Optimal. Leaf size=128 \[ -\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
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Rubi [A]
time = 0.06, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2194, 2193}
\begin {gather*} -\frac {2 b^{5/2} \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 {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 2193
Rule 2194
Rubi steps
\begin {align*} \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b^2 \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b^3 \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {2 b^{5/2} \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 {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 107, normalized size = 0.84 \begin {gather*} -\frac {2 b^{5/2} \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 {2 \left (23 b^2 x^2-11 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 120, normalized size = 0.94
method | result | size |
derivativedivides | \(-\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {2 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(120\) |
default | \(-\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {2 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(120\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 52, normalized size = 0.41 \begin {gather*} -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 144, normalized size = 1.12 \begin {gather*} \left [\frac {15 \, b^{2} x^{3} \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} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, \frac {2 \, {\left (15 \, b^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\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 {7}{2}} \operatorname {atanh}{\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 = 0.39, size = 52, normalized size = 0.41 \begin {gather*} -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.63, size = 822, normalized size = 6.42 \begin {gather*} \frac {4}{5\,x^{5/2}\,\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 )}+\frac {8\,b}{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 {16\,b^2}{\sqrt {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}+\frac {8\,\sqrt {2}\,b^{5/2}\,\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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