Optimal. Leaf size=166 \[ \frac {35 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{9/2}}+\frac {35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}+\frac {35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A] time = 0.11, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ \frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac {35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac {35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}+\frac {35 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{9/2}}-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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Rule 2165
Rule 2168
Rule 2169
Rubi steps
\begin {align*} \int \frac {x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {7 \int \frac {x^{5/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}-\frac {\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x^{3/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{6 b^2}\\ &=-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac {35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac {\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b^3}\\ &=-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac {35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac {35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}-\frac {\left (35 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 b^4}\\ &=\frac {35 \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 )^3}{8 b^{9/2}}-\frac {2 x^{7/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {7 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2}+\frac {35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^3}+\frac {35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^4}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 122, normalized size = 0.73 \[ \frac {35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{8 b^{9/2}}+\frac {\sqrt {x} \left (231 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-280 b x \tanh ^{-1}(\tanh (a+b x))^2+105 \tanh ^{-1}(\tanh (a+b x))^3-48 b^3 x^3\right )}{24 b^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 196, normalized size = 1.18 \[ \left [\frac {105 \, {\left (a^{3} b x + a^{4}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, {\left (b^{6} x + a b^{5}\right )}}, \frac {105 \, {\left (a^{3} b x + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{4} x^{3} - 14 \, a b^{3} x^{2} + 35 \, a^{2} b^{2} x + 105 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, {\left (b^{6} x + a b^{5}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 75, normalized size = 0.45 \[ \frac {{\left ({\left (2 \, x {\left (\frac {4 \, x}{b} - \frac {7 \, a}{b^{2}}\right )} + \frac {35 \, a^{2}}{b^{3}}\right )} x + \frac {105 \, a^{3}}{b^{4}}\right )} \sqrt {x}}{24 \, \sqrt {b x + a}} + \frac {35 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{8 \, b^{\frac {9}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.25, size = 428, normalized size = 2.58 \[ \frac {x^{\frac {7}{2}}}{3 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {7 a \,x^{\frac {5}{2}}}{12 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {35 a^{2} x^{\frac {3}{2}}}{24 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {35 a^{3} \sqrt {x}}{8 b^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {35 a^{3} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{8 b^{\frac {9}{2}}}+\frac {105 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{8 b^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {105 a^{2} \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 )}{8 b^{\frac {9}{2}}}+\frac {35 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}}}{12 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {105 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{8 b^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {105 a \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 )}{8 b^{\frac {9}{2}}}-\frac {7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {5}{2}}}{12 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} x^{\frac {3}{2}}}{24 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {x}}{8 b^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {35 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{8 b^{\frac {9}{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 \[ \int \frac {x^{\frac {7}{2}}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {x^{7/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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