Optimal. Leaf size=84 \[ \frac {x^{-2+m}}{2 (2-m)}-2^{-m} b^2 e^{2 a} x^m (-b x)^{-m} \Gamma (-2+m,-2 b x)-2^{-m} b^2 e^{-2 a} x^m (b x)^{-m} \Gamma (-2+m,2 b x) \]
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Rubi [A]
time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 = {3393, 3388,
2212} \begin {gather*} -e^{2 a} b^2 2^{-m} x^m (-b x)^{-m} \text {Gamma}(m-2,-2 b x)-e^{-2 a} b^2 2^{-m} x^m (b x)^{-m} \text {Gamma}(m-2,2 b x)+\frac {x^{m-2}}{2 (2-m)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2212
Rule 3388
Rule 3393
Rubi steps
\begin {align*} \int x^{-3+m} \sinh ^2(a+b x) \, dx &=-\int \left (\frac {x^{-3+m}}{2}-\frac {1}{2} x^{-3+m} \cosh (2 a+2 b x)\right ) \, dx\\ &=\frac {x^{-2+m}}{2 (2-m)}+\frac {1}{2} \int x^{-3+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^{-3+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{-3+m} \, dx\\ &=\frac {x^{-2+m}}{2 (2-m)}-2^{-m} b^2 e^{2 a} x^m (-b x)^{-m} \Gamma (-2+m,-2 b x)-2^{-m} b^2 e^{-2 a} x^m (b x)^{-m} \Gamma (-2+m,2 b x)\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 77, normalized size = 0.92 \begin {gather*} x^m \left (\frac {1}{(4-2 m) x^2}-2^{-m} b^2 e^{2 a} (-b x)^{-m} \Gamma (-2+m,-2 b x)-2^{-m} b^2 e^{-2 a} (b x)^{-m} \Gamma (-2+m,2 b x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.29, size = 0, normalized size = 0.00 \[\int x^{-3+m} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.10, size = 136, normalized size = 1.62 \begin {gather*} -\frac {4 \, b x \cosh \left ({\left (m - 3\right )} \log \left (x\right )\right ) + {\left (m - 2\right )} \cosh \left ({\left (m - 3\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 - 3\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 - 3\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 - 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m - 3\right )} \log \left (x\right )\right )}{8 \, {\left (b m - 2 \, b\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m - 3} \sinh ^{2}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{m-3}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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