Optimal. Leaf size=250 \[ \frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{64 b c}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{192 b c}-\frac {5}{16} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)} \]
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Rubi [A]
time = 0.19, 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 = {6852, 2320, 12,
272, 45} \begin {gather*} \frac {e^{-4 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{128 b c}-\frac {5 e^{-2 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{64 b c}+\frac {5 e^{2 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{32 b c}-\frac {5 e^{4 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{128 b c}+\frac {e^{6 c (a+b x)} \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x)}{192 b c}-\frac {5}{16} x \sqrt {\sinh ^2(a c+b c x)} \text {csch}(a c+b c x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 45
Rule 272
Rule 2320
Rule 6852
Rubi steps
\begin {align*} \int e^{c (a+b x)} \sinh ^2(a c+b c x)^{5/2} \, dx &=\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \sinh ^5(a c+b c x) \, dx\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{32 x^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {\left (-1+x^2\right )^5}{x^5} \, dx,x,e^{c (a+b x)}\right )}{32 b c}\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {(-1+x)^5}{x^3} \, dx,x,e^{2 c (a+b x)}\right )}{64 b c}\\ &=\frac {\left (\text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\right ) \text {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}\\ &=\frac {e^{-4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}-\frac {5 e^{-2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{64 b c}+\frac {5 e^{2 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{32 b c}-\frac {5 e^{4 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{128 b c}+\frac {e^{6 c (a+b x)} \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}}{192 b c}-\frac {5}{16} x \text {csch}(a c+b c x) \sqrt {\sinh ^2(a c+b c x)}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 106, normalized size = 0.42 \begin {gather*} \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)) \sinh ^2(c (a+b x))^{5/2}}{64 b c} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.37, size = 184, normalized size = 0.74
method | result | size |
default | \(-\frac {-8 \sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \cosh \left (c \left (b x +a \right )\right ) \left (\sinh ^{5}\left (c \left (b x +a \right )\right )\right )-8 \sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \left (\sinh ^{6}\left (c \left (b x +a \right )\right )\right )+10 \sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \cosh \left (c \left (b x +a \right )\right ) \left (\sinh ^{3}\left (c \left (b x +a \right )\right )\right )-15 \cosh \left (c \left (b x +a \right )\right ) \sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\, \sinh \left (c \left (b x +a \right )\right )+15 \ln \left (\cosh \left (c \left (b x +a \right )\right )+\sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}\right ) \sinh \left (c \left (b x +a \right )\right )-8 \sqrt {-\frac {1}{2}+\frac {\cosh \left (2 b c x +2 a c \right )}{2}}}{48 \sinh \left (c \left (b x +a \right )\right ) b c}\) | \(184\) |
risch | \(-\frac {5 x \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{16 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{7 c \left (b x +a \right )}}{192 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{32 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}-\frac {5 \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{64 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-3 c \left (b x +a \right )}}{128 c b \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )}\) | \(326\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 90, normalized size = 0.36 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 218, normalized size = 0.87 \begin {gather*} \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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.45, size = 269, normalized size = 1.08 \begin {gather*} -\frac {120 \, b c x \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 - 4 \, a c\right )} - {\left (2 \, e^{\left (6 \, 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 ) - 15 \, e^{\left (4 \, 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 ) + 60 \, e^{\left (2 \, b c x + 14 \, 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 (-12 \, a c\right )}}{384 \, b c} \end {gather*}
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 \begin {gather*} \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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