Optimal. Leaf size=26 \[ 72 (x-\log (4)) \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right ) \]
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Rubi [F] time = 1.17, 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 {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {72 \left (-25+e^{x/5} (-5+x)\right ) (x-\log (4))}{5 \left (5+e^{x/5}-x\right ) x}+72 \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right )\right ) \, dx\\ &=\frac {72}{5} \int \frac {\left (-25+e^{x/5} (-5+x)\right ) (x-\log (4))}{\left (5+e^{x/5}-x\right ) x} \, dx+72 \int \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right ) \, dx\\ &=72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+\frac {72}{5} \int \left (\frac {(-10+x) (x-\log (4))}{5+e^{x/5}-x}+\frac {(-5+x) (x-\log (4))}{x}\right ) \, dx-72 \int \frac {-25+e^{x/5} (-5+x)}{5 \left (5+e^{x/5}-x\right )} \, dx\\ &=72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )-\frac {72}{5} \int \frac {-25+e^{x/5} (-5+x)}{5+e^{x/5}-x} \, dx+\frac {72}{5} \int \frac {(-10+x) (x-\log (4))}{5+e^{x/5}-x} \, dx+\frac {72}{5} \int \frac {(-5+x) (x-\log (4))}{x} \, dx\\ &=72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+\frac {72}{5} \int \left (-5+x-\log (4)+\frac {5 \log (4)}{x}\right ) \, dx+\frac {72}{5} \int \left (\frac {x^2}{5+e^{x/5}-x}+\frac {10 \log (4)}{5+e^{x/5}-x}-\frac {x (10+\log (4))}{5+e^{x/5}-x}\right ) \, dx-72 \operatorname {Subst}\left (\int \frac {-25+e^x (-5+5 x)}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )\\ &=\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-72 \operatorname {Subst}\left (\int \left (5 (-1+x)+\frac {25 (-2+x) x}{5+e^x-5 x}\right ) \, dx,x,\frac {x}{5}\right )+(144 \log (4)) \int \frac {1}{5+e^{x/5}-x} \, dx-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx\\ &=-\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \operatorname {Subst}\left (\int \frac {(-2+x) x}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \operatorname {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx\\ &=-\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \operatorname {Subst}\left (\int \left (-\frac {2 x}{5+e^x-5 x}+\frac {x^2}{5+e^x-5 x}\right ) \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \operatorname {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx\\ &=-\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \operatorname {Subst}\left (\int \frac {x^2}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+3600 \operatorname {Subst}\left (\int \frac {x}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \operatorname {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.60, size = 23, normalized size = 0.88 \begin {gather*} 72 \, {\left (x - 2 \, \log \relax (2)\right )} \log \left (\frac {15 \, {\left (x - e^{\left (\frac {1}{5} \, x\right )} - 5\right )}}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.29, size = 41, normalized size = 1.58 \begin {gather*} 72 \, x \log \left (15 \, x - 15 \, e^{\left (\frac {1}{5} \, x\right )} - 75\right ) - 72 \, x \log \relax (x) + 144 \, \log \relax (2) \log \relax (x) - 144 \, \log \relax (2) \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 41, normalized size = 1.58
method | result | size |
norman | \(-144 \ln \relax (2) \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )+72 x \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )\) | \(41\) |
risch | \(72 x \ln \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )-72 x \ln \relax (x )-36 i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )+36 i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )^{2}+36 i \pi x \,\mathrm {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )^{2}-36 i \pi x \mathrm {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )^{3}+72 x \ln \relax (5)+72 x \ln \relax (3)+144 \ln \relax (2) \ln \relax (x )-144 \ln \relax (2) \ln \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )\) | \(180\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.48, size = 40, normalized size = 1.54 \begin {gather*} 72 \, {\left (i \, \pi + \log \relax (5) + \log \relax (3)\right )} x - 72 \, {\left (x - 2 \, \log \relax (2)\right )} \log \relax (x) + 72 \, {\left (x - 2 \, \log \relax (2)\right )} \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.73, size = 41, normalized size = 1.58 \begin {gather*} 72\,x\,\ln \left (-\frac {15\,{\left ({\mathrm {e}}^x\right )}^{1/5}-15\,x+75}{x}\right )-72\,\ln \relax (4)\,\ln \left ({\left ({\mathrm {e}}^x\right )}^{1/5}-x+5\right )+72\,\ln \relax (4)\,\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.50, size = 41, normalized size = 1.58 \begin {gather*} 72 x \log {\left (\frac {15 x - 15 e^{\frac {x}{5}} - 75}{x} \right )} + 144 \log {\relax (2 )} \log {\relax (x )} - 144 \log {\relax (2 )} \log {\left (- x + e^{\frac {x}{5}} + 5 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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