Optimal. Leaf size=69 \[ -\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {32 b^3 \sqrt {x}}{5} \]
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Rubi [A] time = 0.04, antiderivative size = 69, 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 = {2168, 30} \[ -\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {32 b^3 \sqrt {x}}{5} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2168
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} (6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^{5/2}} \, dx\\ &=-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} \left (8 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\\ &=-\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} \left (16 b^3\right ) \int \frac {1}{\sqrt {x}} \, dx\\ &=\frac {32 b^3 \sqrt {x}}{5}-\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 57, normalized size = 0.83 \[ \frac {2 \left (-8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-2 b x \tanh ^{-1}(\tanh (a+b x))^2-\tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 x^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 35, normalized size = 0.51 \[ \frac {2 \, {\left (5 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 5 \, a^{2} b x - a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 34, normalized size = 0.49 \[ 2 \, b^{3} \sqrt {x} - \frac {2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x + a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 56, normalized size = 0.81 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{\frac {5}{2}}}+\frac {12 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{\sqrt {x}}+2 b \sqrt {x}\right )}{3}\right )}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 55, normalized size = 0.80 \[ \frac {16}{5} \, {\left (2 \, b^{2} \sqrt {x} - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt {x}}\right )} b - \frac {4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.18, size = 182, normalized size = 2.64 \[ \frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{20\,x^{5/2}}+2\,b^3\,\sqrt {x}+\frac {3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{\sqrt {x}}-\frac {b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,x^{3/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 155.33, size = 70, normalized size = 1.01 \[ \frac {32 b^{3} \sqrt {x}}{5} - \frac {16 b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5 \sqrt {x}} - \frac {4 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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