Optimal. Leaf size=143 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}{b^{9/2}}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {2 x^{7/2}}{7 b} \]
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Rubi [A] time = 0.13, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2159, 2162} \[ \frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}{b^{9/2}}+\frac {2 x^{7/2}}{7 b} \]
Antiderivative was successfully verified.
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Rule 2159
Rule 2162
Rubi steps
\begin {align*} \int \frac {x^{7/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x^{7/2}}{7 b}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac {x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x^{7/2}}{7 b}+\frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac {x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {2 x^{7/2}}{7 b}+\frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3 \int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {2 x^{7/2}}{7 b}+\frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}+\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^4 \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac {2 x^{7/2}}{7 b}+\frac {2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{3 b^3}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{b^4}-\frac {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}}{b^{9/2}}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 129, normalized size = 0.90 \[ \frac {2 \left (-406 b^{5/2} x^{5/2} \tanh ^{-1}(\tanh (a+b x))+350 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2-105 \sqrt {b} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3+105 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+176 b^{7/2} x^{7/2}\right )}{105 b^{9/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 153, normalized size = 1.07 \[ \left [\frac {105 \, a^{3} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}}{105 \, b^{4}}, \frac {2 \, {\left (105 \, a^{3} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (15 \, b^{3} x^{3} - 21 \, a b^{2} x^{2} + 35 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}\right )}}{105 \, b^{4}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 70, normalized size = 0.49 \[ \frac {2 \, a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, b^{6} x^{\frac {7}{2}} - 21 \, a b^{5} x^{\frac {5}{2}} + 35 \, a^{2} b^{4} x^{\frac {3}{2}} - 105 \, a^{3} b^{3} \sqrt {x}\right )}}{105 \, b^{7}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.27, size = 481, normalized size = 3.36 \[ \frac {2 x^{\frac {7}{2}}}{7 b}-\frac {2 x^{\frac {5}{2}} a}{5 b^{2}}-\frac {2 x^{\frac {5}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{5 b^{2}}+\frac {2 a^{2} x^{\frac {3}{2}}}{3 b^{3}}+\frac {4 x^{\frac {3}{2}} a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3 b^{3}}+\frac {2 x^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{3 b^{3}}-\frac {2 \sqrt {x}\, a^{3}}{b^{4}}-\frac {6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{4}}-\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{4}}-\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {x}}{b^{4}}+\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{4}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {8 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {12 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {8 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {2 \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 -a \right )^{4}}{b^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 65, normalized size = 0.45 \[ \frac {2 \, a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, b^{3} x^{\frac {7}{2}} - 21 \, a b^{2} x^{\frac {5}{2}} + 35 \, a^{2} b x^{\frac {3}{2}} - 105 \, a^{3} \sqrt {x}\right )}}{105 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.72, size = 475, normalized size = 3.32 \[ \frac {2\,x^{7/2}}{7\,b}+\frac {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 )}{5\,b^2}+\frac {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}{6\,b^3}+\frac {\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}{4\,b^4}+\frac {\sqrt {2}\,\ln \left (\frac {64\,b^{19/2}\,\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 (\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 )\,\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}}\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}}{16\,b^{9/2}} \]
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 \[ \int \frac {x^{\frac {7}{2}}}{\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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