Optimal. Leaf size=208 \[ \frac{8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.224905, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac{8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3194
Rule 89
Rule 78
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^2}{x^3 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (8 a-7 b)+2 a x}{x^2 (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{4 a f}\\ &=-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left (8 a^2-40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^4 f}\\ &=\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left (8 a^2-40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^2(e+f x)}\right )}{8 a^4 b f}\\ &=-\frac{\left (8 a^2-40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}+\frac{8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{(8 a-7 b) \text{csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\text{csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.423643, size = 117, normalized size = 0.56 \[ -\frac{\text{csch}^2(e+f x) \left (\left (-8 a^2+40 a b-35 b^2\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sinh ^2(e+f x)}{a}+1\right )+3 a \text{csch}^2(e+f x) \left (2 a \text{csch}^2(e+f x)+8 a-7 b\right )\right )}{24 a^3 f \sqrt{a+b \sinh ^2(e+f x)} \left (a \text{csch}^2(e+f x)+b\right )} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.122, size = 73, normalized size = 0.4 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({\frac{ \left ( \cosh \left ( fx+e \right ) \right ) ^{4}}{ \left ({b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{4}+2\,ab \left ( \sinh \left ( fx+e \right ) \right ) ^{2}+{a}^{2} \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{5}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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