3.23.94 \(\int \frac {e^x (3 x^2-6 x^3+3 x^4+(9 x-3 x^2-6 x^3) \log (2)+(-12+9 x+3 x^2) \log ^2(2)) \log (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)})+\log ^{\frac {x}{3}}(\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}) (-x^5+(-5 x^3+2 x^4) \log (2)-x^3 \log ^2(2)+(x^4-x^5+(3 x^3+2 x^4) \log (2)+(-4 x^2-x^3) \log ^2(2)) \log (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}) \log (\log (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)})))}{(-3 x^4+3 x^5+(-9 x^3-6 x^4) \log (2)+(12 x^2+3 x^3) \log ^2(2)) \log (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)})} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^x}{x}-\log ^{\frac {x}{3}}\left (4+x-\frac {5 x}{x-\log (2)}\right ) \]

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Rubi [F]  time = 3.07, 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 {e^x \left (3 x^2-6 x^3+3 x^4+\left (9 x-3 x^2-6 x^3\right ) \log (2)+\left (-12+9 x+3 x^2\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )+\log ^{\frac {x}{3}}\left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \left (-x^5+\left (-5 x^3+2 x^4\right ) \log (2)-x^3 \log ^2(2)+\left (x^4-x^5+\left (3 x^3+2 x^4\right ) \log (2)+\left (-4 x^2-x^3\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )\right )\right )}{\left (-3 x^4+3 x^5+\left (-9 x^3-6 x^4\right ) \log (2)+\left (12 x^2+3 x^3\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(3*x^2 - 6*x^3 + 3*x^4 + (9*x - 3*x^2 - 6*x^3)*Log[2] + (-12 + 9*x + 3*x^2)*Log[2]^2)*Log[(x - x^2 +
(4 + x)*Log[2])/(-x + Log[2])] + Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]^(x/3)*(-x^5 + (-5*x^3 + 2*x^4)*
Log[2] - x^3*Log[2]^2 + (x^4 - x^5 + (3*x^3 + 2*x^4)*Log[2] + (-4*x^2 - x^3)*Log[2]^2)*Log[(x - x^2 + (4 + x)*
Log[2])/(-x + Log[2])]*Log[Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]]))/((-3*x^4 + 3*x^5 + (-9*x^3 - 6*x^4
)*Log[2] + (12*x^2 + 3*x^3)*Log[2]^2)*Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^x (-1+x)}{x^2}+\frac {\log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \left (-x^3+x (-5+2 x) \log (2)-x \log ^2(2)-\left (x^3+x (-3+\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )\right )\right )}{3 \left (x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \left (-x^3+x (-5+2 x) \log (2)-x \log ^2(2)-\left (x^3+x (-3+\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )\right )\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))} \, dx+\int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {e^x}{x}+\frac {1}{3} \int \left (\frac {x \left (-x^2-\log ^2(2)+x \log (4)-\log (32)\right ) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))}-\log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right )\right ) \, dx\\ &=\frac {e^x}{x}+\frac {1}{3} \int \frac {x \left (-x^2-\log ^2(2)+x \log (4)-\log (32)\right ) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))} \, dx-\frac {1}{3} \int \log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right ) \, dx\\ &=\frac {e^x}{x}+\frac {1}{3} \int \left (-\log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )+\frac {\left (-x^2-8 x \log (2)+4 \log ^2(2)\right ) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))}\right ) \, dx-\frac {1}{3} \int \log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right ) \, dx\\ &=\frac {e^x}{x}-\frac {1}{3} \int \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \, dx+\frac {1}{3} \int \frac {\left (-x^2-8 x \log (2)+4 \log ^2(2)\right ) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))} \, dx-\frac {1}{3} \int \log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right ) \, dx\\ &=\frac {e^x}{x}-\frac {1}{3} \int \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \, dx+\frac {1}{3} \int \left (\frac {4 \log ^2(2) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))}+\frac {x^2 \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{-x^3+x (3-\log (2)) \log (2)-4 \log ^2(2)+x^2 (1+\log (4))}+\frac {8 x \log (2) \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{-x^3+x (3-\log (2)) \log (2)-4 \log ^2(2)+x^2 (1+\log (4))}\right ) \, dx-\frac {1}{3} \int \log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right ) \, dx\\ &=\frac {e^x}{x}-\frac {1}{3} \int \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \, dx+\frac {1}{3} \int \frac {x^2 \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{-x^3+x (3-\log (2)) \log (2)-4 \log ^2(2)+x^2 (1+\log (4))} \, dx-\frac {1}{3} \int \log ^{\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )\right ) \, dx+\frac {1}{3} (8 \log (2)) \int \frac {x \log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{-x^3+x (3-\log (2)) \log (2)-4 \log ^2(2)+x^2 (1+\log (4))} \, dx+\frac {1}{3} \left (4 \log ^2(2)\right ) \int \frac {\log ^{-1+\frac {x}{3}}\left (\frac {-x^2+x (1+\log (2))+\log (16)}{-x+\log (2)}\right )}{x^3-x (3-\log (2)) \log (2)+4 \log ^2(2)-x^2 (1+\log (4))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 4.94, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^x \left (3 x^2-6 x^3+3 x^4+\left (9 x-3 x^2-6 x^3\right ) \log (2)+\left (-12+9 x+3 x^2\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )+\log ^{\frac {x}{3}}\left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \left (-x^5+\left (-5 x^3+2 x^4\right ) \log (2)-x^3 \log ^2(2)+\left (x^4-x^5+\left (3 x^3+2 x^4\right ) \log (2)+\left (-4 x^2-x^3\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right ) \log \left (\log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )\right )\right )}{\left (-3 x^4+3 x^5+\left (-9 x^3-6 x^4\right ) \log (2)+\left (12 x^2+3 x^3\right ) \log ^2(2)\right ) \log \left (\frac {x-x^2+(4+x) \log (2)}{-x+\log (2)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^x*(3*x^2 - 6*x^3 + 3*x^4 + (9*x - 3*x^2 - 6*x^3)*Log[2] + (-12 + 9*x + 3*x^2)*Log[2]^2)*Log[(x -
x^2 + (4 + x)*Log[2])/(-x + Log[2])] + Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]^(x/3)*(-x^5 + (-5*x^3 + 2
*x^4)*Log[2] - x^3*Log[2]^2 + (x^4 - x^5 + (3*x^3 + 2*x^4)*Log[2] + (-4*x^2 - x^3)*Log[2]^2)*Log[(x - x^2 + (4
+ x)*Log[2])/(-x + Log[2])]*Log[Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]]))/((-3*x^4 + 3*x^5 + (-9*x^3 -
6*x^4)*Log[2] + (12*x^2 + 3*x^3)*Log[2]^2)*Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]),x]

[Out]

Integrate[(E^x*(3*x^2 - 6*x^3 + 3*x^4 + (9*x - 3*x^2 - 6*x^3)*Log[2] + (-12 + 9*x + 3*x^2)*Log[2]^2)*Log[(x -
x^2 + (4 + x)*Log[2])/(-x + Log[2])] + Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]^(x/3)*(-x^5 + (-5*x^3 + 2
*x^4)*Log[2] - x^3*Log[2]^2 + (x^4 - x^5 + (3*x^3 + 2*x^4)*Log[2] + (-4*x^2 - x^3)*Log[2]^2)*Log[(x - x^2 + (4
+ x)*Log[2])/(-x + Log[2])]*Log[Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]]))/((-3*x^4 + 3*x^5 + (-9*x^3 -
6*x^4)*Log[2] + (12*x^2 + 3*x^3)*Log[2]^2)*Log[(x - x^2 + (4 + x)*Log[2])/(-x + Log[2])]), x]

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fricas [A]  time = 0.77, size = 40, normalized size = 1.29 \begin {gather*} -\frac {x \log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right )^{\frac {1}{3} \, x} - e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x^3-4*x^2)*log(2)^2+(2*x^4+3*x^3)*log(2)-x^5+x^4)*log(((4+x)*log(2)-x^2+x)/(log(2)-x))*log(log(
((4+x)*log(2)-x^2+x)/(log(2)-x)))-x^3*log(2)^2+(2*x^4-5*x^3)*log(2)-x^5)*exp(1/3*x*log(log(((4+x)*log(2)-x^2+x
)/(log(2)-x))))+((3*x^2+9*x-12)*log(2)^2+(-6*x^3-3*x^2+9*x)*log(2)+3*x^4-6*x^3+3*x^2)*exp(x)*log(((4+x)*log(2)
-x^2+x)/(log(2)-x)))/((3*x^3+12*x^2)*log(2)^2+(-6*x^4-9*x^3)*log(2)+3*x^5-3*x^4)/log(((4+x)*log(2)-x^2+x)/(log
(2)-x)),x, algorithm="fricas")

[Out]

-(x*log((x^2 - (x + 4)*log(2) - x)/(x - log(2)))^(1/3*x) - e^x)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, {\left (x^{4} - 2 \, x^{3} + {\left (x^{2} + 3 \, x - 4\right )} \log \relax (2)^{2} + x^{2} - {\left (2 \, x^{3} + x^{2} - 3 \, x\right )} \log \relax (2)\right )} e^{x} \log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right ) - {\left (x^{5} + x^{3} \log \relax (2)^{2} + {\left (x^{5} - x^{4} + {\left (x^{3} + 4 \, x^{2}\right )} \log \relax (2)^{2} - {\left (2 \, x^{4} + 3 \, x^{3}\right )} \log \relax (2)\right )} \log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right ) \log \left (\log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right )\right ) - {\left (2 \, x^{4} - 5 \, x^{3}\right )} \log \relax (2)\right )} \log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right )^{\frac {1}{3} \, x}}{3 \, {\left (x^{5} - x^{4} + {\left (x^{3} + 4 \, x^{2}\right )} \log \relax (2)^{2} - {\left (2 \, x^{4} + 3 \, x^{3}\right )} \log \relax (2)\right )} \log \left (\frac {x^{2} - {\left (x + 4\right )} \log \relax (2) - x}{x - \log \relax (2)}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x^3-4*x^2)*log(2)^2+(2*x^4+3*x^3)*log(2)-x^5+x^4)*log(((4+x)*log(2)-x^2+x)/(log(2)-x))*log(log(
((4+x)*log(2)-x^2+x)/(log(2)-x)))-x^3*log(2)^2+(2*x^4-5*x^3)*log(2)-x^5)*exp(1/3*x*log(log(((4+x)*log(2)-x^2+x
)/(log(2)-x))))+((3*x^2+9*x-12)*log(2)^2+(-6*x^3-3*x^2+9*x)*log(2)+3*x^4-6*x^3+3*x^2)*exp(x)*log(((4+x)*log(2)
-x^2+x)/(log(2)-x)))/((3*x^3+12*x^2)*log(2)^2+(-6*x^4-9*x^3)*log(2)+3*x^5-3*x^4)/log(((4+x)*log(2)-x^2+x)/(log
(2)-x)),x, algorithm="giac")

[Out]

integrate(1/3*(3*(x^4 - 2*x^3 + (x^2 + 3*x - 4)*log(2)^2 + x^2 - (2*x^3 + x^2 - 3*x)*log(2))*e^x*log((x^2 - (x
+ 4)*log(2) - x)/(x - log(2))) - (x^5 + x^3*log(2)^2 + (x^5 - x^4 + (x^3 + 4*x^2)*log(2)^2 - (2*x^4 + 3*x^3)*
log(2))*log((x^2 - (x + 4)*log(2) - x)/(x - log(2)))*log(log((x^2 - (x + 4)*log(2) - x)/(x - log(2)))) - (2*x^
4 - 5*x^3)*log(2))*log((x^2 - (x + 4)*log(2) - x)/(x - log(2)))^(1/3*x))/((x^5 - x^4 + (x^3 + 4*x^2)*log(2)^2
- (2*x^4 + 3*x^3)*log(2))*log((x^2 - (x + 4)*log(2) - x)/(x - log(2)))), x)

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maple [C]  time = 0.62, size = 152, normalized size = 4.90

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{x}-\left (-\ln \left (\ln \relax (2)-x \right )+\ln \left (\left (4+x \right ) \ln \relax (2)-x^{2}+x \right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\left (4+x \right ) \ln \relax (2)-x^{2}+x \right )}{\ln \relax (2)-x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (4+x \right ) \ln \relax (2)-x^{2}+x \right )}{\ln \relax (2)-x}\right )+\mathrm {csgn}\left (\frac {i}{\ln \relax (2)-x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (4+x \right ) \ln \relax (2)-x^{2}+x \right )}{\ln \relax (2)-x}\right )+\mathrm {csgn}\left (i \left (\left (4+x \right ) \ln \relax (2)-x^{2}+x \right )\right )\right )}{2}\right )^{\frac {x}{3}}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-x^3-4*x^2)*ln(2)^2+(2*x^4+3*x^3)*ln(2)-x^5+x^4)*ln(((4+x)*ln(2)-x^2+x)/(ln(2)-x))*ln(ln(((4+x)*ln(2)-
x^2+x)/(ln(2)-x)))-x^3*ln(2)^2+(2*x^4-5*x^3)*ln(2)-x^5)*exp(1/3*x*ln(ln(((4+x)*ln(2)-x^2+x)/(ln(2)-x))))+((3*x
^2+9*x-12)*ln(2)^2+(-6*x^3-3*x^2+9*x)*ln(2)+3*x^4-6*x^3+3*x^2)*exp(x)*ln(((4+x)*ln(2)-x^2+x)/(ln(2)-x)))/((3*x
^3+12*x^2)*ln(2)^2+(-6*x^4-9*x^3)*ln(2)+3*x^5-3*x^4)/ln(((4+x)*ln(2)-x^2+x)/(ln(2)-x)),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [A]  time = 1.07, size = 42, normalized size = 1.35 \begin {gather*} -\frac {x {\left (\log \left (x^{2} - x {\left (\log \relax (2) + 1\right )} - 4 \, \log \relax (2)\right ) - \log \left (x - \log \relax (2)\right )\right )}^{\frac {1}{3} \, x} - e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x^3-4*x^2)*log(2)^2+(2*x^4+3*x^3)*log(2)-x^5+x^4)*log(((4+x)*log(2)-x^2+x)/(log(2)-x))*log(log(
((4+x)*log(2)-x^2+x)/(log(2)-x)))-x^3*log(2)^2+(2*x^4-5*x^3)*log(2)-x^5)*exp(1/3*x*log(log(((4+x)*log(2)-x^2+x
)/(log(2)-x))))+((3*x^2+9*x-12)*log(2)^2+(-6*x^3-3*x^2+9*x)*log(2)+3*x^4-6*x^3+3*x^2)*exp(x)*log(((4+x)*log(2)
-x^2+x)/(log(2)-x)))/((3*x^3+12*x^2)*log(2)^2+(-6*x^4-9*x^3)*log(2)+3*x^5-3*x^4)/log(((4+x)*log(2)-x^2+x)/(log
(2)-x)),x, algorithm="maxima")

[Out]

-(x*(log(x^2 - x*(log(2) + 1) - 4*log(2)) - log(x - log(2)))^(1/3*x) - e^x)/x

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mupad [B]  time = 3.39, size = 40, normalized size = 1.29 \begin {gather*} \frac {{\mathrm {e}}^x}{x}-{\mathrm {e}}^{\frac {x\,\ln \left (\ln \left (-\frac {x+4\,\ln \relax (2)+x\,\ln \relax (2)-x^2}{x-\ln \relax (2)}\right )\right )}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*log(log(-(x + log(2)*(x + 4) - x^2)/(x - log(2)))))/3)*(x^3*log(2)^2 + log(2)*(5*x^3 - 2*x^4) + x^
5 - log(-(x + log(2)*(x + 4) - x^2)/(x - log(2)))*log(log(-(x + log(2)*(x + 4) - x^2)/(x - log(2))))*(log(2)*(
3*x^3 + 2*x^4) - log(2)^2*(4*x^2 + x^3) + x^4 - x^5)) - exp(x)*log(-(x + log(2)*(x + 4) - x^2)/(x - log(2)))*(
log(2)^2*(9*x + 3*x^2 - 12) - log(2)*(3*x^2 - 9*x + 6*x^3) + 3*x^2 - 6*x^3 + 3*x^4))/(log(-(x + log(2)*(x + 4)
- x^2)/(x - log(2)))*(log(2)*(9*x^3 + 6*x^4) + 3*x^4 - 3*x^5 - log(2)^2*(12*x^2 + 3*x^3))),x)

[Out]

exp(x)/x - exp((x*log(log(-(x + 4*log(2) + x*log(2) - x^2)/(x - log(2)))))/3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x**3-4*x**2)*ln(2)**2+(2*x**4+3*x**3)*ln(2)-x**5+x**4)*ln(((4+x)*ln(2)-x**2+x)/(ln(2)-x))*ln(ln
(((4+x)*ln(2)-x**2+x)/(ln(2)-x)))-x**3*ln(2)**2+(2*x**4-5*x**3)*ln(2)-x**5)*exp(1/3*x*ln(ln(((4+x)*ln(2)-x**2+
x)/(ln(2)-x))))+((3*x**2+9*x-12)*ln(2)**2+(-6*x**3-3*x**2+9*x)*ln(2)+3*x**4-6*x**3+3*x**2)*exp(x)*ln(((4+x)*ln
(2)-x**2+x)/(ln(2)-x)))/((3*x**3+12*x**2)*ln(2)**2+(-6*x**4-9*x**3)*ln(2)+3*x**5-3*x**4)/ln(((4+x)*ln(2)-x**2+
x)/(ln(2)-x)),x)

[Out]

Timed out

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