Optimal. Leaf size=176 \[ -\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {15}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \]
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Rubi [A]
time = 0.09, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps
used = 6, 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 {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {15}{4 \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^{3/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {15 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {15 \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {15 \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {15}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {(15 b) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {15}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}
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Mathematica [A]
time = 0.21, size = 141, normalized size = 0.80 \begin {gather*} -\frac {15 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {2}{\sqrt {x} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7 b \sqrt {x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {b \sqrt {x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.62, size = 181, normalized size = 1.03
method | result | size |
derivativedivides | \(-\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}-\frac {7 b^{2} x^{\frac {3}{2}}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b \sqrt {x}\, a}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {15 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(181\) |
default | \(-\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}-\frac {7 b^{2} x^{\frac {3}{2}}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b \sqrt {x}\, a}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {15 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\) | \(181\) |
risch | \(\text {Expression too large to display}\) | \(14445\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.48, size = 73, normalized size = 0.41 \begin {gather*} -\frac {15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} - \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 214, normalized size = 1.22 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\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} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\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 {3}{2}} \operatorname {atanh}^{3}{\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 = 59, normalized size = 0.34 \begin {gather*} -\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2}{a^{3} \sqrt {x}} - \frac {7 \, b^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.17, size = 1077, normalized size = 6.12 \begin {gather*} \frac {x\,\left (\frac {12\,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 )}^3}+\frac {3\,b\,\left (16\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-16\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+32\,b\,x\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 )}^4}\right )-\frac {16\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-16\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+32\,b\,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 {15\,\sqrt {2}\,\sqrt {b}\,\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}}-\frac {8\,b\,\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 )}^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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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