Optimal. Leaf size=169 \[ -\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}+\frac {1}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x) \]
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Rubi [A]
time = 0.21, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps
used = 14, 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}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)+\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x)-\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2} \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 ^3(a+b x) \sinh (a+b x)}{x^4} \, dx &=\int \left (\frac {\sinh (2 a+2 b x)}{4 x^4}+\frac {\sinh (4 a+4 b x)}{8 x^4}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sinh (4 a+4 b x)}{x^4} \, dx+\frac {1}{4} \int \frac {\sinh (2 a+2 b x)}{x^4} \, dx\\ &=-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}+\frac {1}{6} b \int \frac {\cosh (2 a+2 b x)}{x^3} \, dx+\frac {1}{6} b \int \frac {\cosh (4 a+4 b x)}{x^3} \, dx\\ &=-\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}+\frac {1}{6} b^2 \int \frac {\sinh (2 a+2 b x)}{x^2} \, dx+\frac {1}{3} b^2 \int \frac {\sinh (4 a+4 b x)}{x^2} \, dx\\ &=-\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {1}{3} b^3 \int \frac {\cosh (2 a+2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3\right ) \int \frac {\cosh (4 a+4 b x)}{x} \, dx\\ &=-\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {1}{3} \left (b^3 \cosh (2 a)\right ) \int \frac {\cosh (2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3 \cosh (4 a)\right ) \int \frac {\cosh (4 b x)}{x} \, dx+\frac {1}{3} \left (b^3 \sinh (2 a)\right ) \int \frac {\sinh (2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3 \sinh (4 a)\right ) \int \frac {\sinh (4 b x)}{x} \, dx\\ &=-\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}+\frac {1}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)-\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x)\\ \end {align*}
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Mathematica [A]
time = 0.32, size = 150, normalized size = 0.89 \begin {gather*} -\frac {2 b x \cosh (2 (a+b x))+2 b x \cosh (4 (a+b x))-8 b^3 x^3 \cosh (2 a) \text {Chi}(2 b x)-32 b^3 x^3 \cosh (4 a) \text {Chi}(4 b x)+2 \sinh (2 (a+b x))+4 b^2 x^2 \sinh (2 (a+b x))+\sinh (4 (a+b x))+8 b^2 x^2 \sinh (4 (a+b x))-8 b^3 x^3 \sinh (2 a) \text {Shi}(2 b x)-32 b^3 x^3 \sinh (4 a) \text {Shi}(4 b x)}{24 x^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 4.70, size = 246, normalized size = 1.46
method | result | size |
risch | \(\frac {b^{2} {\mathrm e}^{-4 b x -4 a}}{6 x}-\frac {b \,{\mathrm e}^{-4 b x -4 a}}{24 x^{2}}+\frac {{\mathrm e}^{-4 b x -4 a}}{48 x^{3}}-\frac {2 b^{3} {\mathrm e}^{-4 a} \expIntegral \left (1, 4 b x \right )}{3}+\frac {b^{2} {\mathrm e}^{-2 b x -2 a}}{12 x}-\frac {b \,{\mathrm e}^{-2 b x -2 a}}{24 x^{2}}+\frac {{\mathrm e}^{-2 b x -2 a}}{24 x^{3}}-\frac {b^{3} {\mathrm e}^{-2 a} \expIntegral \left (1, 2 b x \right )}{6}-\frac {{\mathrm e}^{2 b x +2 a}}{24 x^{3}}-\frac {b \,{\mathrm e}^{2 b x +2 a}}{24 x^{2}}-\frac {b^{2} {\mathrm e}^{2 b x +2 a}}{12 x}-\frac {b^{3} {\mathrm e}^{2 a} \expIntegral \left (1, -2 b x \right )}{6}-\frac {{\mathrm e}^{4 b x +4 a}}{48 x^{3}}-\frac {b \,{\mathrm e}^{4 b x +4 a}}{24 x^{2}}-\frac {b^{2} {\mathrm e}^{4 b x +4 a}}{6 x}-\frac {2 b^{3} {\mathrm e}^{4 a} \expIntegral \left (1, -4 b x \right )}{3}\) | \(246\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.35, size = 59, normalized size = 0.35 \begin {gather*} 4 \, b^{3} e^{\left (-4 \, a\right )} \Gamma \left (-3, 4 \, b x\right ) + b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) + b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) + 4 \, b^{3} e^{\left (4 \, a\right )} \Gamma \left (-3, -4 \, b x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 261, normalized size = 1.54 \begin {gather*} -\frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + 2 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b x \cosh \left (b x + a\right )^{2} + {\left (6 \, b x \cosh \left (b x + a\right )^{2} + b x\right )} \sinh \left (b x + a\right )^{2} - 8 \, {\left (b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 2 \, {\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} + {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 8 \, {\left (b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{12 \, x^{3}} \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 ^{3}{\left (a + b x \right )}}{x^{4}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 236, normalized size = 1.40 \begin {gather*} \frac {32 \, b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + 8 \, b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 8 \, b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 32 \, b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 8 \, b^{2} x^{2} e^{\left (4 \, b x + 4 \, a\right )} - 4 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 4 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} + 8 \, b^{2} x^{2} e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, b x e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{48 \, x^{3}} \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 )}^3\,\mathrm {sinh}\left (a+b\,x\right )}{x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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