Integrand size = 20, antiderivative size = 82 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\frac {1}{4 a^2 \left (1-a^2 x^2\right )}-\frac {x \text {arctanh}(a x)}{2 a \left (1-a^2 x^2\right )}-\frac {\text {arctanh}(a x)^2}{4 a^2}+\frac {\text {arctanh}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )} \]
1/4/a^2/(-a^2*x^2+1)-1/2*x*arctanh(a*x)/a/(-a^2*x^2+1)-1/4*arctanh(a*x)^2/ a^2+1/2*arctanh(a*x)^2/a^2/(-a^2*x^2+1)
Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.52 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\frac {1-2 a x \text {arctanh}(a x)+\left (1+a^2 x^2\right ) \text {arctanh}(a x)^2}{4 a^2-4 a^4 x^2} \]
Time = 0.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6556, 6518, 241}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6556 |
\(\displaystyle \frac {\text {arctanh}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2}dx}{a}\) |
\(\Big \downarrow \) 6518 |
\(\displaystyle \frac {\text {arctanh}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac {-\frac {1}{2} a \int \frac {x}{\left (1-a^2 x^2\right )^2}dx+\frac {x \text {arctanh}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^2}{4 a}}{a}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {\text {arctanh}(a x)^2}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\frac {x \text {arctanh}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{4 a \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^2}{4 a}}{a}\) |
ArcTanh[a*x]^2/(2*a^2*(1 - a^2*x^2)) - (-1/4*1/(a*(1 - a^2*x^2)) + (x*ArcT anh[a*x])/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^2/(4*a))/a
3.3.68.3.1 Defintions of rubi rules used
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Sy mbol] :> Simp[x*((a + b*ArcTanh[c*x])^p/(2*d*(d + e*x^2))), x] + (Simp[(a + b*ArcTanh[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x] - Simp[b*c*(p/2) Int[x*( (a + b*ArcTanh[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q _.), x_Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p/(2*e*(q + 1))), x] + Simp[b*(p/(2*c*(q + 1))) Int[(d + e*x^2)^q*(a + b*ArcTanh[c* x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && NeQ[q, -1]
Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(-\frac {a^{2} x^{2} \operatorname {arctanh}\left (a x \right )^{2}+a^{2} x^{2}-2 a x \,\operatorname {arctanh}\left (a x \right )+\operatorname {arctanh}\left (a x \right )^{2}}{4 \left (a^{2} x^{2}-1\right ) a^{2}}\) | \(52\) |
risch | \(-\frac {\left (a^{2} x^{2}+1\right ) \ln \left (a x +1\right )^{2}}{16 a^{2} \left (a^{2} x^{2}-1\right )}+\frac {\left (x^{2} \ln \left (-a x +1\right ) a^{2}+2 a x +\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )}{8 a^{2} \left (a x -1\right ) \left (a x +1\right )}-\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{2}+4 a x \ln \left (-a x +1\right )+\ln \left (-a x +1\right )^{2}+4}{16 a^{2} \left (a x -1\right ) \left (a x +1\right )}\) | \(143\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{2 \left (a^{2} x^{2}-1\right )}+\frac {\operatorname {arctanh}\left (a x \right )}{4 a x -4}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}+\frac {\operatorname {arctanh}\left (a x \right )}{4 a x +4}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}+\frac {\ln \left (a x -1\right )^{2}}{16}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (a x +1\right )^{2}}{16}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x -1\right )}+\frac {1}{8 a x +8}}{a^{2}}\) | \(153\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{2 \left (a^{2} x^{2}-1\right )}+\frac {\operatorname {arctanh}\left (a x \right )}{4 a x -4}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}+\frac {\operatorname {arctanh}\left (a x \right )}{4 a x +4}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}+\frac {\ln \left (a x -1\right )^{2}}{16}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (a x +1\right )^{2}}{16}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x -1\right )}+\frac {1}{8 a x +8}}{a^{2}}\) | \(153\) |
parts | \(-\frac {\operatorname {arctanh}\left (a x \right )^{2}}{2 a^{2} \left (a^{2} x^{2}-1\right )}+\frac {\frac {\operatorname {arctanh}\left (a x \right )}{4 a x -4}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}+\frac {\operatorname {arctanh}\left (a x \right )}{4 a x +4}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (a x +1\right )^{2}}{16}+\frac {\ln \left (a x -1\right )^{2}}{16}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x -1\right )}+\frac {1}{8 a x +8}}{a^{2}}\) | \(157\) |
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]
1/16*(4*a*x*log(-(a*x + 1)/(a*x - 1)) - (a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 4)/(a^4*x^2 - a^2)
\[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\int \frac {x \operatorname {atanh}^{2}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (71) = 142\).
Time = 0.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.78 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\frac {{\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )}{4 \, a} + \frac {{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} - \frac {\operatorname {artanh}\left (a x\right )^{2}}{2 \, {\left (a^{2} x^{2} - 1\right )} a^{2}} \]
1/4*(2*x/(a^2*x^2 - 1) - log(a*x + 1)/a + log(a*x - 1)/a)*arctanh(a*x)/a + 1/16*((a^2*x^2 - 1)*log(a*x + 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 - 4)/(a^4*x^2 - a^2) - 1/2*arctanh(a* x)^2/((a^2*x^2 - 1)*a^2)
Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.71 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {1}{32} \, {\left ({\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 2 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} - \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {2 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}} + \frac {2 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{3}}\right )} a \]
-1/32*(((a*x + 1)/((a*x - 1)*a^3) + (a*x - 1)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1))^2 - 2*((a*x + 1)/((a*x - 1)*a^3) - (a*x - 1)/((a*x + 1)*a^3 ))*log(-(a*x + 1)/(a*x - 1)) + 2*(a*x + 1)/((a*x - 1)*a^3) + 2*(a*x - 1)/( (a*x + 1)*a^3))*a
Time = 3.73 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.41 \[ \int \frac {x \text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx=\ln \left (1-a\,x\right )\,\left (\frac {\frac {x}{2}-\frac {1}{2\,a}}{4\,a-4\,a^3\,x^2}+\frac {\frac {x}{2}+\frac {1}{2\,a}}{4\,a-4\,a^3\,x^2}+\ln \left (a\,x+1\right )\,\left (\frac {1}{8\,a^2}+\frac {1}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}\right )\right )-{\ln \left (1-a\,x\right )}^2\,\left (\frac {1}{16\,a^2}+\frac {1}{2\,a^2\,\left (4\,a^2\,x^2-4\right )}\right )-\frac {1}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}-{\ln \left (a\,x+1\right )}^2\,\left (\frac {1}{8\,a^3\,\left (a\,x^2-\frac {1}{a}\right )}+\frac {1}{16\,a^2}\right )+\frac {x\,\ln \left (a\,x+1\right )}{4\,a^2\,\left (a\,x^2-\frac {1}{a}\right )} \]
log(1 - a*x)*((x/2 - 1/(2*a))/(4*a - 4*a^3*x^2) + (x/2 + 1/(2*a))/(4*a - 4 *a^3*x^2) + log(a*x + 1)*(1/(8*a^2) + 1/(2*a^2*(2*a^2*x^2 - 2)))) - log(1 - a*x)^2*(1/(16*a^2) + 1/(2*a^2*(4*a^2*x^2 - 4))) - 1/(2*a^2*(2*a^2*x^2 - 2)) - log(a*x + 1)^2*(1/(8*a^3*(a*x^2 - 1/a)) + 1/(16*a^2)) + (x*log(a*x + 1))/(4*a^2*(a*x^2 - 1/a))