Optimal. Leaf size=224 \[ \frac {35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {35 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \]
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Rubi [A] time = 0.15, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac {35 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \]
Antiderivative was successfully verified.
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Rule 2161
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{4} (5 b) \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {1}{8} \left (35 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {\left (35 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {\left (35 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (35 b^2\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (35 b^2\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {35 b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac {5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac {1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 133, normalized size = 0.59 \[ \frac {80 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+39 b x \tanh ^{-1}(\tanh (a+b x))^2-6 \tanh ^{-1}(\tanh (a+b x))^3-8 b^3 x^3}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac {35 b^2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 255, normalized size = 1.14 \[ \left [\frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, \frac {105 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt {b x + a}}{12 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 93, normalized size = 0.42 \[ \frac {35 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4}} + \frac {2 \, {\left (9 \, {\left (b x + a\right )} b^{2} + a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4}} + \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 13 \, \sqrt {b x + a} a b^{2}}{4 \, a^{4} b^{2} x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 157, normalized size = 0.70 \[ 2 b^{2} \left (\frac {3}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {\frac {\frac {11 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8}+\left (-\frac {13 \arctanh \left (\tanh \left (b x +a \right )\right )}{8}+\frac {13 b x}{8}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {35 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4}}\right ) \]
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 \[ \int \frac {1}{x^{3} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.47, size = 1514, normalized size = 6.76 \[ \text {result too large to display} \]
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 \[ \int \frac {1}{x^{3} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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