Optimal. Leaf size=72 \[ \frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
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Rubi [A]
time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2202, 2198}
\begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2198
Rule 2202
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{9/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(2 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 48, normalized size = 0.67 \begin {gather*} \frac {2 \left (7 b x-5 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}{35 x^{7/2} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.13, size = 59, normalized size = 0.82
method | result | size |
derivativedivides | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}\) | \(59\) |
default | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}\) | \(59\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 34, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (2 \, b^{2} x^{2} - 3 \, a b x - 5 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{35 \, a^{2} x^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 45, normalized size = 0.62 \begin {gather*} \frac {2 \, {\left (2 \, b^{3} x^{3} - a b^{2} x^{2} - 8 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{2} x^{\frac {7}{2}}} \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 = 0.41, size = 59, normalized size = 0.82 \begin {gather*} \frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{7}}{a^{2}} - \frac {7 \, \sqrt {2} b^{7}}{a}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{35 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.54, size = 228, normalized size = 3.17 \begin {gather*} \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 {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{7}-\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 )}{7}-\frac {6\,b\,x}{35}+\frac {4\,b^2\,x^2}{35\,\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 )}+\frac {16\,b^3\,x^3}{35\,{\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}\right )}{x^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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