Optimal. Leaf size=110 \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]
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Rubi [A]
time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2202, 2198}
\begin {gather*} \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 2198
Rule 2202
Rubi steps
\begin {align*} \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{9/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(4 b) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 66, normalized size = 0.60 \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (35 b^2 x^2-42 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2\right )}{105 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.13, size = 105, normalized size = 0.95
method | result | size |
derivativedivides | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
default | \(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(105\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 45, normalized size = 0.41 \begin {gather*} -\frac {2 \, {\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 45, normalized size = 0.41 \begin {gather*} -\frac {2 \, {\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 138, normalized size = 1.25 \begin {gather*} \frac {32 \, {\left (70 \, b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} + 35 \, a b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 21 \, a^{2} b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 7 \, a^{3} b^{\frac {7}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{4} b^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.59, size = 234, normalized size = 2.13 \begin {gather*} \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 {32\,b^2\,x^2}{105\,{\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 {128\,b^3\,x^3}{105\,{\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 {4\,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 )}-\frac {2}{7}\right )}{x^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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