Optimal. Leaf size=169 \[ \frac{1}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Chi}(4 b x)+\frac{1}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)+\frac{4}{3} b^3 \sinh (4 a) \text{Shi}(4 b x)-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.29719, antiderivative size = 169, normalized size of antiderivative = 1., 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 = {5448, 3297, 3303, 3298, 3301} \[ \frac{1}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Chi}(4 b x)+\frac{1}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)+\frac{4}{3} b^3 \sinh (4 a) \text{Shi}(4 b x)-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5448
Rule 3297
Rule 3303
Rule 3298
Rule 3301
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*}
Mathematica [A] time = 0.534986, size = 150, normalized size = 0.89 \[ -\frac{-8 b^3 x^3 \cosh (2 a) \text{Chi}(2 b x)-32 b^3 x^3 \cosh (4 a) \text{Chi}(4 b x)-8 b^3 x^3 \sinh (2 a) \text{Shi}(2 b x)-32 b^3 x^3 \sinh (4 a) \text{Shi}(4 b x)+4 b^2 x^2 \sinh (2 (a+b x))+8 b^2 x^2 \sinh (4 (a+b x))+2 \sinh (2 (a+b x))+\sinh (4 (a+b x))+2 b x \cosh (2 (a+b x))+2 b x \cosh (4 (a+b x))}{24 x^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.072, size = 246, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}{{\rm e}^{-4\,bx-4\,a}}}{6\,x}}-{\frac{b{{\rm e}^{-4\,bx-4\,a}}}{24\,{x}^{2}}}+{\frac{{{\rm e}^{-4\,bx-4\,a}}}{48\,{x}^{3}}}-{\frac{2\,{b}^{3}{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{3}}+{\frac{{b}^{2}{{\rm e}^{-2\,bx-2\,a}}}{12\,x}}-{\frac{b{{\rm e}^{-2\,bx-2\,a}}}{24\,{x}^{2}}}+{\frac{{{\rm e}^{-2\,bx-2\,a}}}{24\,{x}^{3}}}-{\frac{{b}^{3}{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{6}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{48\,{x}^{3}}}-{\frac{b{{\rm e}^{4\,bx+4\,a}}}{24\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{4\,bx+4\,a}}}{6\,x}}-{\frac{2\,{b}^{3}{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{3}}-{\frac{{{\rm e}^{2\,bx+2\,a}}}{24\,{x}^{3}}}-{\frac{b{{\rm e}^{2\,bx+2\,a}}}{24\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{2\,bx+2\,a}}}{12\,x}}-{\frac{{b}^{3}{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{6}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.38147, size = 80, normalized size = 0.47 \begin{align*} 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{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.85127, size = 647, normalized size = 3.83 \begin{align*} -\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{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.15062, size = 319, normalized size = 1.89 \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]