Optimal. Leaf size=157 \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{3 b^2}{16 c d^3 (c x+1)}-\frac{b^2}{16 c d^3 (c x+1)^2}+\frac{3 b^2 \tanh ^{-1}(c x)}{16 c d^3} \]
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Rubi [A] time = 0.177749, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (c x+1)^2}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{3 b^2}{16 c d^3 (c x+1)}-\frac{b^2}{16 c d^3 (c x+1)^2}+\frac{3 b^2 \tanh ^{-1}(c x)}{16 c d^3} \]
Antiderivative was successfully verified.
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Rule 5928
Rule 5926
Rule 627
Rule 44
Rule 207
Rule 5948
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac{b \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 d^2 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{4 d^2 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{4 d^2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d}\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 d^3}-\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 d^3}+\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 d^3}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac{b^2 \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 d^3}+\frac{b^2 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 d^3}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac{b^2 \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{4 d^3}+\frac{b^2 \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{4 d^3}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac{b^2 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 d^3}+\frac{b^2 \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 d^3}\\ &=-\frac{b^2}{16 c d^3 (1+c x)^2}-\frac{3 b^2}{16 c d^3 (1+c x)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}-\frac{b^2 \int \frac{1}{-1+c^2 x^2} \, dx}{16 d^3}-\frac{b^2 \int \frac{1}{-1+c^2 x^2} \, dx}{8 d^3}\\ &=-\frac{b^2}{16 c d^3 (1+c x)^2}-\frac{3 b^2}{16 c d^3 (1+c x)}+\frac{3 b^2 \tanh ^{-1}(c x)}{16 c d^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}\\ \end{align*}
Mathematica [A] time = 0.136996, size = 183, normalized size = 1.17 \[ \frac{-8 a^2-4 a b-b^2}{16 c d^3 (c x+1)^2}+\frac{\left (-4 a b-3 b^2\right ) \log (1-c x)}{32 c d^3}+\frac{\left (4 a b+3 b^2\right ) \log (c x+1)}{32 c d^3}-\frac{b (4 a+3 b)}{16 c d^3 (c x+1)}-\frac{b \tanh ^{-1}(c x) (4 a+b c x+2 b)}{4 c d^3 (c x+1)^2}+\frac{b^2 \left (c^2 x^2+2 c x-3\right ) \tanh ^{-1}(c x)^2}{8 c d^3 (c x+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.062, size = 398, normalized size = 2.5 \begin{align*} -{\frac{{a}^{2}}{2\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{2\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{8\,c{d}^{3}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{4\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{4\,c{d}^{3} \left ( cx+1 \right ) }}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{8\,c{d}^{3}}}-{\frac{{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{32\,c{d}^{3}}}+{\frac{{b}^{2}\ln \left ( cx-1 \right ) }{16\,c{d}^{3}}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{b}^{2}}{16\,c{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{b}^{2}\ln \left ( cx+1 \right ) }{16\,c{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{32\,c{d}^{3}}}-{\frac{3\,{b}^{2}\ln \left ( cx-1 \right ) }{32\,c{d}^{3}}}-{\frac{{b}^{2}}{16\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,{b}^{2}}{16\,c{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,{b}^{2}\ln \left ( cx+1 \right ) }{32\,c{d}^{3}}}-{\frac{ab{\it Artanh} \left ( cx \right ) }{c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{ab\ln \left ( cx-1 \right ) }{8\,c{d}^{3}}}-{\frac{ab}{4\,c{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{ab}{4\,c{d}^{3} \left ( cx+1 \right ) }}+{\frac{ab\ln \left ( cx+1 \right ) }{8\,c{d}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.03299, size = 539, normalized size = 3.43 \begin{align*} -\frac{1}{8} \,{\left (c{\left (\frac{2 \,{\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac{\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac{\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} + \frac{8 \, \operatorname{artanh}\left (c x\right )}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}}\right )} a b - \frac{1}{32} \,{\left (4 \, c{\left (\frac{2 \,{\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac{\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac{\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} \operatorname{artanh}\left (c x\right ) + \frac{{\left ({\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right )^{2} +{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 6 \, c x -{\left (3 \, c^{2} x^{2} + 6 \, c x + 2 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 3\right )} \log \left (c x + 1\right ) + 3 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 8\right )} c^{2}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}}\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (c x\right )^{2}}{2 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} - \frac{a^{2}}{2 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.98649, size = 338, normalized size = 2.15 \begin{align*} -\frac{2 \,{\left (4 \, a b + 3 \, b^{2}\right )} c x -{\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x - 3 \, b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} + 16 \, a^{2} + 16 \, a b + 8 \, b^{2} -{\left ({\left (4 \, a b + 3 \, b^{2}\right )} c^{2} x^{2} + 2 \,{\left (4 \, a b + b^{2}\right )} c x - 12 \, a b - 5 \, b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{32 \,{\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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