Optimal. Leaf size=250 \[ \frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x)}{64 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x)}{32 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x)}{192 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 x \text {csch}(a c+b c x)}{16 \sqrt {\text {csch}^2(a c+b c x)}} \]
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Rubi [A] time = 0.20, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6720, 2282, 12, 266, 43} \[ \frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x)}{64 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x)}{32 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x)}{192 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 x \text {csch}(a c+b c x)}{16 \sqrt {\text {csch}^2(a c+b c x)}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 266
Rule 2282
Rule 6720
Rubi steps
\begin {align*} \int \frac {e^{c (a+b x)}}{\text {csch}^2(a c+b c x)^{5/2}} \, dx &=\frac {\text {csch}(a c+b c x) \int e^{c (a+b x)} \sinh ^5(a c+b c x) \, dx}{\sqrt {\text {csch}^2(a c+b c x)}}\\ &=\frac {\text {csch}(a c+b c x) \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{32 x^5} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\text {csch}^2(a c+b c x)}}\\ &=\frac {\text {csch}(a c+b c x) \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{x^5} \, dx,x,e^{c (a+b x)}\right )}{32 b c \sqrt {\text {csch}^2(a c+b c x)}}\\ &=\frac {\text {csch}(a c+b c x) \operatorname {Subst}\left (\int \frac {(-1+x)^5}{x^3} \, dx,x,e^{2 c (a+b x)}\right )}{64 b c \sqrt {\text {csch}^2(a c+b c x)}}\\ &=\frac {\text {csch}(a c+b c x) \operatorname {Subst}\left (\int \left (10-\frac {1}{x^3}+\frac {5}{x^2}-\frac {10}{x}-5 x+x^2\right ) \, dx,x,e^{2 c (a+b x)}\right )}{64 b c \sqrt {\text {csch}^2(a c+b c x)}}\\ &=\frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x)}{64 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x)}{32 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x)}{128 b c \sqrt {\text {csch}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x)}{192 b c \sqrt {\text {csch}^2(a c+b c x)}}-\frac {5 x \text {csch}(a c+b c x)}{16 \sqrt {\text {csch}^2(a c+b c x)}}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 106, normalized size = 0.42 \[ \frac {\left (\frac {1}{2} e^{-4 c (a+b x)}-5 e^{-2 c (a+b x)}+10 e^{2 c (a+b x)}-\frac {5}{2} e^{4 c (a+b x)}+\frac {1}{3} e^{6 c (a+b x)}-20 b c x\right ) \text {csch}^5(c (a+b x))}{64 b c \text {csch}^2(c (a+b x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 2.05, size = 218, normalized size = 0.87 \[ \frac {5 \, \cosh \left (b c x + a c\right )^{5} + 25 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} - 3\right )} \sinh \left (b c x + a c\right )^{3} - 45 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{3} - 27 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (\cosh \left (b c x + a c\right )^{4} - 24 \, b c x - 9 \, \cosh \left (b c x + a c\right )^{2} - 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 278, normalized size = 1.11 \[ -\frac {{\left (120 \, b c x e^{\left (-a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-4 \, b c x - 5 \, a c\right )} - {\left (2 \, e^{\left (6 \, b c x + 20 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 15 \, e^{\left (4 \, b c x + 18 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + 60 \, e^{\left (2 \, b c x + 16 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-15 \, a c\right )}\right )} e^{\left (a c\right )}}{384 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.54, size = 326, normalized size = 1.30 \[ -\frac {5 x \,{\mathrm e}^{c \left (b x +a \right )}}{16 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}}+\frac {{\mathrm e}^{7 c \left (b x +a \right )}}{192 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}}-\frac {5 \,{\mathrm e}^{5 c \left (b x +a \right )}}{128 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}}+\frac {5 \,{\mathrm e}^{3 c \left (b x +a \right )}}{32 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-c \left (b x +a \right )}}{64 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}}+\frac {{\mathrm e}^{-3 c \left (b x +a \right )}}{128 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 90, normalized size = 0.36 \[ \frac {{\left (2 \, e^{\left (10 \, b c x + 10 \, a c\right )} - 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} + 60 \, e^{\left (6 \, b c x + 6 \, a c\right )} - 30 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 3\right )} e^{\left (-4 \, b c x - 4 \, a c\right )}}{384 \, b c} - \frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left (\frac {1}{{\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int \frac {e^{b c x}}{\left (\operatorname {csch}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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