Optimal. Leaf size=125 \[ -\frac {1}{2} b^2 \sinh (2 a) \text {Chi}(2 b x)+b^2 \sinh (4 a) \text {Chi}(4 b x)-\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x)+\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}+\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x} \]
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Rubi [A] time = 0.23, antiderivative size = 125, 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 = {5448, 3297, 3303, 3298, 3301} \[ -\frac {1}{2} b^2 \sinh (2 a) \text {Chi}(2 b x)+b^2 \sinh (4 a) \text {Chi}(4 b x)-\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x)+\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}+\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x} \]
Antiderivative was successfully verified.
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Rule 3297
Rule 3298
Rule 3301
Rule 3303
Rule 5448
Rubi steps
\begin {align*} \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^3} \, dx &=\int \left (-\frac {\sinh (2 a+2 b x)}{4 x^3}+\frac {\sinh (4 a+4 b x)}{8 x^3}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sinh (4 a+4 b x)}{x^3} \, dx-\frac {1}{4} \int \frac {\sinh (2 a+2 b x)}{x^3} \, dx\\ &=\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}-\frac {1}{4} b \int \frac {\cosh (2 a+2 b x)}{x^2} \, dx+\frac {1}{4} b \int \frac {\cosh (4 a+4 b x)}{x^2} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x}+\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}-\frac {1}{2} b^2 \int \frac {\sinh (2 a+2 b x)}{x} \, dx+b^2 \int \frac {\sinh (4 a+4 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x}+\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}-\frac {1}{2} \left (b^2 \cosh (2 a)\right ) \int \frac {\sinh (2 b x)}{x} \, dx+\left (b^2 \cosh (4 a)\right ) \int \frac {\sinh (4 b x)}{x} \, dx-\frac {1}{2} \left (b^2 \sinh (2 a)\right ) \int \frac {\cosh (2 b x)}{x} \, dx+\left (b^2 \sinh (4 a)\right ) \int \frac {\cosh (4 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x}-\frac {1}{2} b^2 \text {Chi}(2 b x) \sinh (2 a)+b^2 \text {Chi}(4 b x) \sinh (4 a)+\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}-\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x)\\ \end {align*}
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Mathematica [A] time = 0.56, size = 113, normalized size = 0.90 \[ b^2 \sinh (4 a) \text {Chi}(4 b x)+b^2 \sinh (a) (-\cosh (a)) \text {Chi}(2 b x)-\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x)+\frac {\sinh (2 (a+b x))+2 b x \cosh (2 (a+b x))}{8 x^2}-\frac {\sinh (4 (a+b x))+4 b x \cosh (4 (a+b x))}{16 x^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.57, size = 229, normalized size = 1.83 \[ -\frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} - b x \cosh \left (b x + a\right )^{2} + \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + {\left (6 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) + {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) + {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{4 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 168, normalized size = 1.34 \[ \frac {16 \, b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - 8 \, b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 8 \, b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 16 \, b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 4 \, b x e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b x e^{\left (2 \, b x + 2 \, a\right )} + 4 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 4 \, b x e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{32 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.62, size = 178, normalized size = 1.42 \[ -\frac {b \,{\mathrm e}^{-4 b x -4 a}}{8 x}+\frac {{\mathrm e}^{-4 b x -4 a}}{32 x^{2}}+\frac {b^{2} {\mathrm e}^{-4 a} \Ei \left (1, 4 b x \right )}{2}+\frac {b \,{\mathrm e}^{-2 b x -2 a}}{8 x}-\frac {{\mathrm e}^{-2 b x -2 a}}{16 x^{2}}-\frac {b^{2} {\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{4}+\frac {{\mathrm e}^{2 b x +2 a}}{16 x^{2}}+\frac {b \,{\mathrm e}^{2 b x +2 a}}{8 x}+\frac {b^{2} {\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{4}-\frac {{\mathrm e}^{4 b x +4 a}}{32 x^{2}}-\frac {b \,{\mathrm e}^{4 b x +4 a}}{8 x}-\frac {b^{2} {\mathrm e}^{4 a} \Ei \left (1, -4 b x \right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 60, normalized size = 0.48 \[ b^{2} e^{\left (-4 \, a\right )} \Gamma \left (-2, 4 \, b x\right ) - \frac {1}{2} \, b^{2} e^{\left (-2 \, a\right )} \Gamma \left (-2, 2 \, b x\right ) + \frac {1}{2} \, b^{2} e^{\left (2 \, a\right )} \Gamma \left (-2, -2 \, b x\right ) - b^{2} e^{\left (4 \, a\right )} \Gamma \left (-2, -4 \, b x\right ) \]
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 \[ \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^3} \,d x \]
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 \[ \int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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