Optimal. Leaf size=212 \[ \frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
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Rubi [A] time = 0.15, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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Rule 2161
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {1}{x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {1}{6} b \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {1}{8} b^2 \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {1}{16} \left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (5 b^3\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 117, normalized size = 0.55 \[ \frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}+\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-26 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+33 b^2 x^2\right )}{24 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 145, normalized size = 0.68 \[ \left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{4} x^{3}}, -\frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{4} x^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 84, normalized size = 0.40 \[ -\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 200, normalized size = 0.94 \[ 2 b^{3} \left (\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b^{3} x^{3}}+\frac {\frac {10 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b^{2} x^{2}}+\frac {10 \left (\frac {6 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b x}-\frac {6 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right )}}{-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x}\right ) \]
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 \[ \int \frac {1}{x^{4} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.75, size = 1086, normalized size = 5.12 \[ \text {result too large to display} \]
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 \[ \int \frac {1}{x^{4} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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