Optimal. Leaf size=311 \[ \frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {25 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c \left (e^{2 c (a+b x)}+1\right )}-\frac {55 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{6 b c \left (e^{2 c (a+b x)}+1\right )^2}+\frac {26 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{3 b c \left (e^{2 c (a+b x)}+1\right )^3}-\frac {4 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^4}-\frac {15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c} \]
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Rubi [A] time = 0.90, antiderivative size = 311, 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, 203} \[ \frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {25 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c \left (e^{2 c (a+b x)}+1\right )}-\frac {55 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{6 b c \left (e^{2 c (a+b x)}+1\right )^2}+\frac {26 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{3 b c \left (e^{2 c (a+b x)}+1\right )^3}-\frac {4 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^4}-\frac {15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c} \]
Antiderivative was successfully verified.
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Rule 203
Rule 385
Rule 390
Rule 1157
Rule 1814
Rule 2282
Rule 6720
Rubi steps
\begin {align*} \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{5/2} \, dx &=\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \tanh ^5(a c+b c x) \, dx\\ &=\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(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 (\coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {\left (2 \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {4 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {4 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {4 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{6 b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {4 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{4 b c \left (1+e^{2 c (a+b x)}\right )}-\frac {\left (15 \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {4 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac {26 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(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)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{4 b c \left (1+e^{2 c (a+b x)}\right )}-\frac {15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{4 b c}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 133, normalized size = 0.43 \[ \frac {\left (e^{c (a+b x)} \left (157 e^{2 c (a+b x)}+187 e^{4 c (a+b x)}+123 e^{6 c (a+b x)}+12 e^{8 c (a+b x)}+33\right )-45 \left (e^{2 c (a+b x)}+1\right )^4 \tan ^{-1}\left (e^{c (a+b x)}\right )\right ) \sqrt {\tanh ^2(c (a+b x))} \coth (c (a+b x))}{12 b c \left (e^{2 c (a+b x)}+1\right )^4} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 1226, normalized size = 3.94 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 184, normalized size = 0.59 \[ -\frac {45 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 12 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {75 \, e^{\left (7 \, b c x + 7 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 21 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{4}}}{12 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.83, size = 324, normalized size = 1.04 \[ \frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}+\frac {\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}+115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}+21\right )}{12 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{3} c b}+\frac {15 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )}{8 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}-\frac {15 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )}{8 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 145, normalized size = 0.47 \[ -\frac {15 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{4 \, b c} + \frac {12 \, e^{\left (9 \, b c x + 9 \, a c\right )} + 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} + 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
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 \[ \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \]
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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