Optimal. Leaf size=90 \[ -e^{2 a} b 2^{m-1} \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-m-1,-\frac {2 b}{x}\right )+e^{-2 a} b 2^{m-1} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-m-1,\frac {2 b}{x}\right )-\frac {x (e x)^m}{2 (m+1)} \]
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Rubi [A] time = 0.16, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5350, 3312, 3307, 2181} \[ -e^{2 a} b 2^{m-1} \left (-\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,-\frac {2 b}{x}\right )+e^{-2 a} b 2^{m-1} \left (\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,\frac {2 b}{x}\right )-\frac {x (e x)^m}{2 (m+1)} \]
Antiderivative was successfully verified.
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Rule 2181
Rule 3307
Rule 3312
Rule 5350
Rubi steps
\begin {align*} \int (e x)^m \sinh ^2\left (a+\frac {b}{x}\right ) \, dx &=-\left (\left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \operatorname {Subst}\left (\int x^{-2-m} \sinh ^2(a+b x) \, dx,x,\frac {1}{x}\right )\right )\\ &=\left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \operatorname {Subst}\left (\int \left (\frac {x^{-2-m}}{2}-\frac {1}{2} x^{-2-m} \cosh (2 a+2 b x)\right ) \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x (e x)^m}{2 (1+m)}-\frac {1}{2} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \operatorname {Subst}\left (\int x^{-2-m} \cosh (2 a+2 b x) \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x (e x)^m}{2 (1+m)}-\frac {1}{4} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \operatorname {Subst}\left (\int e^{-i (2 i a+2 i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )-\frac {1}{4} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \operatorname {Subst}\left (\int e^{i (2 i a+2 i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x (e x)^m}{2 (1+m)}-2^{-1+m} b e^{2 a} \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {2 b}{x}\right )+2^{-1+m} b e^{-2 a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {2 b}{x}\right )\\ \end {align*}
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Mathematica [A] time = 0.26, size = 88, normalized size = 0.98 \[ -\frac {(e x)^m \left (b 2^m (m+1) (\sinh (a)+\cosh (a))^2 \left (-\frac {b}{x}\right )^m \Gamma \left (-m-1,-\frac {2 b}{x}\right )-b 2^m (m+1) (\cosh (a)-\sinh (a))^2 \left (\frac {b}{x}\right )^m \Gamma \left (-m-1,\frac {2 b}{x}\right )+x\right )}{2 (m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \sinh \left (\frac {a x + b}{x}\right )^{2}, x\right ) \]
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 \[ \int \left (e x\right )^{m} \sinh \left (a + \frac {b}{x}\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\sinh ^{2}\left (a +\frac {b}{x}\right )\right )\, dx \]
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 \[ \frac {1}{4} \, e^{m} \int e^{\left (m \log \relax (x) + 2 \, a + \frac {2 \, b}{x}\right )}\,{d x} + \frac {1}{4} \, e^{m} \int e^{\left (m \log \relax (x) - 2 \, a - \frac {2 \, b}{x}\right )}\,{d x} - \frac {\left (e x\right )^{m + 1}}{2 \, e {\left (m + 1\right )}} \]
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 \[ \int {\mathrm {sinh}\left (a+\frac {b}{x}\right )}^2\,{\left (e\,x\right )}^m \,d x \]
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 \[ \int \left (e x\right )^{m} \sinh ^{2}{\left (a + \frac {b}{x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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