Integrand size = 55, antiderivative size = 23 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=5+e^{\frac {4+2 x}{x}} \left (\frac {26}{5}+x^2\right ) \log (x) \]
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Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 30, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 2241, 2245, 2237, 2634, 6618, 6610, 2240, 2258} \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=e^{\frac {4}{x}+2} x^2 \log (x)+\frac {26}{5} e^{\frac {4}{x}+2} \log (x) \]
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Rule 12
Rule 14
Rule 2237
Rule 2240
Rule 2241
Rule 2245
Rule 2258
Rule 2634
Rule 6610
Rule 6618
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {26 e^{2+\frac {4}{x}}}{x}+5 e^{2+\frac {4}{x}} x-20 e^{2+\frac {4}{x}} \log (x)-\frac {104 e^{2+\frac {4}{x}} \log (x)}{x^2}+10 e^{2+\frac {4}{x}} x \log (x)\right ) \, dx \\ & = 2 \int e^{2+\frac {4}{x}} x \log (x) \, dx-4 \int e^{2+\frac {4}{x}} \log (x) \, dx+\frac {26}{5} \int \frac {e^{2+\frac {4}{x}}}{x} \, dx-\frac {104}{5} \int \frac {e^{2+\frac {4}{x}} \log (x)}{x^2} \, dx+\int e^{2+\frac {4}{x}} x \, dx \\ & = \frac {1}{2} e^{2+\frac {4}{x}} x^2-\frac {26}{5} e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)+2 \int e^{2+\frac {4}{x}} \, dx-2 \int \left (\frac {1}{2} e^{2+\frac {4}{x}} (4+x)-\frac {8 e^2 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+4 \int e^2 \left (e^{4/x}-\frac {4 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+\frac {104}{5} \int -\frac {e^{2+\frac {4}{x}}}{4 x} \, dx \\ & = 2 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-\frac {26}{5} e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-\frac {26}{5} \int \frac {e^{2+\frac {4}{x}}}{x} \, dx+8 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx+\left (4 e^2\right ) \int \left (e^{4/x}-\frac {4 \text {Ei}\left (\frac {4}{x}\right )}{x}\right ) \, dx+\left (16 e^2\right ) \int \frac {\text {Ei}\left (\frac {4}{x}\right )}{x} \, dx-\int e^{2+\frac {4}{x}} (4+x) \, dx \\ & = 2 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-8 e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)+\left (4 e^2\right ) \int e^{4/x} \, dx-\left (16 e^2\right ) \int \frac {\text {Ei}\left (\frac {4}{x}\right )}{x} \, dx-\left (16 e^2\right ) \text {Subst}\left (\int \frac {\text {Ei}(4 x)}{x} \, dx,x,\frac {1}{x}\right )-\int \left (4 e^{2+\frac {4}{x}}+e^{2+\frac {4}{x}} x\right ) \, dx \\ & = 6 e^{2+\frac {4}{x}} x+\frac {1}{2} e^{2+\frac {4}{x}} x^2-8 e^2 \text {Ei}\left (\frac {4}{x}\right )-16 e^2 \left (E_1\left (-\frac {4}{x}\right )+\text {Ei}\left (\frac {4}{x}\right )\right ) \log \left (\frac {1}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-4 \int e^{2+\frac {4}{x}} \, dx+\left (16 e^2\right ) \int \frac {e^{4/x}}{x} \, dx+\left (16 e^2\right ) \text {Subst}\left (\int \frac {E_1(-4 x)}{x} \, dx,x,\frac {1}{x}\right )+\left (16 e^2\right ) \text {Subst}\left (\int \frac {\text {Ei}(4 x)}{x} \, dx,x,\frac {1}{x}\right )-\int e^{2+\frac {4}{x}} x \, dx \\ & = 2 e^{2+\frac {4}{x}} x-24 e^2 \text {Ei}\left (\frac {4}{x}\right )-\frac {64 e^2 \, _3F_3\left (1,1,1;2,2,2;\frac {4}{x}\right )}{x}-8 e^2 \log ^2\left (-\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+16 e^2 \gamma \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-2 \int e^{2+\frac {4}{x}} \, dx-16 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx-\left (16 e^2\right ) \text {Subst}\left (\int \frac {E_1(-4 x)}{x} \, dx,x,\frac {1}{x}\right ) \\ & = -8 e^2 \text {Ei}\left (\frac {4}{x}\right )+\frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x)-8 \int \frac {e^{2+\frac {4}{x}}}{x} \, dx \\ & = \frac {26}{5} e^{2+\frac {4}{x}} \log (x)+e^{2+\frac {4}{x}} x^2 \log (x) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=\frac {1}{5} e^{2+\frac {4}{x}} \left (26+5 x^2\right ) \log (x) \]
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Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\left (5 x^{2}+26\right ) {\mathrm e}^{\frac {4+2 x}{x}} \ln \left (x \right )}{5}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {4+2 x}{x}} \ln \left (x \right ) x^{2}+\frac {26 \,{\mathrm e}^{\frac {4+2 x}{x}} \ln \left (x \right )}{5}\) | \(30\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {4+2 x}{x}} \ln \left (x \right )+\frac {26 x \,{\mathrm e}^{\frac {4+2 x}{x}} \ln \left (x \right )}{5}}{x}\) | \(37\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=\frac {1}{5} \, {\left (5 \, x^{2} + 26\right )} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \left (x\right ) \]
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Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=\frac {\left (5 x^{2} \log {\left (x \right )} + 26 \log {\left (x \right )}\right ) e^{\frac {2 x + 4}{x}}}{5} \]
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\[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=\int { \frac {2 \, {\left (5 \, x^{3} - 10 \, x^{2} - 52\right )} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \left (x\right ) + {\left (5 \, x^{3} + 26 \, x\right )} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )}}{5 \, x^{2}} \,d x } \]
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Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=x^{2} e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \left (x\right ) + \frac {26}{5} \, e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} \log \left (x\right ) \]
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Time = 11.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {4+2 x}{x}} \left (26 x+5 x^3\right )+e^{\frac {4+2 x}{x}} \left (-104-20 x^2+10 x^3\right ) \log (x)}{5 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {4}{x}+2}\,\ln \left (x\right )\,\left (5\,x^2+26\right )}{5} \]
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