Optimal. Leaf size=319 \[ -\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{4 b c}+\frac {e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{4 b c \left (1-e^{2 c (a+b x)}\right )}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{6 b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^4} \]
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Rubi [A] time = 0.91, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6720, 2282, 390, 1814, 1157, 385, 207} \[ -\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{4 b c}+\frac {e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{4 b c \left (1-e^{2 c (a+b x)}\right )}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{6 b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^4} \]
Antiderivative was successfully verified.
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Rule 207
Rule 385
Rule 390
Rule 1157
Rule 1814
Rule 2282
Rule 6720
Rubi steps
\begin {align*} \int e^{c (a+b x)} \coth ^2(a c+b c x)^{5/2} \, dx &=\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \int e^{c (a+b x)} \coth ^5(a c+b c x) \, dx\\ &=\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^5}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \left (1+\frac {2 \left (1+10 x^4+5 x^8\right )}{\left (-1+x^2\right )^5}\right ) \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}+\frac {\left (2 \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {1+10 x^4+5 x^8}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {4 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4}+\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {8+120 x^2+40 x^4+40 x^6}{\left (-1+x^2\right )^4} \, dx,x,e^{c (a+b x)}\right )}{4 b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {4 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}+\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {160+480 x^2+240 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{24 b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {4 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {55 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {240+960 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{96 b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {4 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {55 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {25 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right )}+\frac {\left (15 \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{4 b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {4 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {55 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {25 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right )}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{4 b c}\\ \end {align*}
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Mathematica [A] time = 10.28, size = 164, normalized size = 0.51 \[ \frac {\left (66 e^{c (a+b x)}-314 e^{3 c (a+b x)}+374 e^{5 c (a+b x)}-246 e^{7 c (a+b x)}+24 e^{9 c (a+b x)}+45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (1-e^{c (a+b x)}\right )-45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (e^{c (a+b x)}+1\right )\right ) \tanh (c (a+b x)) \sqrt {\coth ^2(c (a+b x))}}{24 b c \left (e^{2 c (a+b x)}-1\right )^4} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 1617, normalized size = 5.07 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.07, size = 181, normalized size = 0.57 \[ \frac {\frac {24 \, e^{\left (b c x + a c\right )}}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac {45 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} + \frac {45 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac {2 \, {\left (75 \, e^{\left (7 \, b c x + 7 \, a c\right )} - 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} + 109 \, e^{\left (3 \, b c x + 3 \, a c\right )} - 21 \, e^{\left (b c x + a c\right )}\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}}{24 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.98, size = 320, normalized size = 1.00 \[ \frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {\sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}-115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}-21\right )}{12 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{3} c b}+\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{8 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left (1+{\mathrm e}^{c \left (b x +a \right )}\right )}{8 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 167, normalized size = 0.52 \[ -\frac {15 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{8 \, b c} + \frac {15 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{8 \, b c} + \frac {12 \, e^{\left (9 \, b c x + 9 \, a c\right )} - 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} - 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
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 \[ \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {coth}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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