Optimal. Leaf size=184 \[ -\frac{1}{16} b^2 \sinh (a) \text{Chi}(b x)-\frac{9}{32} b^2 \sinh (3 a) \text{Chi}(3 b x)+\frac{25}{32} b^2 \sinh (5 a) \text{Chi}(5 b x)-\frac{1}{16} b^2 \cosh (a) \text{Shi}(b x)-\frac{9}{32} b^2 \cosh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Shi}(5 b x)+\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}+\frac{b \cosh (a+b x)}{16 x}+\frac{3 b \cosh (3 a+3 b x)}{32 x}-\frac{5 b \cosh (5 a+5 b x)}{32 x} \]
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Rubi [A] time = 0.339063, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5448, 3297, 3303, 3298, 3301} \[ -\frac{1}{16} b^2 \sinh (a) \text{Chi}(b x)-\frac{9}{32} b^2 \sinh (3 a) \text{Chi}(3 b x)+\frac{25}{32} b^2 \sinh (5 a) \text{Chi}(5 b x)-\frac{1}{16} b^2 \cosh (a) \text{Shi}(b x)-\frac{9}{32} b^2 \cosh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Shi}(5 b x)+\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}+\frac{b \cosh (a+b x)}{16 x}+\frac{3 b \cosh (3 a+3 b x)}{32 x}-\frac{5 b \cosh (5 a+5 b x)}{32 x} \]
Antiderivative was successfully verified.
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Rule 5448
Rule 3297
Rule 3303
Rule 3298
Rule 3301
Rubi steps
\begin{align*} \int \frac{\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^3} \, dx &=\int \left (-\frac{\sinh (a+b x)}{8 x^3}-\frac{\sinh (3 a+3 b x)}{16 x^3}+\frac{\sinh (5 a+5 b x)}{16 x^3}\right ) \, dx\\ &=-\left (\frac{1}{16} \int \frac{\sinh (3 a+3 b x)}{x^3} \, dx\right )+\frac{1}{16} \int \frac{\sinh (5 a+5 b x)}{x^3} \, dx-\frac{1}{8} \int \frac{\sinh (a+b x)}{x^3} \, dx\\ &=\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}-\frac{1}{16} b \int \frac{\cosh (a+b x)}{x^2} \, dx-\frac{1}{32} (3 b) \int \frac{\cosh (3 a+3 b x)}{x^2} \, dx+\frac{1}{32} (5 b) \int \frac{\cosh (5 a+5 b x)}{x^2} \, dx\\ &=\frac{b \cosh (a+b x)}{16 x}+\frac{3 b \cosh (3 a+3 b x)}{32 x}-\frac{5 b \cosh (5 a+5 b x)}{32 x}+\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}-\frac{1}{16} b^2 \int \frac{\sinh (a+b x)}{x} \, dx-\frac{1}{32} \left (9 b^2\right ) \int \frac{\sinh (3 a+3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2\right ) \int \frac{\sinh (5 a+5 b x)}{x} \, dx\\ &=\frac{b \cosh (a+b x)}{16 x}+\frac{3 b \cosh (3 a+3 b x)}{32 x}-\frac{5 b \cosh (5 a+5 b x)}{32 x}+\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}-\frac{1}{16} \left (b^2 \cosh (a)\right ) \int \frac{\sinh (b x)}{x} \, dx-\frac{1}{32} \left (9 b^2 \cosh (3 a)\right ) \int \frac{\sinh (3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2 \cosh (5 a)\right ) \int \frac{\sinh (5 b x)}{x} \, dx-\frac{1}{16} \left (b^2 \sinh (a)\right ) \int \frac{\cosh (b x)}{x} \, dx-\frac{1}{32} \left (9 b^2 \sinh (3 a)\right ) \int \frac{\cosh (3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2 \sinh (5 a)\right ) \int \frac{\cosh (5 b x)}{x} \, dx\\ &=\frac{b \cosh (a+b x)}{16 x}+\frac{3 b \cosh (3 a+3 b x)}{32 x}-\frac{5 b \cosh (5 a+5 b x)}{32 x}-\frac{1}{16} b^2 \text{Chi}(b x) \sinh (a)-\frac{9}{32} b^2 \text{Chi}(3 b x) \sinh (3 a)+\frac{25}{32} b^2 \text{Chi}(5 b x) \sinh (5 a)+\frac{\sinh (a+b x)}{16 x^2}+\frac{\sinh (3 a+3 b x)}{32 x^2}-\frac{\sinh (5 a+5 b x)}{32 x^2}-\frac{1}{16} b^2 \cosh (a) \text{Shi}(b x)-\frac{9}{32} b^2 \cosh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Shi}(5 b x)\\ \end{align*}
Mathematica [A] time = 0.438801, size = 164, normalized size = 0.89 \[ \frac{-2 b^2 x^2 \sinh (a) \text{Chi}(b x)-9 b^2 x^2 \sinh (3 a) \text{Chi}(3 b x)+25 b^2 x^2 \sinh (5 a) \text{Chi}(5 b x)-2 b^2 x^2 \cosh (a) \text{Shi}(b x)-9 b^2 x^2 \cosh (3 a) \text{Shi}(3 b x)+25 b^2 x^2 \cosh (5 a) \text{Shi}(5 b x)+2 \sinh (a+b x)+\sinh (3 (a+b x))-\sinh (5 (a+b x))+2 b x \cosh (a+b x)+3 b x \cosh (3 (a+b x))-5 b x \cosh (5 (a+b x))}{32 x^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.097, size = 257, normalized size = 1.4 \begin{align*} -{\frac{5\,b{{\rm e}^{-5\,bx-5\,a}}}{64\,x}}+{\frac{{{\rm e}^{-5\,bx-5\,a}}}{64\,{x}^{2}}}+{\frac{25\,{b}^{2}{{\rm e}^{-5\,a}}{\it Ei} \left ( 1,5\,bx \right ) }{64}}+{\frac{3\,b{{\rm e}^{-3\,bx-3\,a}}}{64\,x}}-{\frac{{{\rm e}^{-3\,bx-3\,a}}}{64\,{x}^{2}}}-{\frac{9\,{b}^{2}{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{64}}+{\frac{b{{\rm e}^{-bx-a}}}{32\,x}}-{\frac{{{\rm e}^{-bx-a}}}{32\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{32}}+{\frac{{{\rm e}^{bx+a}}}{32\,{x}^{2}}}+{\frac{b{{\rm e}^{bx+a}}}{32\,x}}+{\frac{{b}^{2}{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{32}}+{\frac{{{\rm e}^{3\,bx+3\,a}}}{64\,{x}^{2}}}+{\frac{3\,b{{\rm e}^{3\,bx+3\,a}}}{64\,x}}+{\frac{9\,{b}^{2}{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{64}}-{\frac{{{\rm e}^{5\,bx+5\,a}}}{64\,{x}^{2}}}-{\frac{5\,b{{\rm e}^{5\,bx+5\,a}}}{64\,x}}-{\frac{25\,{b}^{2}{{\rm e}^{5\,a}}{\it Ei} \left ( 1,-5\,bx \right ) }{64}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.32324, size = 119, normalized size = 0.65 \begin{align*} \frac{25}{32} \, b^{2} e^{\left (-5 \, a\right )} \Gamma \left (-2, 5 \, b x\right ) - \frac{9}{32} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) - \frac{1}{16} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) + \frac{1}{16} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) + \frac{9}{32} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) - \frac{25}{32} \, b^{2} e^{\left (5 \, a\right )} \Gamma \left (-2, -5 \, b x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.81288, size = 857, normalized size = 4.66 \begin{align*} -\frac{10 \, b x \cosh \left (b x + a\right )^{5} + 50 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 6 \, b x \cosh \left (b x + a\right )^{3} + 2 \, \sinh \left (b x + a\right )^{5} + 2 \,{\left (10 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 4 \, b x \cosh \left (b x + a\right ) + 2 \,{\left (50 \, b x \cosh \left (b x + a\right )^{3} - 9 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 25 \,{\left (b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) - b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, 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 )} \cosh \left (3 \, a\right ) + 2 \,{\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 (5 \, \cosh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right ) - 25 \,{\left (b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) + b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, 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 ) + 2 \,{\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 )}{64 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.1854, size = 323, normalized size = 1.76 \begin{align*} \frac{25 \, b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - 9 \, b^{2} x^{2}{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + 2 \, 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 )} - 25 \, b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{2} x^{2}{\rm Ei}\left (b x\right ) e^{a} - 5 \, b x e^{\left (5 \, b x + 5 \, a\right )} + 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b x e^{\left (b x + a\right )} + 2 \, b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} - 5 \, b x e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (5 \, b x + 5 \, a\right )} + e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} - 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}}{64 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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