Integrand size = 13, antiderivative size = 88 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b^2 n^2 x}{1-4 b^2 n^2}-\frac {2 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac {x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2} \]
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Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5630, 8} \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac {2 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac {2 b^2 n^2 x}{1-4 b^2 n^2} \]
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Rule 8
Rule 5630
Rubi steps \begin{align*} \text {integral}& = -\frac {2 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac {x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac {\left (2 b^2 n^2\right ) \int 1 \, dx}{1-4 b^2 n^2} \\ & = \frac {2 b^2 n^2 x}{1-4 b^2 n^2}-\frac {2 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac {x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {x \left (-1+4 b^2 n^2+\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-2 b n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{-2+8 b^2 n^2} \]
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Time = 2.71 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66
method | result | size |
parallelrisch | \(-\frac {x \left (4 b^{2} n^{2}-2 b n \sinh \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+\cosh \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-1\right )}{8 b^{2} n^{2}-2}\) | \(58\) |
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Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 \, b n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (4 \, b^{2} n^{2} - 1\right )} x}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} \]
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\[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int \sinh ^{2}{\left (a - \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {1}{2 n} \\\int \sinh ^{2}{\left (a + \frac {\log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {1}{2 n} \\\frac {2 b^{2} n^{2} x \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} - \frac {2 b^{2} n^{2} x \cosh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} + \frac {2 b n x \sinh {\left (a + b \log {\left (c x^{n} \right )} \right )} \cosh {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} - \frac {x \sinh ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} - 1} & \text {otherwise} \end {cases} \]
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Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}}{4 \, {\left (2 \, b n + 1\right )}} - \frac {1}{2} \, x - \frac {x e^{\left (-2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b}\right )} {\left (x^{n}\right )}^{2 \, b}} \]
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Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.92 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b c^{2 \, b} n x x^{2 \, b n} e^{\left (2 \, a\right )}}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} - \frac {2 \, b^{2} n^{2} x}{4 \, b^{2} n^{2} - 1} - \frac {c^{2 \, b} x x^{2 \, b n} e^{\left (2 \, a\right )}}{4 \, {\left (4 \, b^{2} n^{2} - 1\right )}} - \frac {b n x e^{\left (-2 \, a\right )}}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} + \frac {x}{2 \, {\left (4 \, b^{2} n^{2} - 1\right )}} - \frac {x e^{\left (-2 \, a\right )}}{4 \, {\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} \]
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Time = 1.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{8\,b\,n+4}-\frac {x\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (8\,b\,n-4\right )}-\frac {x}{2} \]
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