Optimal. Leaf size=86 \[ -\frac{e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \text{Gamma}(m+2,-2 b x)}{b^2}-\frac{e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \text{Gamma}(m+2,2 b x)}{b^2}-\frac{x^{m+2}}{2 (m+2)} \]
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Rubi [A] time = 0.139944, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3312, 3307, 2181} \[ -\frac{e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \text{Gamma}(m+2,-2 b x)}{b^2}-\frac{e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \text{Gamma}(m+2,2 b x)}{b^2}-\frac{x^{m+2}}{2 (m+2)} \]
Antiderivative was successfully verified.
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Rule 3312
Rule 3307
Rule 2181
Rubi steps
\begin{align*} \int x^{1+m} \sinh ^2(a+b x) \, dx &=-\int \left (\frac{x^{1+m}}{2}-\frac{1}{2} x^{1+m} \cosh (2 a+2 b x)\right ) \, dx\\ &=-\frac{x^{2+m}}{2 (2+m)}+\frac{1}{2} \int x^{1+m} \cosh (2 a+2 b x) \, dx\\ &=-\frac{x^{2+m}}{2 (2+m)}+\frac{1}{4} \int e^{-i (2 i a+2 i b x)} x^{1+m} \, dx+\frac{1}{4} \int e^{i (2 i a+2 i b x)} x^{1+m} \, dx\\ &=-\frac{x^{2+m}}{2 (2+m)}-\frac{2^{-4-m} e^{2 a} x^m (-b x)^{-m} \Gamma (2+m,-2 b x)}{b^2}-\frac{2^{-4-m} e^{-2 a} x^m (b x)^{-m} \Gamma (2+m,2 b x)}{b^2}\\ \end{align*}
Mathematica [A] time = 0.130026, size = 79, normalized size = 0.92 \[ \frac{1}{16} x^m \left (-\frac{e^{2 a} 2^{-m} (-b x)^{-m} \text{Gamma}(m+2,-2 b x)}{b^2}-\frac{e^{-2 a} 2^{-m} (b x)^{-m} \text{Gamma}(m+2,2 b x)}{b^2}-\frac{8 x^2}{m+2}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{x}^{1+m} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.7728, size = 427, normalized size = 4.97 \begin{align*} -\frac{4 \, b x \cosh \left ({\left (m + 1\right )} \log \left (x\right )\right ) +{\left (m + 2\right )} \cosh \left ({\left (m + 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 2, 2 \, b x\right ) -{\left (m + 2\right )} \cosh \left ({\left (m + 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 2, -2 \, b x\right ) -{\left (m + 2\right )} \Gamma \left (m + 2, 2 \, b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) +{\left (m + 2\right )} \Gamma \left (m + 2, -2 \, b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 1\right )} \log \left (x\right )\right )}{8 \,{\left (b m + 2 \, b\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \sinh ^{2}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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