Optimal. Leaf size=110 \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]
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Rubi [A] time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]
Antiderivative was successfully verified.
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Rule 2167
Rule 2171
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(4 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx}{11 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{99 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 66, normalized size = 0.60 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-154 b x \tanh ^{-1}(\tanh (a+b x))+63 \tanh ^{-1}(\tanh (a+b x))^2+99 b^2 x^2\right )}{693 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 67, normalized size = 0.61 \[ -\frac {2 \, {\left (8 \, b^{5} x^{5} - 4 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{3} + 113 \, a^{3} b^{2} x^{2} + 161 \, a^{4} b x + 63 \, a^{5}\right )} \sqrt {b x + a}}{693 \, a^{3} x^{\frac {11}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 78, normalized size = 0.71 \[ -\frac {\sqrt {2} {\left (\frac {99 \, \sqrt {2} b^{11}}{a} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{11}}{a^{3}} - \frac {11 \, \sqrt {2} b^{11}}{a^{2}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{693 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.30, size = 105, normalized size = 0.95 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 45, normalized size = 0.41 \[ -\frac {2 \, {\left (8 \, b^{3} x^{3} - 20 \, a b^{2} x^{2} + 35 \, a^{2} b x + 63 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{693 \, a^{3} x^{\frac {11}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.66, size = 353, normalized size = 3.21 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {23\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{99}-\frac {226\,b^2\,x^2}{693}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{22}+\frac {4\,b^3\,x^3}{231\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {32\,b^4\,x^4}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^5\,x^5}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{11/2}} \]
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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