Optimal. Leaf size=193 \[ \frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )}-\frac {3 \text {ArcTan}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c} \]
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Rubi [A]
time = 0.20, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6852, 2320,
398, 1172, 12, 294, 209} \begin {gather*} -\frac {3 \text {ArcTan}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac {3 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 )}-\frac {2 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 )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 209
Rule 294
Rule 398
Rule 1172
Rule 2320
Rule 6852
Rubi steps
\begin {align*} \int e^{c (a+b x)} \tanh ^2(a c+b c x)^{3/2} \, dx &=\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \tanh ^3(a c+b c x) \, dx\\ &=\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{\left (1+x^2\right )^3} \, 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 ) \text {Subst}\left (\int \left (1-\frac {2 \left (1+3 x^4\right )}{\left (1+x^2\right )^3}\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 ) \text {Subst}\left (\int \frac {1+3 x^4}{\left (1+x^2\right )^3} \, 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 {2 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 )^2}+\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \text {Subst}\left (\int -\frac {12 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{2 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 {2 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 )^2}-\frac {\left (6 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, 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 {2 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 )^2}+\frac {3 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 )}-\frac {\left (3 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, 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 {2 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 )^2}+\frac {3 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 )}-\frac {3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 104, normalized size = 0.54 \begin {gather*} \frac {\left (e^{c (a+b x)} \left (2+5 e^{2 c (a+b x)}+e^{4 c (a+b x)}\right )-3 \left (1+e^{2 c (a+b x)}\right )^2 \text {ArcTan}\left (e^{c (a+b x)}\right )\right ) \coth (c (a+b x)) \sqrt {\tanh ^2(c (a+b x))}}{b c \left (1+e^{2 c (a+b x)}\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains complex when optimal does not.
time = 4.72, size = 301, normalized size = 1.56
method | result | size |
risch | \(\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) b c}+\frac {\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )} \left (3 \,{\mathrm e}^{2 c \left (b x +a \right )}+1\right )}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}+\frac {3 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}-\frac {3 i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )}{2 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}\) | \(301\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.48, size = 90, normalized size = 0.47 \begin {gather*} -\frac {3 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} + 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 458 vs.
\(2 (179) = 358\).
time = 0.35, size = 458, normalized size = 2.37 \begin {gather*} \frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \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} + 1\right )} \sinh \left (b c x + a c\right )^{3} + 5 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 3 \, {\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \, {\left (3 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (\cosh \left (b c x + a c\right )^{3} + \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \arctan \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) + {\left (5 \, \cosh \left (b c x + a c\right )^{4} + 15 \, \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right ) + 2 \, \cosh \left (b c x + a c\right )}{b c \cosh \left (b c x + a c\right )^{4} + 4 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + b c \sinh \left (b c x + a c\right )^{4} + 2 \, b c \cosh \left (b c x + a c\right )^{2} + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )^{2} + b c + 4 \, {\left (b c \cosh \left (b c x + a c\right )^{3} + b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\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 = 0.43, size = 129, normalized size = 0.67 \begin {gather*} -\frac {3 \, \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 ) - e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {3 \, 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 ) + 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 )}^{2}}}{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 = -0.01 \begin {gather*} \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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