Optimal. Leaf size=24 \[ \frac {4}{125 \left (2-e^x\right ) x^5 (4+\log (5+x))} \]
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Rubi [F] time = 19.50, 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 {-800-168 x+e^x \left (400+164 x+16 x^2\right )+\left (-200-40 x+e^x \left (100+40 x+4 x^2\right )\right ) \log (5+x)}{40000 x^6+8000 x^7+e^x \left (-40000 x^6-8000 x^7\right )+e^{2 x} \left (10000 x^6+2000 x^7\right )+\left (20000 x^6+4000 x^7+e^x \left (-20000 x^6-4000 x^7\right )+e^{2 x} \left (5000 x^6+1000 x^7\right )\right ) \log (5+x)+\left (2500 x^6+500 x^7+e^x \left (-2500 x^6-500 x^7\right )+e^{2 x} \left (625 x^6+125 x^7\right )\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 {4 \left (-200-42 x+e^x \left (100+41 x+4 x^2\right )+(5+x) \left (-10+e^x (5+x)\right ) \log (5+x)\right )}{125 \left (2-e^x\right )^2 x^6 (5+x) (4+\log (5+x))^2} \, dx\\ &=\frac {4}{125} \int \frac {-200-42 x+e^x \left (100+41 x+4 x^2\right )+(5+x) \left (-10+e^x (5+x)\right ) \log (5+x)}{\left (2-e^x\right )^2 x^6 (5+x) (4+\log (5+x))^2} \, dx\\ &=\frac {4}{125} \int \left (\frac {2}{\left (-2+e^x\right )^2 x^5 (4+\log (5+x))}+\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^6 (5+x) (4+\log (5+x))^2}\right ) \, dx\\ &=\frac {4}{125} \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^6 (5+x) (4+\log (5+x))^2} \, dx+\frac {8}{125} \int \frac {1}{\left (-2+e^x\right )^2 x^5 (4+\log (5+x))} \, dx\\ &=\frac {4}{125} \int \left (\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{5 \left (-2+e^x\right ) x^6 (4+\log (5+x))^2}-\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{25 \left (-2+e^x\right ) x^5 (4+\log (5+x))^2}+\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{125 \left (-2+e^x\right ) x^4 (4+\log (5+x))^2}-\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{625 \left (-2+e^x\right ) x^3 (4+\log (5+x))^2}+\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{3125 \left (-2+e^x\right ) x^2 (4+\log (5+x))^2}-\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{15625 \left (-2+e^x\right ) x (4+\log (5+x))^2}+\frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{15625 \left (-2+e^x\right ) (5+x) (4+\log (5+x))^2}\right ) \, dx+\frac {8}{125} \int \frac {1}{\left (-2+e^x\right )^2 x^5 (4+\log (5+x))} \, dx\\ &=-\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x (4+\log (5+x))^2} \, dx}{1953125}+\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) (5+x) (4+\log (5+x))^2} \, dx}{1953125}+\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^2 (4+\log (5+x))^2} \, dx}{390625}-\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^3 (4+\log (5+x))^2} \, dx}{78125}+\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^4 (4+\log (5+x))^2} \, dx}{15625}-\frac {4 \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^5 (4+\log (5+x))^2} \, dx}{3125}+\frac {4}{625} \int \frac {100+41 x+4 x^2+25 \log (5+x)+10 x \log (5+x)+x^2 \log (5+x)}{\left (-2+e^x\right ) x^6 (4+\log (5+x))^2} \, dx+\frac {8}{125} \int \frac {1}{\left (-2+e^x\right )^2 x^5 (4+\log (5+x))} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 1.79, size = 22, normalized size = 0.92 \begin {gather*} -\frac {4}{125 \left (-2+e^x\right ) x^5 (4+\log (5+x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.10, size = 34, normalized size = 1.42 \begin {gather*} -\frac {4}{125 \, {\left (4 \, x^{5} e^{x} - 8 \, x^{5} + {\left (x^{5} e^{x} - 2 \, x^{5}\right )} \log \left (x + 5\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 36, normalized size = 1.50 \begin {gather*} -\frac {4}{125 \, {\left (x^{5} e^{x} \log \left (x + 5\right ) + 4 \, x^{5} e^{x} - 2 \, x^{5} \log \left (x + 5\right ) - 8 \, x^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 20, normalized size = 0.83
method | result | size |
risch | \(-\frac {4}{125 \left ({\mathrm e}^{x}-2\right ) x^{5} \left (\ln \left (5+x \right )+4\right )}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 34, normalized size = 1.42 \begin {gather*} -\frac {4}{125 \, {\left (4 \, x^{5} e^{x} - 8 \, x^{5} + {\left (x^{5} e^{x} - 2 \, x^{5}\right )} \log \left (x + 5\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.23, size = 19, normalized size = 0.79 \begin {gather*} -\frac {4}{125\,x^5\,\left ({\mathrm {e}}^x-2\right )\,\left (\ln \left (x+5\right )+4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.40, size = 36, normalized size = 1.50 \begin {gather*} - \frac {4}{- 250 x^{5} \log {\left (x + 5 \right )} - 1000 x^{5} + \left (125 x^{5} \log {\left (x + 5 \right )} + 500 x^{5}\right ) e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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