Optimal. Leaf size=128 \[ \frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3} \]
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Rubi [A]
time = 0.05, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2199, 2200,
2196} \begin {gather*} \frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{7/2}}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2196
Rule 2199
Rule 2200
Rubi steps
\begin {align*} \int \frac {x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}-\frac {\left (15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{4 b^2}\\ &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}+\frac {\left (15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b^3}\\ &=\frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 104, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {x} \left (8 b^2 x^2-25 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2\right )}{4 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}\right )}{4 b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs.
\(2(102)=204\).
time = 0.13, size = 261, normalized size = 2.04
method | result | size |
derivativedivides | \(\frac {x^{\frac {5}{2}}}{2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {5 a \,x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 a^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}}-\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{2 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{2 b^{\frac {7}{2}}}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}}\) | \(261\) |
default | \(\frac {x^{\frac {5}{2}}}{2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {5 a \,x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 a^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}}-\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{2 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{2 b^{\frac {7}{2}}}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}}\) | \(261\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 175, normalized size = 1.37 \begin {gather*} \left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 63, normalized size = 0.49 \begin {gather*} \frac {{\left (x {\left (\frac {2 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} - \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x}}{4 \, \sqrt {b x + a}} - \frac {15 \, a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{4 \, b^{\frac {7}{2}}} \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 \frac {x^{5/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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