Optimal. Leaf size=167 \[ \frac {5 b^4 \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4} \]
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Rubi [A]
time = 0.09, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2199, 2194,
2192} \begin {gather*} \frac {5 b^4 \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2192
Rule 2194
Rule 2199
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^5} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{8} (5 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx\\ &=-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{16} \left (5 b^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\\ &=-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{64} \left (5 b^3\right ) \int \frac {1}{x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {1}{128} \left (5 b^4\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {\left (5 b^4\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5 b^4 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 134, normalized size = 0.80 \begin {gather*} \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {-b x+\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (15 b^3 x^3+10 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+8 b x \tanh ^{-1}(\tanh (a+b x))^2-48 \tanh ^{-1}(\tanh (a+b x))^3\right )}{192 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 169, normalized size = 1.01
method | result | size |
default | \(2 b^{4} \left (\frac {-\frac {5 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {73 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{384}+\left (\frac {55 \arctanh \left (\tanh \left (b x +a \right )\right )}{384}-\frac {55 b x}{384}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {5 a^{2}}{128}-\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{64}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{128}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{4} x^{4}}+\frac {5 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{128 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right )\) | \(169\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 167, normalized size = 1.00 \begin {gather*} \left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \end {gather*}
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 \begin {gather*} \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{5}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 108, normalized size = 0.65 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}\right )}}{a b^{4} x^{4}}\right )}}{384 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.85, size = 1069, normalized size = 6.40 \begin {gather*} \frac {5\,b^3\,\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}}}{32\,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 )}-\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 (\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\,x^4\,\left (4\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+8\,b\,x\right )}-\frac {59\,b^2\,\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 (\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 )}{48\,x^2\,\left (2\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+4\,b\,x\right )}+\frac {17\,b\,\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 (\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}{16\,x^3\,\left (3\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x\right )}+\frac {\sqrt {2}\,b^4\,\ln \left (\frac {\left (2\,\sqrt {2}\,a-\sqrt {2}\,\left (2\,a-\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 )+2\,b\,x\right )+\sqrt {2}\,b\,x+\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}}\,\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{}\mathrm {i}\right )\,\left ({\left (2\,a-\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 )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\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 )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\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 )+2\,b\,x\right )\right )\,1024{}\mathrm {i}}{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}}\right )\,5{}\mathrm {i}}{64\,{\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 )}^{3/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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