Optimal. Leaf size=319 \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}} \]
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Rubi [A]
time = 1.22, 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 = {6852, 2320,
398, 1828, 1171, 393, 213} \begin {gather*} \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 213
Rule 393
Rule 398
Rule 1171
Rule 1828
Rule 2320
Rule 6852
Rubi steps
\begin {align*} \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx &=\frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^5(a c+b c x) \, dx}{\sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {(2 \tanh (a c+b c x)) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \text {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}+\frac {(15 \tanh (a c+b c x)) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{4 b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}}\\ \end {align*}
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Mathematica [A]
time = 10.48, size = 164, normalized size = 0.51 \begin {gather*} \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 (-1+e^{2 c (a+b x)}\right )^4 \log \left (1-e^{c (a+b x)}\right )-45 \left (-1+e^{2 c (a+b x)}\right )^4 \log \left (1+e^{c (a+b x)}\right )\right ) \tanh (c (a+b x))}{24 b c \left (-1+e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(c (a+b x))}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 6.55, size = 320, normalized size = 1.00
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}-\frac {{\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}-115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}-21\right )}{12 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{3} \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, b c}+\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}-\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}\) | \(320\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 167, normalized size = 0.52 \begin {gather*} -\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 )}} \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. 1617 vs.
\(2 (281) = 562\).
time = 0.37, size = 1617, normalized size = 5.07 \begin {gather*} \text {Too large to display} \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 = 0.52, size = 215, normalized size = 0.67 \begin {gather*} \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 ) - 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 ) + 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 )} \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 )\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4}}}{24 \, 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 \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {tanh}\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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