Integrand size = 51, antiderivative size = 19 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=11+e^x+\log \left (\left (\left (16+\frac {5}{x}\right )^2+x\right )^2\right ) \]
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Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6873, 6874, 2225, 1601} \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=2 \log \left (x^3+256 x^2+160 x+25\right )+e^x-4 \log (x) \]
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Rule 1601
Rule 2225
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{x \left (25+160 x+256 x^2+x^3\right )} \, dx \\ & = \int \left (e^x+\frac {2 \left (-50-160 x+x^3\right )}{x \left (25+160 x+256 x^2+x^3\right )}\right ) \, dx \\ & = 2 \int \frac {-50-160 x+x^3}{x \left (25+160 x+256 x^2+x^3\right )} \, dx+\int e^x \, dx \\ & = e^x+2 \int \left (-\frac {2}{x}+\frac {160+512 x+3 x^2}{25+160 x+256 x^2+x^3}\right ) \, dx \\ & = e^x-4 \log (x)+2 \int \frac {160+512 x+3 x^2}{25+160 x+256 x^2+x^3} \, dx \\ & = e^x-4 \log (x)+2 \log \left (25+160 x+256 x^2+x^3\right ) \\ \end{align*}
Time = 1.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^x-4 \log (x)+2 \log \left (25+160 x+256 x^2+x^3\right ) \]
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Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26
method | result | size |
norman | \({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\) | \(24\) |
risch | \({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\) | \(24\) |
parallelrisch | \({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\) | \(24\) |
parts | \({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\) | \(24\) |
default | \({\mathrm e}^{x}+\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\left (256 \textit {\_R1}^{2}+160 \textit {\_R1} +25\right ) {\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\left (-\textit {\_R}^{2}-256 \textit {\_R} -160\right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )-4 \ln \left (x \right )-320 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )-25 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-160 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-256 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R1}^{2} {\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )\) | \(311\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} - 4 \log {\left (x \right )} + 2 \log {\left (x^{3} + 256 x^{2} + 160 x + 25 \right )} \]
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Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]
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Time = 8.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=2\,\ln \left (x^3+256\,x^2+160\,x+25\right )+{\mathrm {e}}^x-4\,\ln \left (x\right ) \]
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