Integrand size = 42, antiderivative size = 26 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=\log \left (\frac {2 x^3}{3 \left (2-4 e^{-1-16 x^2} x^2\right )}\right ) \]
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\[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=\int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{x}-\frac {4 x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2}\right ) \, dx \\ & = 3 \log (x)-4 \int \frac {x \left (-1+16 x^2\right )}{e^{1+16 x^2}-2 x^2} \, dx \\ & = 3 \log (x)-2 \text {Subst}\left (\int \frac {-1+16 x}{e^{1+16 x}-2 x} \, dx,x,x^2\right ) \\ & = 3 \log (x)-2 \text {Subst}\left (\int \left (-\frac {1}{e^{1+16 x}-2 x}+\frac {16 x}{e^{1+16 x}-2 x}\right ) \, dx,x,x^2\right ) \\ & = 3 \log (x)+2 \text {Subst}\left (\int \frac {1}{e^{1+16 x}-2 x} \, dx,x,x^2\right )-32 \text {Subst}\left (\int \frac {x}{e^{1+16 x}-2 x} \, dx,x,x^2\right ) \\ \end{align*}
Time = 2.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 x^2+3 \log (x)-\log \left (e^{1+16 x^2}-2 x^2\right ) \]
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Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(16 x^{2}+3 \ln \left (x \right )-\ln \left (x^{2}-\frac {{\mathrm e}^{16 x^{2}+1}}{2}\right )\) | \(28\) |
risch | \(3 \ln \left (x \right )+16 x^{2}+1-\ln \left (-2 x^{2}+{\mathrm e}^{16 x^{2}+1}\right )\) | \(29\) |
norman | \(16 x^{2}+3 \ln \left (x \right )-\ln \left (2 x^{2}-{\mathrm e}^{16 x^{2}+1}\right )\) | \(30\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} - \log \left (-2 \, x^{2} + e^{\left (16 \, x^{2} + 1\right )}\right ) + 3 \, \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 x^{2} + 3 \log {\left (x \right )} - \log {\left (- 2 x^{2} + e^{16 x^{2} + 1} \right )} \]
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Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} - \log \left (-{\left (2 \, x^{2} - e^{\left (16 \, x^{2} + 1\right )}\right )} e^{\left (-1\right )}\right ) + 3 \, \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=16 \, x^{2} + \frac {3}{2} \, \log \left (16 \, x^{2}\right ) - \log \left (16 \, x^{2} - 8 \, e^{\left (16 \, x^{2} + 1\right )}\right ) + 1 \]
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Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {3 e^{1+16 x^2}-2 x^2-64 x^4}{e^{1+16 x^2} x-2 x^3} \, dx=3\,\ln \left (x\right )-\ln \left (2\,x^2-\mathrm {e}\,{\mathrm {e}}^{16\,x^2}\right )+16\,x^2 \]
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