Optimal. Leaf size=157 \[ -\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac{5 b^2}{16 c^2 d^3 (c x+1)}+\frac{b^2}{16 c^2 d^3 (c x+1)^2}+\frac{5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.21377, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {37, 5938, 5926, 627, 44, 207, 5948} \[ -\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac{5 b^2}{16 c^2 d^3 (c x+1)}+\frac{b^2}{16 c^2 d^3 (c x+1)^2}+\frac{5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 37
Rule 5938
Rule 5926
Rule 627
Rule 44
Rule 207
Rule 5948
Rubi steps
\begin{align*} \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-(2 b c) \int \left (\frac{a+b \tanh ^{-1}(c x)}{4 c^2 d^3 (1+c x)^3}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{8 c^2 d^3 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{8 c^2 d^3 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c d^3}-\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c d^3}+\frac{(3 b) \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c d^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 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 c d^3}+\frac{\left (3 b^2\right ) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c d^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac{b^2 \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{4 c d^3}+\frac{\left (3 b^2\right ) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{4 c d^3}\\ &=\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\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 c d^3}+\frac{\left (3 b^2\right ) \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c d^3}\\ &=\frac{b^2}{16 c^2 d^3 (1+c x)^2}-\frac{5 b^2}{16 c^2 d^3 (1+c x)}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac{b^2 \int \frac{1}{-1+c^2 x^2} \, dx}{16 c d^3}-\frac{\left (3 b^2\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{8 c d^3}\\ &=\frac{b^2}{16 c^2 d^3 (1+c x)^2}-\frac{5 b^2}{16 c^2 d^3 (1+c x)}+\frac{5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3}+\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}\\ \end{align*}
Mathematica [A] time = 0.308307, size = 150, normalized size = 0.96 \[ \frac{-2 \left (16 a^2+12 a b+5 b^2\right ) (c x+1)+2 \left (8 a^2+4 a b+b^2\right )-b (12 a+5 b) (c x+1)^2 \log (1-c x)+b (12 a+5 b) (c x+1)^2 \log (c x+1)-8 b \tanh ^{-1}(c x) (a (8 c x+4)+b (3 c x+2))+4 b^2 \left (3 c^2 x^2-2 c x-1\right ) \tanh ^{-1}(c x)^2}{32 c^2 d^3 (c x+1)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.066, size = 460, normalized size = 2.9 \begin{align*}{\frac{{a}^{2}}{2\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{{a}^{2}}{{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{2\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{{c}^{2}{d}^{3} \left ( cx+1 \right ) }}-{\frac{3\,{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{8\,{c}^{2}{d}^{3}}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{4\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,{b}^{2}{\it Artanh} \left ( cx \right ) }{4\,{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{8\,{c}^{2}{d}^{3}}}-{\frac{3\,{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{32\,{c}^{2}{d}^{3}}}+{\frac{3\,{b}^{2}\ln \left ( cx-1 \right ) }{16\,{c}^{2}{d}^{3}}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{3\,{b}^{2}}{16\,{c}^{2}{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{3\,{b}^{2}\ln \left ( cx+1 \right ) }{16\,{c}^{2}{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{3\,{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{32\,{c}^{2}{d}^{3}}}-{\frac{5\,{b}^{2}\ln \left ( cx-1 \right ) }{32\,{c}^{2}{d}^{3}}}+{\frac{{b}^{2}}{16\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{5\,{b}^{2}}{16\,{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{5\,{b}^{2}\ln \left ( cx+1 \right ) }{32\,{c}^{2}{d}^{3}}}+{\frac{ab{\it Artanh} \left ( cx \right ) }{{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-2\,{\frac{ab{\it Artanh} \left ( cx \right ) }{{c}^{2}{d}^{3} \left ( cx+1 \right ) }}-{\frac{3\,ab\ln \left ( cx-1 \right ) }{8\,{c}^{2}{d}^{3}}}+{\frac{ab}{4\,{c}^{2}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{3\,ab}{4\,{c}^{2}{d}^{3} \left ( cx+1 \right ) }}+{\frac{3\,ab\ln \left ( cx+1 \right ) }{8\,{c}^{2}{d}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [B] time = 1.02855, size = 579, normalized size = 3.69 \begin{align*} -\frac{{\left (2 \, c x + 1\right )} b^{2} \operatorname{artanh}\left (c x\right )^{2}}{2 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} - \frac{1}{8} \,{\left (c{\left (\frac{2 \,{\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac{3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac{3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac{8 \,{\left (2 \, c x + 1\right )} \operatorname{artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} a b - \frac{1}{32} \,{\left (4 \, c{\left (\frac{2 \,{\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac{3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac{3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} \operatorname{artanh}\left (c x\right ) + \frac{{\left (3 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + 3 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 10 \, c x -{\left (5 \, c^{2} x^{2} + 10 \, c x + 6 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 5\right )} \log \left (c x + 1\right ) + 5 \,{\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 8\right )} c^{2}}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}}\right )} b^{2} - \frac{{\left (2 \, c x + 1\right )} a^{2}}{2 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.97724, size = 354, normalized size = 2.25 \begin{align*} -\frac{2 \,{\left (16 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} c x -{\left (3 \, b^{2} c^{2} x^{2} - 2 \, b^{2} c x - 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 (12 \, a b + 5 \, b^{2}\right )} c^{2} x^{2} - 2 \,{\left (4 \, a b + b^{2}\right )} c x - 4 \, a b - 3 \, b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{32 \,{\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} x}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac{b^{2} x \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 x \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.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2} x}{{\left (c d x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]