Optimal. Leaf size=103 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b x (16 a+5 b)+\frac{b^2 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^2 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]
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Rubi [A] time = 0.194466, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3217, 1259, 1805, 453, 206} \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b x (16 a+5 b)+\frac{b^2 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^2 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]
Antiderivative was successfully verified.
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Rule 3217
Rule 1259
Rule 1805
Rule 453
Rule 206
Rubi steps
\begin{align*} \int \text{csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-2 a x^2+(a+b) x^4\right )^2}{x^2 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-6 a^2+\left (18 a^2+b^2\right ) x^2-6 (3 a-b) (a+b) x^4+6 (a+b)^2 x^6}{x^2 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac{\operatorname{Subst}\left (\int \frac{24 a^2-3 \left (16 a^2-3 b^2\right ) x^2+24 (a+b)^2 x^4}{x^2 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{24 d}\\ &=\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-48 a^2+3 \left (16 a^2+16 a b+5 b^2\right ) x^2}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=-\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{(b (16 a+5 b)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{1}{16} b (16 a+5 b) x-\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 0.306401, size = 77, normalized size = 0.75 \[ \frac{b ((96 a+45 b) \sinh (2 (c+d x))-192 a c-192 a d x-9 b \sinh (4 (c+d x))+b \sinh (6 (c+d x))-60 b c-60 b d x)-192 a^2 \coth (c+d x)}{192 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.038, size = 91, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{a}^{2}{\rm coth} \left (dx+c\right )+2\,ab \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/2\,dx-c/2 \right ) +{b}^{2} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.08813, size = 197, normalized size = 1.91 \begin{align*} -\frac{1}{4} \, a b{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{384} \, b^{2}{\left (\frac{{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac{120 \,{\left (d x + c\right )}}{d} + \frac{45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac{2 \, a^{2}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.70794, size = 562, normalized size = 5.46 \begin{align*} \frac{b^{2} \cosh \left (d x + c\right )^{7} + 7 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 10 \, b^{2} \cosh \left (d x + c\right )^{5} + 5 \,{\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 10 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 6 \,{\left (16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} +{\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 100 \, b^{2} \cosh \left (d x + c\right )^{3} + 18 \,{\left (16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \,{\left (128 \, a^{2} + 32 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 24 \,{\left ({\left (16 \, a b + 5 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \sinh \left (d x + c\right )}{384 \, d \sinh \left (d x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.29338, size = 273, normalized size = 2.65 \begin{align*} -\frac{{\left (16 \, a b + 5 \, b^{2}\right )}{\left (d x + c\right )}}{16 \, d} + \frac{{\left (352 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 110 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 96 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 45 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} - \frac{2 \, a^{2}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} + \frac{b^{2} d^{2} e^{\left (6 \, d x + 6 \, c\right )} - 9 \, b^{2} d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 96 \, a b d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )}}{384 \, d^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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