Integrand size = 60, antiderivative size = 21 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {e^{2+2 x} \log ^2(4+x)}{5+2 x} \]
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\[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{2+2 x} \log (4+x) \left (5+2 x+2 \left (8+6 x+x^2\right ) \log (4+x)\right )}{(4+x) (5+2 x)^2} \, dx \\ & = 2 \int \frac {e^{2+2 x} \log (4+x) \left (5+2 x+2 \left (8+6 x+x^2\right ) \log (4+x)\right )}{(4+x) (5+2 x)^2} \, dx \\ & = 2 \int \left (\frac {e^{2+2 x} \log (4+x)}{(4+x) (5+2 x)}+\frac {2 e^{2+2 x} (2+x) \log ^2(4+x)}{(5+2 x)^2}\right ) \, dx \\ & = 2 \int \frac {e^{2+2 x} \log (4+x)}{(4+x) (5+2 x)} \, dx+4 \int \frac {e^{2+2 x} (2+x) \log ^2(4+x)}{(5+2 x)^2} \, dx \\ & = -\frac {2 \operatorname {ExpIntegralEi}(2 (4+x)) \log (4+x)}{3 e^6}+\frac {2 \operatorname {ExpIntegralEi}(5+2 x) \log (4+x)}{3 e^3}-2 \int \frac {-\operatorname {ExpIntegralEi}(2 (4+x))+e^3 \operatorname {ExpIntegralEi}(5+2 x)}{3 e^6 (4+x)} \, dx+4 \int \left (-\frac {e^{2+2 x} \log ^2(4+x)}{2 (5+2 x)^2}+\frac {e^{2+2 x} \log ^2(4+x)}{2 (5+2 x)}\right ) \, dx \\ & = -\frac {2 \operatorname {ExpIntegralEi}(2 (4+x)) \log (4+x)}{3 e^6}+\frac {2 \operatorname {ExpIntegralEi}(5+2 x) \log (4+x)}{3 e^3}-2 \int \frac {e^{2+2 x} \log ^2(4+x)}{(5+2 x)^2} \, dx+2 \int \frac {e^{2+2 x} \log ^2(4+x)}{5+2 x} \, dx-\frac {2 \int \frac {-\operatorname {ExpIntegralEi}(2 (4+x))+e^3 \operatorname {ExpIntegralEi}(5+2 x)}{4+x} \, dx}{3 e^6} \\ & = -\frac {2 \operatorname {ExpIntegralEi}(2 (4+x)) \log (4+x)}{3 e^6}+\frac {2 \operatorname {ExpIntegralEi}(5+2 x) \log (4+x)}{3 e^3}-2 \int \frac {e^{2+2 x} \log ^2(4+x)}{(5+2 x)^2} \, dx+2 \int \frac {e^{2+2 x} \log ^2(4+x)}{5+2 x} \, dx-\frac {2 \int \left (\frac {e^3 \operatorname {ExpIntegralEi}(5+2 x)}{4+x}+\frac {\operatorname {ExpIntegralEi}(8+2 x)}{-4-x}\right ) \, dx}{3 e^6} \\ & = -\frac {2 \operatorname {ExpIntegralEi}(2 (4+x)) \log (4+x)}{3 e^6}+\frac {2 \operatorname {ExpIntegralEi}(5+2 x) \log (4+x)}{3 e^3}-2 \int \frac {e^{2+2 x} \log ^2(4+x)}{(5+2 x)^2} \, dx+2 \int \frac {e^{2+2 x} \log ^2(4+x)}{5+2 x} \, dx-\frac {2 \int \frac {\operatorname {ExpIntegralEi}(8+2 x)}{-4-x} \, dx}{3 e^6}-\frac {2 \int \frac {\operatorname {ExpIntegralEi}(5+2 x)}{4+x} \, dx}{3 e^3} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {2 e^{2+2 x} \log ^2(4+x)}{10+4 x} \]
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Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {{\mathrm e}^{2+2 x} \ln \left (4+x \right )^{2}}{5+2 x}\) | \(21\) |
parallelrisch | \(\frac {{\mathrm e}^{2+2 x} \ln \left (4+x \right )^{2}}{5+2 x}\) | \(21\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {e^{\left (2 \, x + 2\right )} \log \left (x + 4\right )^{2}}{2 \, x + 5} \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {e^{2 x + 2} \log {\left (x + 4 \right )}^{2}}{2 x + 5} \]
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {e^{\left (2 \, x + 2\right )} \log \left (x + 4\right )^{2}}{2 \, x + 5} \]
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Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\frac {e^{\left (2 \, x + 8\right )} \log \left (x + 4\right )^{2}}{2 \, {\left (x + 4\right )} e^{6} - 3 \, e^{6}} \]
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Timed out. \[ \int \frac {e^{2+2 x} (10+4 x) \log (4+x)+e^{2+2 x} \left (32+24 x+4 x^2\right ) \log ^2(4+x)}{100+105 x+36 x^2+4 x^3} \, dx=\int \frac {{\mathrm {e}}^{2\,x+2}\,\left (4\,x^2+24\,x+32\right )\,{\ln \left (x+4\right )}^2+{\mathrm {e}}^{2\,x+2}\,\left (4\,x+10\right )\,\ln \left (x+4\right )}{4\,x^3+36\,x^2+105\,x+100} \,d x \]
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