Optimal. Leaf size=186 \[ -\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A]
time = 0.07, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps
used = 5, 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 {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2198
Rule 2202
Rubi steps
\begin {align*} \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {(8 b) \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {\left (16 b^2\right ) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (64 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (128 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 100, normalized size = 0.54 \begin {gather*} \frac {2 \left (-5 b^4 x^4+60 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+90 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-20 b x \tanh ^{-1}(\tanh (a+b x))^3+3 \tanh ^{-1}(\tanh (a+b x))^4\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.13, size = 196, normalized size = 1.05
method | result | size |
derivativedivides | \(-\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {16 b \left (-\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(196\) |
default | \(-\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {16 b \left (-\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\) | \(196\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 67, normalized size = 0.36 \begin {gather*} -\frac {2 \, {\left (128 \, b^{5} x^{5} + 320 \, a b^{4} x^{4} + 240 \, a^{2} b^{3} x^{3} + 40 \, a^{3} b^{2} x^{2} - 5 \, a^{4} b x + 3 \, a^{5}\right )}}{15 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} x^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.45, size = 82, normalized size = 0.44 \begin {gather*} -\frac {2 \, {\left (128 \, b^{4} x^{4} + 192 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x + 3 \, a^{4}\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \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 = 0.43, size = 173, normalized size = 0.93 \begin {gather*} -\frac {2 \, \sqrt {x} {\left (\frac {11 \, b^{4} x}{a^{5}} + \frac {12 \, b^{3}}{a^{4}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {4 \, {\left (45 \, b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} - 240 \, a b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 490 \, a^{2} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 320 \, a^{3} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 73 \, a^{4} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5} a^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.86, size = 466, normalized size = 2.51 \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 {4}{5\,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 )}+\frac {256\,x^2}{5\,{\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 {64\,x}{15\,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}-\frac {2048\,b\,x^3}{5\,{\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 {8192\,b^2\,x^4}{15\,{\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}\right )}{x^{9/2}-\frac {x^{7/2}\,\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 )}{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 )}^2}{4\,b^2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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