) + 1) - 2*x + x^2 +
exp(x)*(4*x - 2*x^2 + 2*x^3))/(log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x - 8) + 2))*(exp(x)*(4*x
^2 - 4*x^3) - 2*x^2) - exp(x)*(2*x^4 - 2*x^5) + x^4 + log(exp(2*x)/(exp(2*x)*(8*x^2 - 16*x + 8) + exp(x)*(8*x
- 8) + 2))^2*(exp(x)*(2*x - 2) + 1)), x)

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sympy [A]  time = 0.34, size = 37, normalized size = 1.28 \begin {gather*} - \frac {x}{- x^{2} + \log {\left (\frac {e^{2 x}}{\left (8 x - 8\right ) e^{x} + \left (8 x^{2} - 16 x + 8\right ) e^{2 x} + 2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x+2)*exp(x)-1)*ln(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x-8)*exp(x)+2))+(-2*x**3+2*x**2-4*x)
*exp(x)-x**2+2*x)/(((2*x-2)*exp(x)+1)*ln(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x-8)*exp(x)+2))**2+((-4*x**3+
4*x**2)*exp(x)-2*x**2)*ln(exp(x)**2/((8*x**2-16*x+8)*exp(x)**2+(8*x-8)*exp(x)+2))+(2*x**5-2*x**4)*exp(x)+x**4)
,x)

[Out]

-x/(-x**2 + log(exp(2*x)/((8*x - 8)*exp(x) + (8*x**2 - 16*x + 8)*exp(2*x) + 2)))

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