Optimal. Leaf size=30 \[ -\left (\left (\frac {e^x}{x}-2 x\right ) \log (4)\right )+\frac {2 \log (x)}{3+\log (5-x)} \]
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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 {-30 x+6 x^2+e^x \left (-45+54 x-9 x^2\right ) \log (4)+\left (-90 x^2+18 x^3\right ) \log (4)+\left (-10 x+2 x^2+e^x \left (-30+36 x-6 x^2\right ) \log (4)+\left (-60 x^2+12 x^3\right ) \log (4)\right ) \log (5-x)+\left (e^x \left (-5+6 x-x^2\right ) \log (4)+\left (-10 x^2+2 x^3\right ) \log (4)\right ) \log ^2(5-x)-2 x^2 \log (x)}{-45 x^2+9 x^3+\left (-30 x^2+6 x^3\right ) \log (5-x)+\left (-5 x^2+x^3\right ) \log ^2(5-x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 (-5+x) \left (6 x^2 \log (4)+x \left (2-3 e^x \log (4)\right )+e^x \log (64)\right )-2 (-5+x) \left (x-3 e^x x \log (4)+6 x^2 \log (4)+e^x \log (64)\right ) \log (5-x)-(-5+x) \left (-e^x (-1+x)+2 x^2\right ) \log (4) \log ^2(5-x)+2 x^2 \log (x)}{(5-x) x^2 (3+\log (5-x))^2} \, dx\\ &=\int \left (\frac {6}{x (3+\log (5-x))^2}+\frac {18 \log (4)}{(3+\log (5-x))^2}+\frac {2 \log (5-x)}{x (3+\log (5-x))^2}+\frac {12 \log (4) \log (5-x)}{(3+\log (5-x))^2}+\frac {2 \log (4) \log ^2(5-x)}{(3+\log (5-x))^2}-\frac {e^x (-1+x) \left (\log (262144)+\log (4096) \log (5-x)+\log (4) \log ^2(5-x)\right )}{x^2 (3+\log (5-x))^2}-\frac {2 \log (x)}{(-5+x) (3+\log (5-x))^2}\right ) \, dx\\ &=2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \int \frac {\log (x)}{(-5+x) (3+\log (5-x))^2} \, dx+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx+(2 \log (4)) \int \frac {\log ^2(5-x)}{(3+\log (5-x))^2} \, dx+(12 \log (4)) \int \frac {\log (5-x)}{(3+\log (5-x))^2} \, dx+(18 \log (4)) \int \frac {1}{(3+\log (5-x))^2} \, dx-\int \frac {e^x (-1+x) \left (\log (262144)+\log (4096) \log (5-x)+\log (4) \log ^2(5-x)\right )}{x^2 (3+\log (5-x))^2} \, dx\\ &=2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(2 \log (4)) \text {Subst}\left (\int \frac {\log ^2(x)}{(3+\log (x))^2} \, dx,x,5-x\right )-(12 \log (4)) \text {Subst}\left (\int \frac {\log (x)}{(3+\log (x))^2} \, dx,x,5-x\right )-(18 \log (4)) \text {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )-\int \frac {e^x (-1+x) (\log (64)+\log (4) \log (5-x))}{x^2 (3+\log (5-x))} \, dx\\ &=\frac {18 (5-x) \log (4)}{3+\log (5-x)}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-\log (4) \int \frac {e^x (-1+x)}{x^2} \, dx-(2 \log (4)) \text {Subst}\left (\int \left (1+\frac {9}{(3+\log (x))^2}-\frac {6}{3+\log (x)}\right ) \, dx,x,5-x\right )-(12 \log (4)) \text {Subst}\left (\int \left (-\frac {3}{(3+\log (x))^2}+\frac {1}{3+\log (x)}\right ) \, dx,x,5-x\right )-(18 \log (4)) \text {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)+\frac {18 (5-x) \log (4)}{3+\log (5-x)}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \text {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )-(18 \log (4)) \text {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )+(36 \log (4)) \text {Subst}\left (\int \frac {1}{(3+\log (x))^2} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)-\frac {18 \text {Ei}(3+\log (5-x)) \log (4)}{e^3}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \text {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )+(36 \log (4)) \text {Subst}\left (\int \frac {1}{3+\log (x)} \, dx,x,5-x\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)-\frac {18 \text {Ei}(3+\log (5-x)) \log (4)}{e^3}+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx-(18 \log (4)) \text {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )+(36 \log (4)) \text {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (5-x)\right )\\ &=-\frac {e^x \log (4)}{x}+2 x \log (4)+2 \int \frac {\log (5-x)}{x (3+\log (5-x))^2} \, dx-2 \text {Subst}\left (\int \frac {\log (5-x)}{x (3+\log (x))^2} \, dx,x,5-x\right )+6 \int \frac {1}{x (3+\log (5-x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.15, size = 30, normalized size = 1.00 \begin {gather*} -\frac {e^x \log (4)}{x}+2 x \log (4)+\frac {2 \log (x)}{3+\log (5-x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 6.91, size = 33, normalized size = 1.10
method | result | size |
risch | \(\frac {2 \ln \left (2\right ) \left (2 x^{2}-{\mathrm e}^{x}\right )}{x}+\frac {2 \ln \left (x \right )}{\ln \left (5-x \right )+3}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.55, size = 55, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left (6 \, x^{2} \log \left (2\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right ) + {\left (2 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right )\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 55, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left (6 \, x^{2} \log \left (2\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right ) + {\left (2 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right )\right )} \log \left (-x + 5\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.21, size = 27, normalized size = 0.90 \begin {gather*} 4 x \log {\left (2 \right )} + \frac {2 \log {\left (x \right )}}{\log {\left (5 - x \right )} + 3} - \frac {2 e^{x} \log {\left (2 \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 59, normalized size = 1.97 \begin {gather*} \frac {2 \, {\left (2 \, x^{2} \log \left (2\right ) \log \left (-x + 5\right ) + 6 \, x^{2} \log \left (2\right ) - e^{x} \log \left (2\right ) \log \left (-x + 5\right ) - 3 \, e^{x} \log \left (2\right ) + x \log \left (x\right )\right )}}{x \log \left (-x + 5\right ) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.31, size = 60, normalized size = 2.00 \begin {gather*} 4\,x\,\ln \left (2\right )-\frac {10}{x}+\frac {\frac {2\,\left (x\,\ln \left (x\right )-3\,x+15\right )}{x}-\frac {2\,\ln \left (5-x\right )\,\left (x-5\right )}{x}}{\ln \left (5-x\right )+3}-\frac {2\,{\mathrm {e}}^x\,\ln \left (2\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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