Optimal. Leaf size=35 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
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Rubi [A] time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2167} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
Antiderivative was successfully verified.
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Rule 2167
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 34, normalized size = 0.97 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{5/2} \left (5 b x-5 \tanh ^{-1}(\tanh (a+b x))\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 31, normalized size = 0.89 \[ -\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{5 \, a x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 33, normalized size = 0.94 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{6}}{5 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} a {\left | b \right |}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 29, normalized size = 0.83 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 15, normalized size = 0.43 \[ -\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}}}{5 \, a x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.51, size = 332, normalized size = 9.49 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}+\frac {4\,b^2\,x^2}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}-\frac {4\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-5\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x}\right )}{x^{5/2}} \]
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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