Optimal. Leaf size=177 \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{5/2}}+\frac {3 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b^2}+\frac {x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{32 b}-\frac {1}{8} x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]
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Rubi [A] time = 0.10, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{5/2}}+\frac {3 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b^2}+\frac {x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{32 b}-\frac {1}{8} x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]
Antiderivative was successfully verified.
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Rule 2165
Rule 2169
Rubi steps
\begin {align*} \int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx &=\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {1}{8} \left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx\\ &=-\frac {1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {1}{16} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac {x^{3/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {\left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{64 b}\\ &=-\frac {1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac {3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{64 b^2}+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac {\left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \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}{128 b^2}\\ &=\frac {3 \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 )^4}{64 b^{5/2}}-\frac {1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac {3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{64 b^2}+\frac {1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 122, normalized size = 0.69 \[ \frac {\sqrt {b} \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (11 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+11 b x \tanh ^{-1}(\tanh (a+b x))^2-3 \tanh ^{-1}(\tanh (a+b x))^3-3 b^3 x^3\right )+3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{64 b^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 163, normalized size = 0.92 \[ \left [\frac {3 \, a^{4} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{128 \, b^{3}}, -\frac {3 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{64 \, b^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.72, size = 147, normalized size = 0.83 \[ \frac {1}{384} \, \sqrt {2} {\left (8 \, \sqrt {2} {\left (\sqrt {b x + a} {\left (2 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {x} - \frac {3 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}}\right )} a + \sqrt {2} {\left ({\left (2 \, {\left (4 \, {\left (6 \, x + \frac {a}{b}\right )} x - \frac {5 \, a^{2}}{b^{2}}\right )} x + \frac {15 \, a^{3}}{b^{3}}\right )} \sqrt {b x + a} \sqrt {x} + \frac {15 \, a^{4} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {7}{2}}}\right )} b\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.24, size = 471, normalized size = 2.66 \[ \frac {x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{4 b}-\frac {a \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{8 b^{2}}+\frac {a^{2} \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{32 b^{2}}+\frac {3 a^{3} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b^{2}}+\frac {3 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{4}}{64 b^{\frac {5}{2}}}+\frac {3 a^{3} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{16 b^{\frac {5}{2}}}+\frac {9 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b^{2}}+\frac {9 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{32 b^{\frac {5}{2}}}+\frac {a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{16 b^{2}}+\frac {9 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b^{2}}+\frac {3 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{16 b^{\frac {5}{2}}}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{8 b^{2}}+\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{32 b^{2}}+\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{64 b^{2}}+\frac {3 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{64 b^{\frac {5}{2}}} \]
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 \[ \int x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}\,{d x} \]
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 \[ \int x^{3/2}\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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