Optimal. Leaf size=251 \[ \frac {b \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )}{1-a^2}-\frac {b \text {Li}_2\left (1-\frac {2 b x}{(1-a) (a+b x+1)}\right )}{1-a^2}-\frac {2 b \log \left (\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)}{1-a^2}+\frac {2 b \log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)}{1-a^2}+\frac {b \text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{2 (1-a)}-\frac {b \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )}{2 (a+1)}-\frac {\tanh ^{-1}(a+b x)^2}{x}+\frac {b \log \left (\frac {2}{-a-b x+1}\right ) \tanh ^{-1}(a+b x)}{1-a}+\frac {b \log \left (\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)}{a+1} \]
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Rubi [A] time = 0.70, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 15, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {6109, 371, 706, 31, 633, 6741, 6121, 6688, 12, 6725, 5920, 2402, 2315, 2447, 5918} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{1-a^2}-\frac {b \text {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (a+b x+1)}\right )}{1-a^2}+\frac {b \text {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{2 (1-a)}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )}{2 (a+1)}-\frac {2 b \log \left (\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)}{1-a^2}+\frac {2 b \log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \tanh ^{-1}(a+b x)}{1-a^2}-\frac {\tanh ^{-1}(a+b x)^2}{x}+\frac {b \log \left (\frac {2}{-a-b x+1}\right ) \tanh ^{-1}(a+b x)}{1-a}+\frac {b \log \left (\frac {2}{a+b x+1}\right ) \tanh ^{-1}(a+b x)}{a+1} \]
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 371
Rule 633
Rule 706
Rule 2315
Rule 2402
Rule 2447
Rule 5918
Rule 5920
Rule 6109
Rule 6121
Rule 6688
Rule 6725
Rule 6741
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a+b x)^2}{x^2} \, dx &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+(2 b) \int \frac {\tanh ^{-1}(a+b x)}{x \left (1-(a+b x)^2\right )} \, dx\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+(2 b) \int \frac {\tanh ^{-1}(a+b x)}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+2 \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+2 \operatorname {Subst}\left (\int \frac {b \tanh ^{-1}(x)}{(-a+x) \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+(2 b) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{(-a+x) \left (1-x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+(2 b) \operatorname {Subst}\left (\int \left (\frac {\tanh ^{-1}(x)}{\left (-1+a^2\right ) (a-x)}+\frac {\tanh ^{-1}(x)}{2 (-1+a) (-1+x)}-\frac {\tanh ^{-1}(x)}{2 (1+a) (1+x)}\right ) \, dx,x,a+b x\right )\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}-\frac {b \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{-1+x} \, dx,x,a+b x\right )}{1-a}-\frac {b \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{1+x} \, dx,x,a+b x\right )}{1+a}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{a-x} \, dx,x,a+b x\right )}{1-a^2}\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{1-a}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1+a}-\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1-a^2}+\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}-\frac {b \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,a+b x\right )}{1-a}-\frac {b \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,a+b x\right )}{1+a}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,a+b x\right )}{1-a^2}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 (a-x)}{(-1+a) (1+x)}\right )}{1-x^2} \, dx,x,a+b x\right )}{1-a^2}\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{1-a}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1+a}-\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1-a^2}+\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}-\frac {b \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}+\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a-b x}\right )}{1-a}-\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+a+b x}\right )}{1+a}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+a+b x}\right )}{1-a^2}\\ &=-\frac {\tanh ^{-1}(a+b x)^2}{x}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{1-a}+\frac {b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1+a}-\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )}{1-a^2}+\frac {2 b \tanh ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}+\frac {b \text {Li}_2\left (1-\frac {2}{1-a-b x}\right )}{2 (1-a)}-\frac {b \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{2 (1+a)}+\frac {b \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )}{1-a^2}-\frac {b \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )}{1-a^2}\\ \end {align*}
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Mathematica [C] time = 1.44, size = 208, normalized size = 0.83 \[ \frac {-\left (\left (a^3+a^2 b x+b x \left (\sqrt {1-a^2} e^{\tanh ^{-1}(a)}-1\right )-a\right ) \tanh ^{-1}(a+b x)^2\right )+a b x \text {Li}_2\left (e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )+a b x \tanh ^{-1}(a+b x) \left (-2 \log \left (1-e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )+2 \tanh ^{-1}(a)-i \pi \right )+a b x \left (i \pi \left (\log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )-\log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )\right )+2 \tanh ^{-1}(a) \left (\log \left (1-e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )-\log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )\right )\right )}{a \left (a^2-1\right ) x} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 342, normalized size = 1.36 \[ -\frac {\arctanh \left (b x +a \right )^{2}}{x}+\frac {2 b \arctanh \left (b x +a \right ) \ln \left (b x +a -1\right )}{2 a -2}-\frac {2 b \arctanh \left (b x +a \right ) \ln \left (b x +a +1\right )}{2+2 a}-\frac {2 b \arctanh \left (b x +a \right ) \ln \left (b x \right )}{\left (a -1\right ) \left (1+a \right )}-\frac {b \dilog \left (\frac {b x +a -1}{a -1}\right )}{\left (a -1\right ) \left (1+a \right )}-\frac {b \ln \left (b x \right ) \ln \left (\frac {b x +a -1}{a -1}\right )}{\left (a -1\right ) \left (1+a \right )}+\frac {b \dilog \left (\frac {b x +a +1}{1+a}\right )}{\left (a -1\right ) \left (1+a \right )}+\frac {b \ln \left (b x \right ) \ln \left (\frac {b x +a +1}{1+a}\right )}{\left (a -1\right ) \left (1+a \right )}+\frac {b \ln \left (b x +a +1\right )^{2}}{4+4 a}-\frac {b \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (b x +a +1\right )}{2 \left (1+a \right )}+\frac {b \ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{2+2 a}+\frac {b \dilog \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{2+2 a}+\frac {b \ln \left (b x +a -1\right )^{2}}{4 a -4}-\frac {b \dilog \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{2 \left (a -1\right )}-\frac {b \ln \left (b x +a -1\right ) \ln \left (\frac {1}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{2 \left (a -1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 244, normalized size = 0.97 \[ \frac {1}{4} \, b^{2} {\left (\frac {{\left (a - 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a - 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a + 1\right )} \log \left (b x + a - 1\right )^{2}}{a^{2} b - b} - \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{a^{2} b - b} + \frac {4 \, {\left (\log \left (\frac {b x}{a + 1} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a + 1}\right )\right )}}{a^{2} b - b} - \frac {4 \, {\left (\log \left (\frac {b x}{a - 1} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a - 1}\right )\right )}}{a^{2} b - b}\right )} - b {\left (\frac {\log \left (b x + a + 1\right )}{a + 1} - \frac {\log \left (b x + a - 1\right )}{a - 1} + \frac {2 \, \log \relax (x)}{a^{2} - 1}\right )} \operatorname {artanh}\left (b x + a\right ) - \frac {\operatorname {artanh}\left (b x + a\right )^{2}}{x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atanh}\left (a+b\,x\right )}^2}{x^2} \,d x \]
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 \[ \int \frac {\operatorname {atanh}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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