Integrand size = 59, antiderivative size = 18 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3 e^x \log \left (4+\frac {e^{5 x}}{4}+x\right ) \]
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Time = 0.53 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6874, 2225, 2634} \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3 e^x \log \left (\frac {1}{4} \left (4 x+e^{5 x}+16\right )\right ) \]
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Rule 2225
Rule 2634
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {12 e^x (19+5 x)}{16+e^{5 x}+4 x}+3 e^x \left (5+\log \left (4+\frac {e^{5 x}}{4}+x\right )\right )\right ) \, dx \\ & = 3 \int e^x \left (5+\log \left (4+\frac {e^{5 x}}{4}+x\right )\right ) \, dx-12 \int \frac {e^x (19+5 x)}{16+e^{5 x}+4 x} \, dx \\ & = 3 \int \left (5 e^x+e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )\right ) \, dx-12 \int \left (\frac {19 e^x}{16+e^{5 x}+4 x}+\frac {5 e^x x}{16+e^{5 x}+4 x}\right ) \, dx \\ & = 3 \int e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right ) \, dx+15 \int e^x \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx \\ & = 15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )-3 \int \frac {e^x \left (4+5 e^{5 x}\right )}{e^{5 x}+4 (4+x)} \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx \\ & = 15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )-3 \int \left (5 e^x-\frac {4 e^x (19+5 x)}{16+e^{5 x}+4 x}\right ) \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx \\ & = 15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )+12 \int \frac {e^x (19+5 x)}{16+e^{5 x}+4 x} \, dx-15 \int e^x \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx \\ & = 3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )+12 \int \left (\frac {19 e^x}{16+e^{5 x}+4 x}+\frac {5 e^x x}{16+e^{5 x}+4 x}\right ) \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx \\ & = 3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3 e^x \log \left (4+\frac {e^{5 x}}{4}+x\right ) \]
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Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83
method | result | size |
risch | \(3 \ln \left (\frac {{\mathrm e}^{5 x}}{4}+x +4\right ) {\mathrm e}^{x}\) | \(15\) |
parallelrisch | \(3 \ln \left (\frac {{\mathrm e}^{5 x}}{4}+x +4\right ) {\mathrm e}^{x}\) | \(15\) |
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Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3 \, e^{x} \log \left (x + \frac {1}{4} \, e^{\left (5 \, x\right )} + 4\right ) \]
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Timed out. \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=\text {Timed out} \]
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Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=-6 \, e^{x} \log \left (2\right ) + 3 \, e^{x} \log \left (4 \, x + e^{\left (5 \, x\right )} + 16\right ) \]
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Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3 \, e^{x} \log \left (x + \frac {1}{4} \, e^{\left (5 \, x\right )} + 4\right ) \]
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Time = 0.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {12 e^x+15 e^{6 x}+\left (3 e^{6 x}+e^x (48+12 x)\right ) \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )}{16+e^{5 x}+4 x} \, dx=3\,{\mathrm {e}}^x\,\ln \left (x+\frac {{\mathrm {e}}^{5\,x}}{4}+4\right ) \]
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