Optimal. Leaf size=119 \[ -\frac {b \cosh (a+b x)}{8 x}-\frac {3 b \cosh (3 a+3 b x)}{8 x}+\frac {1}{8} b^2 \text {Chi}(b x) \sinh (a)+\frac {9}{8} b^2 \text {Chi}(3 b x) \sinh (3 a)-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}+\frac {1}{8} b^2 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Shi}(3 b x) \]
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Rubi [A]
time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5556, 3378,
3384, 3379, 3382} \begin {gather*} \frac {1}{8} b^2 \sinh (a) \text {Chi}(b x)+\frac {9}{8} b^2 \sinh (3 a) \text {Chi}(3 b x)+\frac {1}{8} b^2 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Shi}(3 b x)-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}-\frac {b \cosh (a+b x)}{8 x}-\frac {3 b \cosh (3 a+3 b x)}{8 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 3378
Rule 3379
Rule 3382
Rule 3384
Rule 5556
Rubi steps
\begin {align*} \int \frac {\cosh ^2(a+b x) \sinh (a+b x)}{x^3} \, dx &=\int \left (\frac {\sinh (a+b x)}{4 x^3}+\frac {\sinh (3 a+3 b x)}{4 x^3}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\sinh (a+b x)}{x^3} \, dx+\frac {1}{4} \int \frac {\sinh (3 a+3 b x)}{x^3} \, dx\\ &=-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}+\frac {1}{8} b \int \frac {\cosh (a+b x)}{x^2} \, dx+\frac {1}{8} (3 b) \int \frac {\cosh (3 a+3 b x)}{x^2} \, dx\\ &=-\frac {b \cosh (a+b x)}{8 x}-\frac {3 b \cosh (3 a+3 b x)}{8 x}-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}+\frac {1}{8} b^2 \int \frac {\sinh (a+b x)}{x} \, dx+\frac {1}{8} \left (9 b^2\right ) \int \frac {\sinh (3 a+3 b x)}{x} \, dx\\ &=-\frac {b \cosh (a+b x)}{8 x}-\frac {3 b \cosh (3 a+3 b x)}{8 x}-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}+\frac {1}{8} \left (b^2 \cosh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^2 \cosh (3 a)\right ) \int \frac {\sinh (3 b x)}{x} \, dx+\frac {1}{8} \left (b^2 \sinh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^2 \sinh (3 a)\right ) \int \frac {\cosh (3 b x)}{x} \, dx\\ &=-\frac {b \cosh (a+b x)}{8 x}-\frac {3 b \cosh (3 a+3 b x)}{8 x}+\frac {1}{8} b^2 \text {Chi}(b x) \sinh (a)+\frac {9}{8} b^2 \text {Chi}(3 b x) \sinh (3 a)-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{8 x^2}+\frac {1}{8} b^2 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Shi}(3 b x)\\ \end {align*}
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Mathematica [A]
time = 0.18, size = 105, normalized size = 0.88 \begin {gather*} -\frac {b x \cosh (a+b x)+3 b x \cosh (3 (a+b x))-b^2 x^2 \text {Chi}(b x) \sinh (a)-9 b^2 x^2 \text {Chi}(3 b x) \sinh (3 a)+\sinh (a+b x)+\sinh (3 (a+b x))-b^2 x^2 \cosh (a) \text {Shi}(b x)-9 b^2 x^2 \cosh (3 a) \text {Shi}(3 b x)}{8 x^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 4.01, size = 169, normalized size = 1.42
method | result | size |
risch | \(-\frac {3 b \,{\mathrm e}^{-3 b x -3 a}}{16 x}+\frac {{\mathrm e}^{-3 b x -3 a}}{16 x^{2}}+\frac {9 b^{2} {\mathrm e}^{-3 a} \expIntegral \left (1, 3 b x \right )}{16}-\frac {b \,{\mathrm e}^{-b x -a}}{16 x}+\frac {{\mathrm e}^{-b x -a}}{16 x^{2}}+\frac {b^{2} {\mathrm e}^{-a} \expIntegral \left (1, b x \right )}{16}-\frac {{\mathrm e}^{b x +a}}{16 x^{2}}-\frac {b \,{\mathrm e}^{b x +a}}{16 x}-\frac {b^{2} {\mathrm e}^{a} \expIntegral \left (1, -b x \right )}{16}-\frac {{\mathrm e}^{3 b x +3 a}}{16 x^{2}}-\frac {3 b \,{\mathrm e}^{3 b x +3 a}}{16 x}-\frac {9 b^{2} {\mathrm e}^{3 a} \expIntegral \left (1, -3 b x \right )}{16}\) | \(169\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.33, size = 58, normalized size = 0.49 \begin {gather*} \frac {9}{8} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) + \frac {1}{8} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) - \frac {1}{8} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) - \frac {9}{8} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 196, normalized size = 1.65 \begin {gather*} -\frac {6 \, b x \cosh \left (b x + a\right )^{3} + 18 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )^{3} - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) - {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) - b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) - {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) + b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{16 \, x^{2}} \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 \frac {\sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 156, normalized size = 1.31 \begin {gather*} \frac {9 \, b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 9 \, b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + b^{2} x^{2} {\rm Ei}\left (b x\right ) e^{a} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} - b x e^{\left (b x + a\right )} - b x e^{\left (-b x - a\right )} - 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{16 \, x^{2}} \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 \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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