Optimal. Leaf size=124 \[ \frac{1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac{1}{8} b \left (-10 a^2 b^2+15 a^4+3 b^4\right ) \tanh ^{-1}(\cosh (x))-\frac{1}{8} \text{csch}^2(x) (a \cosh (x)+b)^2 \left (b \left (5 a^2-3 b^2\right ) \cosh (x)+2 a \left (2 a^2-b^2\right )\right )+a^5 \log (\sinh (x))-\frac{1}{4} \text{csch}^4(x) (a \cosh (x)+b)^4 (a+b \cosh (x)) \]
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Rubi [A] time = 0.243005, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.727, Rules used = {4392, 2668, 739, 819, 774, 635, 204, 260} \[ \frac{1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac{1}{8} b \left (-10 a^2 b^2+15 a^4+3 b^4\right ) \tanh ^{-1}(\cosh (x))-\frac{1}{8} \text{csch}^2(x) (a \cosh (x)+b)^2 \left (b \left (5 a^2-3 b^2\right ) \cosh (x)+2 a \left (2 a^2-b^2\right )\right )+a^5 \log (\sinh (x))-\frac{1}{4} \text{csch}^4(x) (a \cosh (x)+b)^4 (a+b \cosh (x)) \]
Antiderivative was successfully verified.
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Rule 4392
Rule 2668
Rule 739
Rule 819
Rule 774
Rule 635
Rule 204
Rule 260
Rubi steps
\begin{align*} \int (a \coth (x)+b \text{csch}(x))^5 \, dx &=-\left (i \int (i b+i a \cosh (x))^5 \text{csch}^5(x) \, dx\right )\\ &=-\left (a^5 \operatorname{Subst}\left (\int \frac{(i b+x)^5}{\left (-a^2-x^2\right )^3} \, dx,x,i a \cosh (x)\right )\right )\\ &=-\frac{1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text{csch}^4(x)-\frac{1}{4} a^3 \operatorname{Subst}\left (\int \frac{(i b+x)^3 \left (-4 a^2+3 b^2+i b x\right )}{\left (-a^2-x^2\right )^2} \, dx,x,i a \cosh (x)\right )\\ &=-\frac{1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text{csch}^2(x)-\frac{1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text{csch}^4(x)-\frac{1}{8} a \operatorname{Subst}\left (\int \frac{(i b+x) \left (8 a^4-7 a^2 b^2+3 b^4-i b \left (7 a^2-3 b^2\right ) x\right )}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=\frac{1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac{1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text{csch}^2(x)-\frac{1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text{csch}^4(x)+\frac{1}{8} a \operatorname{Subst}\left (\int \frac{-i a^2 b \left (7 a^2-3 b^2\right )-i b \left (8 a^4-7 a^2 b^2+3 b^4\right )-\left (8 a^4-7 a^2 b^2+3 b^4+b^2 \left (7 a^2-3 b^2\right )\right ) x}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=\frac{1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac{1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text{csch}^2(x)-\frac{1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text{csch}^4(x)-a^5 \operatorname{Subst}\left (\int \frac{x}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )-\frac{1}{8} \left (i a b \left (15 a^4-10 a^2 b^2+3 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=-\frac{1}{8} b \left (15 a^4-10 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\cosh (x))+\frac{1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac{1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text{csch}^2(x)-\frac{1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text{csch}^4(x)+a^5 \log (\sinh (x))\\ \end{align*}
Mathematica [A] time = 0.469392, size = 244, normalized size = 1.97 \[ -\frac{1}{64} \text{csch}^4(x) \left (20 a^2 b^3 \cosh (3 x)+2 b \left (70 a^2 b^2+15 a^4+11 b^4\right ) \cosh (x)+30 a^2 b^3 \log \left (\tanh \left (\frac{x}{2}\right )\right )+10 a^2 b^3 \cosh (4 x) \log \left (\tanh \left (\frac{x}{2}\right )\right )+4 \cosh (2 x) \left (b \left (-10 a^2 b^2+15 a^4+3 b^4\right ) \log \left (\tanh \left (\frac{x}{2}\right )\right )+8 \left (5 a^3 b^2+a^5\right )+8 a^5 \log (\sinh (x))\right )+50 a^4 b \cosh (3 x)-45 a^4 b \log \left (\tanh \left (\frac{x}{2}\right )\right )-15 a^4 b \cosh (4 x) \log \left (\tanh \left (\frac{x}{2}\right )\right )-24 a^5 \log (\sinh (x))-8 a^5 \cosh (4 x) \log (\sinh (x))-16 a^5+80 a b^4-6 b^5 \cosh (3 x)-9 b^5 \log \left (\tanh \left (\frac{x}{2}\right )\right )-3 b^5 \cosh (4 x) \log \left (\tanh \left (\frac{x}{2}\right )\right )\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.028, size = 223, normalized size = 1.8 \begin{align*}{a}^{5}\ln \left ( \sinh \left ( x \right ) \right ) -{\frac{{a}^{5} \left ({\rm coth} \left (x\right ) \right ) ^{2}}{2}}-{\frac{{a}^{5} \left ({\rm coth} \left (x\right ) \right ) ^{4}}{4}}-5\,{\frac{{a}^{4}b \left ( \cosh \left ( x \right ) \right ) ^{3}}{ \left ( \sinh \left ( x \right ) \right ) ^{4}}}+5\,{\frac{{a}^{4}b\cosh \left ( x \right ) }{ \left ( \sinh \left ( x \right ) \right ) ^{4}}}-{\frac{5\,{a}^{4}b{\rm coth} \left (x\right ) \left ({\rm csch} \left (x\right ) \right ) ^{3}}{4}}+{\frac{15\,{a}^{4}b{\rm csch} \left (x\right ){\rm coth} \left (x\right )}{8}}-{\frac{15\,{a}^{4}b{\it Artanh} \left ({{\rm e}^{x}} \right ) }{4}}-{\frac{5\,{a}^{3}{b}^{2} \left ( \cosh \left ( x \right ) \right ) ^{2}}{2\, \left ( \sinh \left ( x \right ) \right ) ^{4}}}-{\frac{5\,{a}^{3}{b}^{2} \left ( \cosh \left ( x \right ) \right ) ^{2}}{2\, \left ( \sinh \left ( x \right ) \right ) ^{2}}}-{\frac{10\,{a}^{2}{b}^{3}\cosh \left ( x \right ) }{3\, \left ( \sinh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{a}^{2}{b}^{3}{\rm coth} \left (x\right ) \left ({\rm csch} \left (x\right ) \right ) ^{3}}{6}}-{\frac{5\,{a}^{2}{b}^{3}{\rm csch} \left (x\right ){\rm coth} \left (x\right )}{4}}+{\frac{5\,{a}^{2}{b}^{3}{\it Artanh} \left ({{\rm e}^{x}} \right ) }{2}}-{\frac{5\,{b}^{4}a \left ( \cosh \left ( x \right ) \right ) ^{2}}{4\, \left ( \sinh \left ( x \right ) \right ) ^{4}}}+{\frac{5\,{b}^{4}a \left ( \cosh \left ( x \right ) \right ) ^{2}}{4\, \left ( \sinh \left ( x \right ) \right ) ^{2}}}-{\frac{{b}^{5}{\rm coth} \left (x\right ) \left ({\rm csch} \left (x\right ) \right ) ^{3}}{4}}+{\frac{3\,{b}^{5}{\rm csch} \left (x\right ){\rm coth} \left (x\right )}{8}}-{\frac{3\,{b}^{5}{\it Artanh} \left ({{\rm e}^{x}} \right ) }{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.07308, size = 446, normalized size = 3.6 \begin{align*} -\frac{5}{2} \, a^{3} b^{2} \coth \left (x\right )^{4} + a^{5}{\left (x + \frac{4 \,{\left (e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \log \left (e^{\left (-x\right )} + 1\right ) + \log \left (e^{\left (-x\right )} - 1\right )\right )} + \frac{5}{8} \, a^{4} b{\left (\frac{2 \,{\left (5 \, e^{\left (-x\right )} + 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} - 3 \, \log \left (e^{\left (-x\right )} + 1\right ) + 3 \, \log \left (e^{\left (-x\right )} - 1\right )\right )} - \frac{1}{8} \, b^{5}{\left (\frac{2 \,{\left (3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + 3 \, \log \left (e^{\left (-x\right )} + 1\right ) - 3 \, \log \left (e^{\left (-x\right )} - 1\right )\right )} + \frac{5}{4} \, a^{2} b^{3}{\left (\frac{2 \,{\left (e^{\left (-x\right )} + 7 \, e^{\left (-3 \, x\right )} + 7 \, e^{\left (-5 \, x\right )} + e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right )\right )} - \frac{20 \, a b^{4}}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.4722, size = 6710, normalized size = 54.11 \begin{align*} \text{result too large to display} \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 = 1.18024, size = 316, normalized size = 2.55 \begin{align*} \frac{1}{16} \,{\left (8 \, a^{5} - 15 \, a^{4} b + 10 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \frac{1}{16} \,{\left (8 \, a^{5} + 15 \, a^{4} b - 10 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) - \frac{3 \, a^{5}{\left (e^{\left (-x\right )} + e^{x}\right )}^{4} + 25 \, a^{4} b{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 10 \, a^{2} b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 3 \, b^{5}{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 8 \, a^{5}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 80 \, a^{3} b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 60 \, a^{4} b{\left (e^{\left (-x\right )} + e^{x}\right )} + 40 \, a^{2} b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )} + 20 \, b^{5}{\left (e^{\left (-x\right )} + e^{x}\right )} - 160 \, a^{3} b^{2} + 80 \, a b^{4}}{4 \,{\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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