Optimal. Leaf size=145 \[ \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {5 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
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Rubi [A] time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {5 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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Rule 2162
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {5 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b}\\ &=-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {5 \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {(5 b) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {5 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (5 b^2\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {5 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}
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Mathematica [A] time = 0.22, size = 120, normalized size = 0.83 \[ \frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}+\frac {b^2 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}+\frac {2 \left (\tanh ^{-1}(\tanh (a+b x))-7 b x\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 184, normalized size = 1.27 \[ \left [\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {15 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 58, normalized size = 0.40 \[ \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} + \frac {2 \, {\left (6 \, b x - a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 128, normalized size = 0.88 \[ -\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}+\frac {4 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {b^{2} \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {5 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 64, normalized size = 0.44 \[ \frac {15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.16, size = 871, normalized size = 6.01 \[ \text {result too large to display} \]
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 \[ \int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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