Integrand size = 58, antiderivative size = 19 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=\left (7+e^{4/3}+22 x+e^x x \log (5)\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(19)=38\).
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.79, number of steps used = 19, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2227, 2225, 2207, 1607} \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=484 x^2+e^{2 x} x^2 \log ^2(5)+44 e^x x^2 \log (5)+44 \left (7+e^{4/3}\right ) x+14 e^x x \log (5)-2 e^{x+\frac {4}{3}} \log (5)+2 e^{x+\frac {4}{3}} (x+1) \log (5) \]
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Rule 1607
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = 44 \left (7+e^{4/3}\right ) x+484 x^2+\log (5) \int e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \, dx+\log ^2(5) \int e^{2 x} \left (2 x+2 x^2\right ) \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2+\log (5) \int \left (14 e^x+102 e^x x+44 e^x x^2+2 e^{\frac {4}{3}+x} (1+x)\right ) \, dx+\log ^2(5) \int e^{2 x} x (2+2 x) \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2+(2 \log (5)) \int e^{\frac {4}{3}+x} (1+x) \, dx+(14 \log (5)) \int e^x \, dx+(44 \log (5)) \int e^x x^2 \, dx+(102 \log (5)) \int e^x x \, dx+\log ^2(5) \int \left (2 e^{2 x} x+2 e^{2 x} x^2\right ) \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2+14 e^x \log (5)+102 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)-(2 \log (5)) \int e^{\frac {4}{3}+x} \, dx-(88 \log (5)) \int e^x x \, dx-(102 \log (5)) \int e^x \, dx+\left (2 \log ^2(5)\right ) \int e^{2 x} x \, dx+\left (2 \log ^2(5)\right ) \int e^{2 x} x^2 \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2-88 e^x \log (5)-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)+e^{2 x} x \log ^2(5)+e^{2 x} x^2 \log ^2(5)+(88 \log (5)) \int e^x \, dx-\log ^2(5) \int e^{2 x} \, dx-\left (2 \log ^2(5)\right ) \int e^{2 x} x \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)-\frac {1}{2} e^{2 x} \log ^2(5)+e^{2 x} x^2 \log ^2(5)+\log ^2(5) \int e^{2 x} \, dx \\ & = 44 \left (7+e^{4/3}\right ) x+484 x^2-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)+e^{2 x} x^2 \log ^2(5) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=\left (7+e^{4/3}+22 x+e^x x \log (5)\right )^2 \]
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Leaf count of result is larger than twice the leaf count of optimal. \(45\) vs. \(2(15)=30\).
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.42
method | result | size |
risch | \({\mathrm e}^{2 x} \ln \left (5\right )^{2} x^{2}+\ln \left (5\right ) \left (2 x \,{\mathrm e}^{\frac {4}{3}}+44 x^{2}+14 x \right ) {\mathrm e}^{x}+44 x \,{\mathrm e}^{\frac {4}{3}}+484 x^{2}+308 x\) | \(46\) |
norman | \(\left (44 \,{\mathrm e}^{\frac {4}{3}}+308\right ) x +\left (2 \ln \left (5\right ) {\mathrm e}^{\frac {4}{3}}+14 \ln \left (5\right )\right ) x \,{\mathrm e}^{x}+{\mathrm e}^{2 x} \ln \left (5\right )^{2} x^{2}+484 x^{2}+44 x^{2} \ln \left (5\right ) {\mathrm e}^{x}\) | \(51\) |
default | \({\mathrm e}^{2 x} \ln \left (5\right )^{2} x^{2}+2 \,{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{\frac {4}{3}} x +44 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+14 x \,{\mathrm e}^{x} \ln \left (5\right )+44 x \,{\mathrm e}^{\frac {4}{3}}+484 x^{2}+308 x\) | \(52\) |
parallelrisch | \({\mathrm e}^{2 x} \ln \left (5\right )^{2} x^{2}+2 \,{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{\frac {4}{3}} x +44 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+14 x \,{\mathrm e}^{x} \ln \left (5\right )+484 x^{2}+\left (44 \,{\mathrm e}^{\frac {4}{3}}+308\right ) x\) | \(52\) |
parts | \({\mathrm e}^{2 x} \ln \left (5\right )^{2} x^{2}+2 \,{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{\frac {4}{3}} x +44 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+14 x \,{\mathrm e}^{x} \ln \left (5\right )+44 x \,{\mathrm e}^{\frac {4}{3}}+484 x^{2}+308 x\) | \(52\) |
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (15) = 30\).
Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.37 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=x^{2} e^{\left (2 \, x\right )} \log \left (5\right )^{2} + 2 \, {\left (22 \, x^{2} + x e^{\frac {4}{3}} + 7 \, x\right )} e^{x} \log \left (5\right ) + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 308 \, x \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (19) = 38\).
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=x^{2} e^{2 x} \log {\left (5 \right )}^{2} + 484 x^{2} + x \left (44 e^{\frac {4}{3}} + 308\right ) + \left (44 x^{2} \log {\left (5 \right )} + 2 x e^{\frac {4}{3}} \log {\left (5 \right )} + 14 x \log {\left (5 \right )}\right ) e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (15) = 30\).
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.32 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=x^{2} e^{\left (2 \, x\right )} \log \left (5\right )^{2} + 2 \, {\left (22 \, x^{2} + x {\left (e^{\frac {4}{3}} + 7\right )}\right )} e^{x} \log \left (5\right ) + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 308 \, x \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (15) = 30\).
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.58 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=x^{2} e^{\left (2 \, x\right )} \log \left (5\right )^{2} + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 2 \, {\left (x e^{\left (x + \frac {4}{3}\right )} + {\left (22 \, x^{2} + 7 \, x\right )} e^{x}\right )} \log \left (5\right ) + 308 \, x \]
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Time = 7.89 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \left (308+44 e^{4/3}+968 x+e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \log (5)+e^{2 x} \left (2 x+2 x^2\right ) \log ^2(5)\right ) \, dx=x\,\left ({\mathrm {e}}^x\,\ln \left (5\right )+22\right )\,\left (22\,x+2\,{\mathrm {e}}^{4/3}+x\,{\mathrm {e}}^x\,\ln \left (5\right )+14\right ) \]
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