Optimal. Leaf size=145 \[ \frac {5 b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {5 b}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \]
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Rubi [A]
time = 0.08, 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 = {2199, 2194,
2193} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 2193
Rule 2194
Rule 2199
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 = 0.15, size = 120, normalized size = 0.83 \begin {gather*} \frac {2 \left (-7 b x+\tanh ^{-1}(\tanh (a+b x))\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {5 b^{3/2} \text {ArcTan}\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 {b^2 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x)) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.59, size = 128, normalized size = 0.88
method | result | size |
derivativedivides | \(\frac {b^{2} \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {5 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}+\frac {4 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}\) | \(128\) |
default | \(\frac {b^{2} \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {5 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}+\frac {4 b}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}\) | \(128\) |
risch | \(\text {Expression too large to display}\) | \(14156\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.48, size = 64, normalized size = 0.44 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 184, normalized size = 1.27 \begin {gather*} \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 ] \end {gather*}
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 \begin {gather*} \int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 58, normalized size = 0.40 \begin {gather*} \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}}} \end {gather*}
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 \begin {gather*} \frac {\frac {32\,b}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}-\frac {80\,b^2\,x}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}}{\sqrt {x}\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}-\frac {8}{3\,x^{3/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {20\,\sqrt {2}\,b^{3/2}\,\ln \left (-\frac {\sqrt {b}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^6+60\,a^2\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4-160\,a^3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3+240\,a^4\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+64\,a^6-12\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^5-192\,a^5\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{2\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}\right )}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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