Integrand size = 16, antiderivative size = 21 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=-5+5 (14+x)^2+x \left (1+e^x \log ^2(5)\right ) \]
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Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2207, 2225} \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=5 x^2+141 x-e^x \log ^2(5)+e^x (x+1) \log ^2(5) \]
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Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = 141 x+5 x^2+\log ^2(5) \int e^x (1+x) \, dx \\ & = 141 x+5 x^2+e^x (1+x) \log ^2(5)-\log ^2(5) \int e^x \, dx \\ & = 141 x+5 x^2-e^x \log ^2(5)+e^x (1+x) \log ^2(5) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=141 x+5 x^2+e^x x \log ^2(5) \]
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Time = 0.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
default | \(141 x +x \ln \left (5\right )^{2} {\mathrm e}^{x}+5 x^{2}\) | \(18\) |
norman | \(141 x +x \ln \left (5\right )^{2} {\mathrm e}^{x}+5 x^{2}\) | \(18\) |
risch | \(141 x +x \ln \left (5\right )^{2} {\mathrm e}^{x}+5 x^{2}\) | \(18\) |
parallelrisch | \(141 x +x \ln \left (5\right )^{2} {\mathrm e}^{x}+5 x^{2}\) | \(18\) |
parts | \(141 x +x \ln \left (5\right )^{2} {\mathrm e}^{x}+5 x^{2}\) | \(18\) |
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=x e^{x} \log \left (5\right )^{2} + 5 \, x^{2} + 141 \, x \]
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Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=5 x^{2} + x e^{x} \log {\left (5 \right )}^{2} + 141 x \]
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Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=x e^{x} \log \left (5\right )^{2} + 5 \, x^{2} + 141 \, x \]
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=x e^{x} \log \left (5\right )^{2} + 5 \, x^{2} + 141 \, x \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \left (141+10 x+e^x (1+x) \log ^2(5)\right ) \, dx=x\,\left (5\,x+{\mathrm {e}}^x\,{\ln \left (5\right )}^2+141\right ) \]
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