Optimal. Leaf size=154 \[ \frac {\cosh (a+b x)}{12 x^3}+\frac {b^2 \cosh (a+b x)}{24 x}-\frac {\cosh (3 a+3 b x)}{12 x^3}-\frac {3 b^2 \cosh (3 a+3 b x)}{8 x}-\frac {1}{24} b^3 \text {Chi}(b x) \sinh (a)+\frac {9}{8} b^3 \text {Chi}(3 b x) \sinh (3 a)+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2}-\frac {1}{24} b^3 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^3 \cosh (3 a) \text {Shi}(3 b x) \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.21, antiderivative size = 154, 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}{24} b^3 \sinh (a) \text {Chi}(b x)+\frac {9}{8} b^3 \sinh (3 a) \text {Chi}(3 b x)-\frac {1}{24} b^3 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^3 \cosh (3 a) \text {Shi}(3 b x)+\frac {b^2 \cosh (a+b x)}{24 x}-\frac {3 b^2 \cosh (3 a+3 b x)}{8 x}+\frac {\cosh (a+b x)}{12 x^3}-\frac {\cosh (3 a+3 b x)}{12 x^3}+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 3378
Rule 3379
Rule 3382
Rule 3384
Rule 5556
Rubi steps
\begin {align*} \int \frac {\cosh (a+b x) \sinh ^2(a+b x)}{x^4} \, dx &=\int \left (-\frac {\cosh (a+b x)}{4 x^4}+\frac {\cosh (3 a+3 b x)}{4 x^4}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {\cosh (a+b x)}{x^4} \, dx\right )+\frac {1}{4} \int \frac {\cosh (3 a+3 b x)}{x^4} \, dx\\ &=\frac {\cosh (a+b x)}{12 x^3}-\frac {\cosh (3 a+3 b x)}{12 x^3}-\frac {1}{12} b \int \frac {\sinh (a+b x)}{x^3} \, dx+\frac {1}{4} b \int \frac {\sinh (3 a+3 b x)}{x^3} \, dx\\ &=\frac {\cosh (a+b x)}{12 x^3}-\frac {\cosh (3 a+3 b x)}{12 x^3}+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2}-\frac {1}{24} b^2 \int \frac {\cosh (a+b x)}{x^2} \, dx+\frac {1}{8} \left (3 b^2\right ) \int \frac {\cosh (3 a+3 b x)}{x^2} \, dx\\ &=\frac {\cosh (a+b x)}{12 x^3}+\frac {b^2 \cosh (a+b x)}{24 x}-\frac {\cosh (3 a+3 b x)}{12 x^3}-\frac {3 b^2 \cosh (3 a+3 b x)}{8 x}+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2}-\frac {1}{24} b^3 \int \frac {\sinh (a+b x)}{x} \, dx+\frac {1}{8} \left (9 b^3\right ) \int \frac {\sinh (3 a+3 b x)}{x} \, dx\\ &=\frac {\cosh (a+b x)}{12 x^3}+\frac {b^2 \cosh (a+b x)}{24 x}-\frac {\cosh (3 a+3 b x)}{12 x^3}-\frac {3 b^2 \cosh (3 a+3 b x)}{8 x}+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2}-\frac {1}{24} \left (b^3 \cosh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^3 \cosh (3 a)\right ) \int \frac {\sinh (3 b x)}{x} \, dx-\frac {1}{24} \left (b^3 \sinh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^3 \sinh (3 a)\right ) \int \frac {\cosh (3 b x)}{x} \, dx\\ &=\frac {\cosh (a+b x)}{12 x^3}+\frac {b^2 \cosh (a+b x)}{24 x}-\frac {\cosh (3 a+3 b x)}{12 x^3}-\frac {3 b^2 \cosh (3 a+3 b x)}{8 x}-\frac {1}{24} b^3 \text {Chi}(b x) \sinh (a)+\frac {9}{8} b^3 \text {Chi}(3 b x) \sinh (3 a)+\frac {b \sinh (a+b x)}{24 x^2}-\frac {b \sinh (3 a+3 b x)}{8 x^2}-\frac {1}{24} b^3 \cosh (a) \text {Shi}(b x)+\frac {9}{8} b^3 \cosh (3 a) \text {Shi}(3 b x)\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.19, size = 138, normalized size = 0.90 \begin {gather*} \frac {2 \cosh (a+b x)+b^2 x^2 \cosh (a+b x)-2 \cosh (3 (a+b x))-9 b^2 x^2 \cosh (3 (a+b x))-b^3 x^3 \text {Chi}(b x) \sinh (a)+27 b^3 x^3 \text {Chi}(3 b x) \sinh (3 a)+b x \sinh (a+b x)-3 b x \sinh (3 (a+b x))-b^3 x^3 \cosh (a) \text {Shi}(b x)+27 b^3 x^3 \cosh (3 a) \text {Shi}(3 b x)}{24 x^3} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A]
time = 4.23, size = 234, normalized size = 1.52
method | result | size |
risch | \(-\frac {3 b^{2} {\mathrm e}^{-3 b x -3 a}}{16 x}+\frac {b \,{\mathrm e}^{-3 b x -3 a}}{16 x^{2}}-\frac {{\mathrm e}^{-3 b x -3 a}}{24 x^{3}}+\frac {9 b^{3} {\mathrm e}^{-3 a} \expIntegral \left (1, 3 b x \right )}{16}+\frac {b^{2} {\mathrm e}^{-b x -a}}{48 x}-\frac {b \,{\mathrm e}^{-b x -a}}{48 x^{2}}+\frac {{\mathrm e}^{-b x -a}}{24 x^{3}}-\frac {b^{3} {\mathrm e}^{-a} \expIntegral \left (1, b x \right )}{48}+\frac {{\mathrm e}^{b x +a}}{24 x^{3}}+\frac {b \,{\mathrm e}^{b x +a}}{48 x^{2}}+\frac {b^{2} {\mathrm e}^{b x +a}}{48 x}+\frac {b^{3} {\mathrm e}^{a} \expIntegral \left (1, -b x \right )}{48}-\frac {{\mathrm e}^{3 b x +3 a}}{24 x^{3}}-\frac {b \,{\mathrm e}^{3 b x +3 a}}{16 x^{2}}-\frac {3 b^{2} {\mathrm e}^{3 b x +3 a}}{16 x}-\frac {9 b^{3} {\mathrm e}^{3 a} \expIntegral \left (1, -3 b x \right )}{16}\) | \(234\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A]
time = 0.33, size = 58, normalized size = 0.38 \begin {gather*} -\frac {27}{8} \, b^{3} e^{\left (-3 \, a\right )} \Gamma \left (-3, 3 \, b x\right ) + \frac {1}{8} \, b^{3} e^{\left (-a\right )} \Gamma \left (-3, b x\right ) - \frac {1}{8} \, b^{3} e^{a} \Gamma \left (-3, -b x\right ) + \frac {27}{8} \, b^{3} e^{\left (3 \, a\right )} \Gamma \left (-3, -3 \, b x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A]
time = 0.47, size = 224, normalized size = 1.45 \begin {gather*} -\frac {6 \, b x \sinh \left (b x + a\right )^{3} + 2 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 6 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) - 27 \, {\left (b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + {\left (b^{3} x^{3} {\rm Ei}\left (b x\right ) - b^{3} x^{3} {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \, {\left (9 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right ) - 27 \, {\left (b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + {\left (b^{3} x^{3} {\rm Ei}\left (b x\right ) + b^{3} x^{3} {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{48 \, x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{4}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A]
time = 0.40, size = 222, normalized size = 1.44 \begin {gather*} \frac {27 \, b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + b^{3} x^{3} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 27 \, b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} - b^{3} x^{3} {\rm Ei}\left (b x\right ) e^{a} - 9 \, b^{2} x^{2} e^{\left (3 \, b x + 3 \, a\right )} + b^{2} x^{2} e^{\left (b x + a\right )} + b^{2} x^{2} e^{\left (-b x - a\right )} - 9 \, b^{2} x^{2} e^{\left (-3 \, b x - 3 \, a\right )} - 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 )} - 2 \, e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________