3.97.89 \(\int \frac {4 x^4+24 x^5-24 x^6+(-8 x^3-48 x^4+48 x^5) \log (4)+(4 x^2+24 x^3-24 x^4) \log ^2(4)+e^x (-12 x^2-20 x^3+(8 x+20 x^2) \log (4))+(-8 x^4+8 x^5+(16 x^3-16 x^4) \log (4)+(-8 x^2+8 x^3) \log ^2(4)+e^x (8 x^2-8 x \log (4))) \log (\frac {e^x-x^2+x^3+(x-x^2) \log (4)}{-x+\log (4)})}{5 x^3-5 x^4+(-10 x^2+10 x^3) \log (4)+(5 x-5 x^2) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx\)

Optimal. Leaf size=34 \[ \frac {4}{5} x^2 \left (1+2 x-\log \left (x-x^2+\frac {e^x}{-x+\log (4)}\right )\right ) \]

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Rubi [F]  time = 5.85, 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 {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*x^4 + 24*x^5 - 24*x^6 + (-8*x^3 - 48*x^4 + 48*x^5)*Log[4] + (4*x^2 + 24*x^3 - 24*x^4)*Log[4]^2 + E^x*(-
12*x^2 - 20*x^3 + (8*x + 20*x^2)*Log[4]) + (-8*x^4 + 8*x^5 + (16*x^3 - 16*x^4)*Log[4] + (-8*x^2 + 8*x^3)*Log[4
]^2 + E^x*(8*x^2 - 8*x*Log[4]))*Log[(E^x - x^2 + x^3 + (x - x^2)*Log[4])/(-x + Log[4])])/(5*x^3 - 5*x^4 + (-10
*x^2 + 10*x^3)*Log[4] + (5*x - 5*x^2)*Log[4]^2 + E^x*(-5*x + 5*Log[4])),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (e^x \left (5 x^2+x (3-5 \log (4))-2 \log (4)\right )+x \left (6 x^4-\log ^2(4)+x^2 \left (-1+12 \log (4)+6 \log ^2(4)\right )-6 x^3 (1+\log (16))+x \left (-6 \log ^2(4)+\log (16)\right )\right )-2 \left (e^x+(-1+x) x (x-\log (4))\right ) (x-\log (4)) \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right )}{5 \left (e^x+(-1+x) x (x-\log (4))\right ) (x-\log (4))} \, dx\\ &=\frac {4}{5} \int \frac {x \left (e^x \left (5 x^2+x (3-5 \log (4))-2 \log (4)\right )+x \left (6 x^4-\log ^2(4)+x^2 \left (-1+12 \log (4)+6 \log ^2(4)\right )-6 x^3 (1+\log (16))+x \left (-6 \log ^2(4)+\log (16)\right )\right )-2 \left (e^x+(-1+x) x (x-\log (4))\right ) (x-\log (4)) \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right )}{\left (e^x+(-1+x) x (x-\log (4))\right ) (x-\log (4))} \, dx\\ &=\frac {4}{5} \int \left (\frac {x^2 \left (x^4+\log ^2(4)-2 x^3 (2+\log (4))+x^2 \left (2+7 \log (4)+\log ^2(4)\right )-x \left (3 \log ^2(4)+\log (64)\right )\right )}{(x-\log (4)) \left (e^x+x^3+x \log (4)-x^2 (1+\log (4))\right )}+\frac {x \left (5 x^2+3 x \left (1-\frac {5 \log (4)}{3}\right )-2 \log (4)-2 x \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+2 \log (4) \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right )}{x-\log (4)}\right ) \, dx\\ &=\frac {4}{5} \int \frac {x^2 \left (x^4+\log ^2(4)-2 x^3 (2+\log (4))+x^2 \left (2+7 \log (4)+\log ^2(4)\right )-x \left (3 \log ^2(4)+\log (64)\right )\right )}{(x-\log (4)) \left (e^x+x^3+x \log (4)-x^2 (1+\log (4))\right )} \, dx+\frac {4}{5} \int \frac {x \left (5 x^2+3 x \left (1-\frac {5 \log (4)}{3}\right )-2 \log (4)-2 x \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+2 \log (4) \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right )}{x-\log (4)} \, dx\\ &=\frac {4}{5} \int \frac {x^2 \left (x^3+x^2 (-4-\log (4))+2 \log (4)+x (2+3 \log (4))-\log (64)\right )}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \frac {x \left (5 x^2+x (3-5 \log (4))-2 \log (4)-2 (x-\log (4)) \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right )}{x-\log (4)} \, dx\\ &=\frac {4}{5} \int \frac {x^2 \left (x^3-\log (4)-x^2 (4+\log (4))+x (2+\log (64))\right )}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \left (\frac {x \left (5 x^2+x (3-5 \log (4))-2 \log (4)\right )}{x-\log (4)}-2 x \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right ) \, dx\\ &=\frac {4}{5} \int \frac {x \left (5 x^2+x (3-5 \log (4))-2 \log (4)\right )}{x-\log (4)} \, dx+\frac {4}{5} \int \left (\frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}+\frac {x^4 (-4-\log (4))}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}+\frac {x^2 \log (4)}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))}+\frac {x^3 (2+\log (64))}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}\right ) \, dx-\frac {8}{5} \int x \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right ) \, dx\\ &=-\frac {4}{5} x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+\frac {4}{5} \int \frac {x^2 \left (e^x (-1+x-\log (4))+(-1+2 x) (x-\log (4))^2\right )}{\left (e^x+(-1+x) x (x-\log (4))\right ) (x-\log (4))} \, dx+\frac {4}{5} \int \left (3 x+5 x^2+\log (4)+\frac {\log ^2(4)}{x-\log (4)}\right ) \, dx+\frac {4}{5} \int \frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx-\frac {1}{5} (4 (4+\log (4))) \int \frac {x^4}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 (2+\log (64))) \int \frac {x^3}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx\\ &=\frac {6 x^2}{5}+\frac {4 x^3}{3}+\frac {4}{5} x \log (4)-\frac {4}{5} x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+\frac {4}{5} \log ^2(4) \log (x-\log (4))+\frac {4}{5} \int \frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \left (\frac {x^2 (-1+x-\log (4))}{x-\log (4)}+\frac {x^2 \left (-x^3+\log (4)+x^2 (4+\log (4))-x (2+\log (64))\right )}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}\right ) \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx-\frac {1}{5} (4 (4+\log (4))) \int \frac {x^4}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 (2+\log (64))) \int \frac {x^3}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx\\ &=\frac {6 x^2}{5}+\frac {4 x^3}{3}+\frac {4}{5} x \log (4)-\frac {4}{5} x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+\frac {4}{5} \log ^2(4) \log (x-\log (4))+\frac {4}{5} \int \frac {x^2 (-1+x-\log (4))}{x-\log (4)} \, dx+\frac {4}{5} \int \frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \frac {x^2 \left (-x^3+\log (4)+x^2 (4+\log (4))-x (2+\log (64))\right )}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx-\frac {1}{5} (4 (4+\log (4))) \int \frac {x^4}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 (2+\log (64))) \int \frac {x^3}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx\\ &=\frac {6 x^2}{5}+\frac {4 x^3}{3}+\frac {4}{5} x \log (4)-\frac {4}{5} x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+\frac {4}{5} \log ^2(4) \log (x-\log (4))+\frac {4}{5} \int \left (-x+x^2-\log (4)-\frac {\log ^2(4)}{x-\log (4)}\right ) \, dx+\frac {4}{5} \int \frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \left (\frac {x^2 \log (4)}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}+\frac {x^4 (4+\log (4))}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}+\frac {x^5}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))}+\frac {x^3 (-2-\log (64))}{e^x+x^3+x \log (4)-x^2 (1+\log (4))}\right ) \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx-\frac {1}{5} (4 (4+\log (4))) \int \frac {x^4}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 (2+\log (64))) \int \frac {x^3}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx\\ &=\frac {4 x^2}{5}+\frac {8 x^3}{5}-\frac {4}{5} x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )+\frac {4}{5} \int \frac {x^5}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {4}{5} \int \frac {x^5}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{e^x+x^3+x \log (4)-x^2 (1+\log (4))} \, dx+\frac {1}{5} (4 \log (4)) \int \frac {x^2}{-e^x-x^3-x \log (4)+x^2 (1+\log (4))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 39, normalized size = 1.15 \begin {gather*} \frac {4}{5} \left (x^2+2 x^3-x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^4 + 24*x^5 - 24*x^6 + (-8*x^3 - 48*x^4 + 48*x^5)*Log[4] + (4*x^2 + 24*x^3 - 24*x^4)*Log[4]^2 +
E^x*(-12*x^2 - 20*x^3 + (8*x + 20*x^2)*Log[4]) + (-8*x^4 + 8*x^5 + (16*x^3 - 16*x^4)*Log[4] + (-8*x^2 + 8*x^3)
*Log[4]^2 + E^x*(8*x^2 - 8*x*Log[4]))*Log[(E^x - x^2 + x^3 + (x - x^2)*Log[4])/(-x + Log[4])])/(5*x^3 - 5*x^4
+ (-10*x^2 + 10*x^3)*Log[4] + (5*x - 5*x^2)*Log[4]^2 + E^x*(-5*x + 5*Log[4])),x]

[Out]

(4*(x^2 + 2*x^3 - x^2*Log[x - x^2 - E^x/(x - Log[4])]))/5

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fricas [A]  time = 0.55, size = 49, normalized size = 1.44 \begin {gather*} \frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-\frac {x^{3} - x^{2} - 2 \, {\left (x^{2} - x\right )} \log \relax (2) + e^{x}}{x - 2 \, \log \relax (2)}\right ) + \frac {4}{5} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^4+16*x^3)*log(2)+8*x^5-8*x^4)*log((e
xp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*log(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+
4*x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log(2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log
(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x^3),x, algorithm="fricas")

[Out]

8/5*x^3 - 4/5*x^2*log(-(x^3 - x^2 - 2*(x^2 - x)*log(2) + e^x)/(x - 2*log(2))) + 4/5*x^2

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giac [A]  time = 0.40, size = 54, normalized size = 1.59 \begin {gather*} \frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-x^{3} + 2 \, x^{2} \log \relax (2) + x^{2} - 2 \, x \log \relax (2) - e^{x}\right ) + \frac {4}{5} \, x^{2} \log \left (x - 2 \, \log \relax (2)\right ) + \frac {4}{5} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^4+16*x^3)*log(2)+8*x^5-8*x^4)*log((e
xp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*log(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+
4*x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log(2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log
(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x^3),x, algorithm="giac")

[Out]

8/5*x^3 - 4/5*x^2*log(-x^3 + 2*x^2*log(2) + x^2 - 2*x*log(2) - e^x) + 4/5*x^2*log(x - 2*log(2)) + 4/5*x^2

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maple [C]  time = 0.22, size = 379, normalized size = 11.15

method	result	size
risch	\(-\frac {4 x^{2} \ln \left (\left (x^{2}-x \right ) \ln \relax (2)-\frac {x^{3}}{2}+\frac {x^{2}}{2}-\frac {{\mathrm e}^{x}}{2}\right )}{5}+\frac {4 x^{2} \ln \left (\ln \relax (2)-\frac {x}{2}\right )}{5}+\frac {8 x^{3}}{5}+\frac {4 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \relax (2)-\frac {x}{2}}\right )^{2}}{5}+\frac {2 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {x}{2}}\right ) \mathrm {csgn}\left (i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \relax (2)-\frac {x}{2}}\right )}{5}-\frac {2 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {x}{2}}\right ) \mathrm {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \relax (2)-\frac {x}{2}}\right )^{2}}{5}+\frac {2 i \pi \,x^{2} \mathrm {csgn}\left (i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \relax (2)-\frac {x}{2}}\right )^{2}}{5}+\frac {2 i \pi \,x^{2} \mathrm {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \relax (2)+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \relax (2)-\frac {x}{2}}\right )^{3}}{5}-\frac {4 i \pi \,x^{2}}{5}+\frac {4 x^{2}}{5}\)	\(379\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x*ln(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*ln(2)^2+2*(-16*x^4+16*x^3)*ln(2)+8*x^5-8*x^4)*ln((exp(x)+2*(-
x^2+x)*ln(2)+x^3-x^2)/(2*ln(2)-x))+(2*(20*x^2+8*x)*ln(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+4*x^2)*ln(2)^
2+2*(48*x^5-48*x^4-8*x^3)*ln(2)-24*x^6+24*x^5+4*x^4)/((10*ln(2)-5*x)*exp(x)+4*(-5*x^2+5*x)*ln(2)^2+2*(10*x^3-1
0*x^2)*ln(2)-5*x^4+5*x^3),x,method=_RETURNVERBOSE)

[Out]

-4/5*x^2*ln((x^2-x)*ln(2)-1/2*x^3+1/2*x^2-1/2*exp(x))+4/5*x^2*ln(ln(2)-1/2*x)+8/5*x^3+4/5*I*Pi*x^2*csgn(I*(-(x
^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x))/(ln(2)-1/2*x))^2+2/5*I*Pi*x^2*csgn(I/(ln(2)-1/2*x))*csgn(I*(-(x^2-x)*l
n(2)+1/2*x^3-1/2*x^2+1/2*exp(x)))*csgn(I*(-(x^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x))/(ln(2)-1/2*x))-2/5*I*Pi*x
^2*csgn(I/(ln(2)-1/2*x))*csgn(I*(-(x^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x))/(ln(2)-1/2*x))^2+2/5*I*Pi*x^2*csgn
(I*(-(x^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x)))*csgn(I*(-(x^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x))/(ln(2)-1/2*
x))^2+2/5*I*Pi*x^2*csgn(I*(-(x^2-x)*ln(2)+1/2*x^3-1/2*x^2+1/2*exp(x))/(ln(2)-1/2*x))^3-4/5*I*Pi*x^2+4/5*x^2

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maxima [A]  time = 0.50, size = 54, normalized size = 1.59 \begin {gather*} \frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-x^{3} + x^{2} {\left (2 \, \log \relax (2) + 1\right )} - 2 \, x \log \relax (2) - e^{x}\right ) + \frac {4}{5} \, x^{2} \log \left (x - 2 \, \log \relax (2)\right ) + \frac {4}{5} \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^4+16*x^3)*log(2)+8*x^5-8*x^4)*log((e
xp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*log(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+
4*x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log(2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log
(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x^3),x, algorithm="maxima")

[Out]

8/5*x^3 - 4/5*x^2*log(-x^3 + x^2*(2*log(2) + 1) - 2*x*log(2) - e^x) + 4/5*x^2*log(x - 2*log(2)) + 4/5*x^2

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mupad [B]  time = 7.03, size = 45, normalized size = 1.32 \begin {gather*} \frac {4\,x^2\,\left (2\,x-\ln \left (-\frac {{\mathrm {e}}^x-x^2+x^3+2\,\ln \relax (2)\,\left (x-x^2\right )}{x-2\,\ln \relax (2)}\right )+1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(12*x^2 - 2*log(2)*(8*x + 20*x^2) + 20*x^3) - 4*log(2)^2*(4*x^2 + 24*x^3 - 24*x^4) + log(-(exp(x)
- x^2 + x^3 + 2*log(2)*(x - x^2))/(x - 2*log(2)))*(exp(x)*(16*x*log(2) - 8*x^2) - 2*log(2)*(16*x^3 - 16*x^4) +
8*x^4 - 8*x^5 + 4*log(2)^2*(8*x^2 - 8*x^3)) + 2*log(2)*(8*x^3 + 48*x^4 - 48*x^5) - 4*x^4 - 24*x^5 + 24*x^6)/(
exp(x)*(5*x - 10*log(2)) - 4*log(2)^2*(5*x - 5*x^2) + 2*log(2)*(10*x^2 - 10*x^3) - 5*x^3 + 5*x^4),x)

[Out]

(4*x^2*(2*x - log(-(exp(x) - x^2 + x^3 + 2*log(2)*(x - x^2))/(x - 2*log(2))) + 1))/5

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sympy [B]  time = 1.28, size = 110, normalized size = 3.24 \begin {gather*} \frac {8 x^{3}}{5} + \frac {4 x^{2}}{5} + \left (- \frac {4 x^{2}}{5} + \frac {16 \log {\relax (2 )}^{2}}{15}\right ) \log {\left (\frac {x^{3} - x^{2} + \left (- 2 x^{2} + 2 x\right ) \log {\relax (2 )} + e^{x}}{- x + 2 \log {\relax (2 )}} \right )} + \frac {16 \log {\relax (2 )}^{2} \log {\left (x - 2 \log {\relax (2 )} \right )}}{15} - \frac {16 \log {\relax (2 )}^{2} \log {\left (x^{3} - 2 x^{2} \log {\relax (2 )} - x^{2} + 2 x \log {\relax (2 )} + e^{x} \right )}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x*ln(2)+8*x**2)*exp(x)+4*(8*x**3-8*x**2)*ln(2)**2+2*(-16*x**4+16*x**3)*ln(2)+8*x**5-8*x**4)*l
n((exp(x)+2*(-x**2+x)*ln(2)+x**3-x**2)/(2*ln(2)-x))+(2*(20*x**2+8*x)*ln(2)-20*x**3-12*x**2)*exp(x)+4*(-24*x**4
+24*x**3+4*x**2)*ln(2)**2+2*(48*x**5-48*x**4-8*x**3)*ln(2)-24*x**6+24*x**5+4*x**4)/((10*ln(2)-5*x)*exp(x)+4*(-
5*x**2+5*x)*ln(2)**2+2*(10*x**3-10*x**2)*ln(2)-5*x**4+5*x**3),x)

[Out]

8*x**3/5 + 4*x**2/5 + (-4*x**2/5 + 16*log(2)**2/15)*log((x**3 - x**2 + (-2*x**2 + 2*x)*log(2) + exp(x))/(-x +
2*log(2))) + 16*log(2)**2*log(x - 2*log(2))/15 - 16*log(2)**2*log(x**3 - 2*x**2*log(2) - x**2 + 2*x*log(2) + e
xp(x))/15

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