Integrand size = 30, antiderivative size = 19 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4+4 e^x \log (3)+\log \left (\frac {16+x^2}{x}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1607, 6857, 457, 78, 2225} \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=\log \left (x^2+16\right )-\log (x)+e^x \log (81) \]
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Rule 78
Rule 457
Rule 1607
Rule 2225
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{x \left (16+x^2\right )} \, dx \\ & = \int \left (\frac {-16+x^2}{x \left (16+x^2\right )}+e^x \log (81)\right ) \, dx \\ & = \log (81) \int e^x \, dx+\int \frac {-16+x^2}{x \left (16+x^2\right )} \, dx \\ & = e^x \log (81)+\frac {1}{2} \text {Subst}\left (\int \frac {-16+x}{x (16+x)} \, dx,x,x^2\right ) \\ & = e^x \log (81)+\frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{x}+\frac {2}{16+x}\right ) \, dx,x,x^2\right ) \\ & = e^x \log (81)-\log (x)+\log \left (16+x^2\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=e^x \log (81)-\log (x)+\log \left (16+x^2\right ) \]
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Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
norman | \(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\) | \(18\) |
risch | \(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\) | \(18\) |
parallelrisch | \(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\) | \(18\) |
parts | \(4 \ln \left (3\right ) {\mathrm e}^{x}+\ln \left (x^{2}+16\right )-\ln \left (x \right )\) | \(18\) |
default | \(\ln \left (x^{2}+16\right )-\ln \left (x \right )+8 i \ln \left (3\right ) {\mathrm e}^{4 i} \operatorname {Ei}_{1}\left (-x +4 i\right )-8 i \ln \left (3\right ) {\mathrm e}^{-4 i} \operatorname {Ei}_{1}\left (-x -4 i\right )+4 \ln \left (3\right ) \left ({\mathrm e}^{x}-2 i {\mathrm e}^{4 i} \operatorname {Ei}_{1}\left (-x +4 i\right )+2 i {\mathrm e}^{-4 i} \operatorname {Ei}_{1}\left (-x -4 i\right )\right )\) | \(79\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 e^{x} \log {\left (3 \right )} - \log {\left (x \right )} + \log {\left (x^{2} + 16 \right )} \]
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Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=4 \, e^{x} \log \left (3\right ) + \log \left (x^{2} + 16\right ) - \log \left (x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16+x^2+e^x \left (64 x+4 x^3\right ) \log (3)}{16 x+x^3} \, dx=\ln \left (x^2+16\right )-\ln \left (x\right )+4\,{\mathrm {e}}^x\,\ln \left (3\right ) \]
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