Optimal. Leaf size=135 \[ -\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{2 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )}{e}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{-2 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 b x^2\right )}{e} \]
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Rubi [A]
time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5448, 5437,
2250} \begin {gather*} -\frac {e^{2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-2 b x^2\right )}{e}-\frac {e^{-2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},2 b x^2\right )}{e}-\frac {(e x)^{m+1}}{2 e (m+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 2250
Rule 5437
Rule 5448
Rubi steps
\begin {align*} \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx &=\int \left (-\frac {1}{2} (e x)^m+\frac {1}{2} (e x)^m \cosh \left (2 a+2 b x^2\right )\right ) \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{2} \int (e x)^m \cosh \left (2 a+2 b x^2\right ) \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{4} \int e^{-2 a-2 b x^2} (e x)^m \, dx+\frac {1}{4} \int e^{2 a+2 b x^2} (e x)^m \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{2 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )}{e}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{-2 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 b x^2\right )}{e}\\ \end {align*}
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Mathematica [A]
time = 0.43, size = 152, normalized size = 1.13 \begin {gather*} -\frac {2^{\frac {1}{2} (-7-m)} x (e x)^m \left (-b^2 x^4\right )^{\frac {1}{2} (-1-m)} \left (2^{\frac {5+m}{2}} \left (-b^2 x^4\right )^{\frac {1+m}{2}}+(1+m) \left (-b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},2 b x^2\right ) (\cosh (2 a)-\sinh (2 a))+(1+m) \left (b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-2 b x^2\right ) (\cosh (2 a)+\sinh (2 a))\right )}{1+m} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 1.44, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\sinh ^{2}\left (x^{2} b +a \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs.
\(2 (111) = 222\).
time = 0.09, size = 274, normalized size = 2.03 \begin {gather*} -\frac {8 \, b x \cosh \left (m \log \left (x \cosh \left (1\right ) + x \sinh \left (1\right )\right )\right ) + {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) + 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) - {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) - 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) + 8 \, b x \sinh \left (m \log \left (x \cosh \left (1\right ) + x \sinh \left (1\right )\right )\right ) - {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) + 2 \, a\right ) + {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) - 2 \, a\right )}{16 \, {\left (b m + 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 \left (e x\right )^{m} \sinh ^{2}{\left (a + b x^{2} \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 {\mathrm {sinh}\left (b\,x^2+a\right )}^2\,{\left (e\,x\right )}^m \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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