Integrand size = 17, antiderivative size = 54 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {1}{4 a \left (1-a^2 x^2\right )}+\frac {x \text {arctanh}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^2}{4 a} \] Output:
-1/4/a/(-a^2*x^2+1)+x*arctanh(a*x)/(-2*a^2*x^2+2)+1/4*arctanh(a*x)^2/a
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {1-2 a x \text {arctanh}(a x)+\left (-1+a^2 x^2\right ) \text {arctanh}(a x)^2}{4 a \left (-1+a^2 x^2\right )} \] Input:
Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^2,x]
Output:
(1 - 2*a*x*ArcTanh[a*x] + (-1 + a^2*x^2)*ArcTanh[a*x]^2)/(4*a*(-1 + a^2*x^ 2))
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {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.
\(\displaystyle \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6518 |
\(\displaystyle -\frac {1}{2} a \int \frac {x}{\left (1-a^2 x^2\right )^2}dx+\frac {x \text {arctanh}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^2}{4 a}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {x \text {arctanh}(a x)}{2 \left (1-a^2 x^2\right )}-\frac {1}{4 a \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^2}{4 a}\) |
Input:
Int[ArcTanh[a*x]/(1 - a^2*x^2)^2,x]
Output:
-1/4*1/(a*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(2*(1 - a^2*x^2)) + ArcTanh[a* x]^2/(4*a)
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]
Time = 0.57 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(-\frac {-a^{2} x^{2} \operatorname {arctanh}\left (a x \right )^{2}-a^{2} x^{2}+2 a x \,\operatorname {arctanh}\left (a x \right )+\operatorname {arctanh}\left (a x \right )^{2}}{4 \left (a^{2} x^{2}-1\right ) a}\) | \(54\) |
risch | \(\frac {\ln \left (a x +1\right )^{2}}{16 a}-\frac {\left (x^{2} \ln \left (-a x +1\right ) a^{2}+2 a x -\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )}{8 \left (a^{2} x^{2}-1\right ) a}+\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{2}+4 a x \ln \left (-a x +1\right )-\ln \left (-a x +1\right )^{2}+4}{16 a \left (a x -1\right ) \left (a x +1\right )}\) | \(124\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x +1\right )}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x -1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{16}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (a x +1\right )^{2}}{16}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x +1\right )}+\frac {1}{8 a x -8}}{a}\) | \(134\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x +1\right )}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x -1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{16}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {\ln \left (a x +1\right )^{2}}{16}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x +1\right )}+\frac {1}{8 a x -8}}{a}\) | \(134\) |
parts | \(-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x +1\right ) a}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4 a}-\frac {\operatorname {arctanh}\left (a x \right )}{4 a \left (a x -1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4 a}-\frac {a \left (\frac {\frac {\ln \left (a x -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{2}}-\frac {-\frac {\ln \left (a x +1\right )^{2}}{4}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{2}}-\frac {1}{2 a^{2} \left (a x -1\right )}+\frac {1}{2 a^{2} \left (a x +1\right )}\right )}{4}\) | \(181\) |
Input:
int(arctanh(a*x)/(-a^2*x^2+1)^2,x,method=_RETURNVERBOSE)
Output:
-1/4*(-a^2*x^2*arctanh(a*x)^2-a^2*x^2+2*a*x*arctanh(a*x)+arctanh(a*x)^2)/( a^2*x^2-1)/a
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{3} x^{2} - a\right )}} \] Input:
integrate(arctanh(a*x)/(-a^2*x^2+1)^2,x, algorithm="fricas")
Output:
-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^3*x^2 - a)
\[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \] Input:
integrate(atanh(a*x)/(-a**2*x**2+1)**2,x)
Output:
Integral(atanh(a*x)/((a*x - 1)**2*(a*x + 1)**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (46) = 92\).
Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.26 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right ) - \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} \] Input:
integrate(arctanh(a*x)/(-a^2*x^2+1)^2,x, algorithm="maxima")
Output:
-1/4*(2*x/(a^2*x^2 - 1) - log(a*x + 1)/a + log(a*x - 1)/a)*arctanh(a*x) - 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/(a^4*x^2 - a^2)
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (46) = 92\).
Time = 1.11 (sec) , antiderivative size = 255, normalized size of antiderivative = 4.72 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {1}{8} \, a^{2} {\left (\frac {{\left (a x - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} - 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{a - \frac {a {\left (\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1\right )}}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}} + 1}\right )}{{\left (a x + 1\right )} a^{4}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{4}}\right )} \] Input:
integrate(arctanh(a*x)/(-a^2*x^2+1)^2,x, algorithm="giac")
Output:
1/8*a^2*((a*x - 1)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/(a - a*(a*((a*x + 1)/ (a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1)) - 1)/(a*((a*x + 1)/(a*x - 1) + 1)/(a - a*(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1)) + 1))/((a*x + 1)*a^ 4) + (a*x - 1)/((a*x + 1)*a^4))
Time = 3.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.96 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {{\ln \left (a\,x+1\right )}^2}{16\,a}-\ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{8\,a}-\frac {x}{2\,\left (2\,a^2\,x^2-2\right )}\right )+\frac {{\ln \left (1-a\,x\right )}^2}{16\,a}+\frac {1}{2\,a\,\left (2\,a^2\,x^2-2\right )}-\frac {x\,\ln \left (a\,x+1\right )}{4\,a\,\left (a\,x^2-\frac {1}{a}\right )} \] Input:
int(atanh(a*x)/(a^2*x^2 - 1)^2,x)
Output:
log(a*x + 1)^2/(16*a) - log(1 - a*x)*(log(a*x + 1)/(8*a) - x/(2*(2*a^2*x^2 - 2))) + log(1 - a*x)^2/(16*a) + 1/(2*a*(2*a^2*x^2 - 2)) - (x*log(a*x + 1 ))/(4*a*(a*x^2 - 1/a))
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {\mathit {atanh} \left (a x \right )^{2} a^{2} x^{2}-\mathit {atanh} \left (a x \right )^{2}-2 \mathit {atanh} \left (a x \right ) a x +a^{2} x^{2}}{4 a \left (a^{2} x^{2}-1\right )} \] Input:
int(atanh(a*x)/(-a^2*x^2+1)^2,x)
Output:
(atanh(a*x)**2*a**2*x**2 - atanh(a*x)**2 - 2*atanh(a*x)*a*x + a**2*x**2)/( 4*a*(a**2*x**2 - 1))