Optimal. Leaf size=148 \[ \frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
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Rubi [A] time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
Antiderivative was successfully verified.
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Rule 2167
Rule 2171
Rubi steps
\begin {align*} \int \frac {1}{x^{9/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(6 b) \int \frac {1}{x^{7/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (24 b^2\right ) \int \frac {1}{x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (16 b^3\right ) \int \frac {1}{x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 82, normalized size = 0.55 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-35 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+21 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+35 b^3 x^3\right )}{35 x^{7/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 45, normalized size = 0.30 \[ \frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{4} x^{\frac {7}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 103, normalized size = 0.70 \[ \frac {64 \, {\left (35 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 21 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 7 \, a^{2} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3}\right )} b^{\frac {7}{2}}}{35 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.30, size = 151, normalized size = 1.02 \[ -\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {12 b \left (-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 55, normalized size = 0.37 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} + a^{3} b x - 5 \, a^{4}\right )}}{35 \, \sqrt {b x + a} a^{4} x^{\frac {7}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.64, size = 287, normalized size = 1.94 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{7\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,b^2\,x^2}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^3\,x^3}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}+\frac {48\,b\,x}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{7/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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