Integrand size = 20, antiderivative size = 57 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {1}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\text {arctanh}(a x)^2}{4 a^3} \] Output:
-1/4/a^3/(-a^2*x^2+1)+1/2*x*arctanh(a*x)/a^2/(-a^2*x^2+1)-1/4*arctanh(a*x) ^2/a^3
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {1-2 a x \text {arctanh}(a x)+\left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}{4 a^3 \left (-1+a^2 x^2\right )} \] Input:
Integrate[(x^2*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^3*(-1 + a^2*x ^2))
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6560, 6510}
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.
\(\displaystyle \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6560 |
\(\displaystyle -\frac {\int \frac {\text {arctanh}(a x)}{1-a^2 x^2}dx}{2 a^2}+\frac {x \text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {1}{4 a^3 \left (1-a^2 x^2\right )}\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle -\frac {\text {arctanh}(a x)^2}{4 a^3}+\frac {x \text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {1}{4 a^3 \left (1-a^2 x^2\right )}\) |
Input:
Int[(x^2*ArcTanh[a*x])/(1 - a^2*x^2)^2,x]
Output:
-1/4*1/(a^3*(1 - a^2*x^2)) + (x*ArcTanh[a*x])/(2*a^2*(1 - a^2*x^2)) - ArcT anh[a*x]^2/(4*a^3)
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (-Si mp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])/(2*c^2*d*(q + 1))), x] + Sim p[1/(2*c^2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x ]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q , -5/2]
Time = 0.60 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\frac {a^{2} x^{2} \operatorname {arctanh}\left (a x \right )^{2}-a^{2} x^{2}+2 a x \,\operatorname {arctanh}\left (a x \right )-\operatorname {arctanh}\left (a x \right )^{2}}{4 \left (a^{2} x^{2}-1\right ) a^{3}}\) | \(55\) |
risch | \(-\frac {\ln \left (a x +1\right )^{2}}{16 a^{3}}+\frac {\left (x^{2} \ln \left (-a x +1\right ) a^{2}-2 a x -\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )}{8 a^{3} \left (a^{2} x^{2}-1\right )}-\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{2}-4 a x \ln \left (-a x +1\right )-\ln \left (-a x +1\right )^{2}-4}{16 a^{3} \left (a x -1\right ) \left (a x +1\right )}\) | \(124\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x +1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x -1\right )}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (a x +1\right )^{2}}{16}+\frac {\ln \left (a x -1\right )^{2}}{16}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x +1\right )}+\frac {1}{8 a x -8}}{a^{3}}\) | \(134\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x +1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4}-\frac {\operatorname {arctanh}\left (a x \right )}{4 \left (a x -1\right )}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{8}+\frac {\ln \left (a x +1\right )^{2}}{16}+\frac {\ln \left (a x -1\right )^{2}}{16}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{8}-\frac {1}{8 \left (a x +1\right )}+\frac {1}{8 a x -8}}{a^{3}}\) | \(134\) |
parts | \(-\frac {\operatorname {arctanh}\left (a x \right )}{4 a^{3} \left (a x +1\right )}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{4 a^{3}}-\frac {\operatorname {arctanh}\left (a x \right )}{4 a^{3} \left (a x -1\right )}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{4 a^{3}}-\frac {a \left (\frac {-\frac {\ln \left (a x +1\right )^{2}}{4}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{4}}-\frac {\frac {\ln \left (a x -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{4}}+\frac {-\frac {1}{2 a^{2} \left (a x -1\right )}+\frac {1}{2 a^{2} \left (a x +1\right )}}{a^{2}}\right )}{4}\) | \(187\) |
Input:
int(x^2*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^3
Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 4}{16 \, {\left (a^{5} x^{2} - a^{3}\right )}} \] Input:
integrate(x^2*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^5*x^2 - a^3)
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \] Input:
integrate(x**2*atanh(a*x)/(-a**2*x**2+1)**2,x)
Output:
Integral(x**2*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. 126 vs. \(2 (49) = 98\).
Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.21 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{a^{4} x^{2} - a^{2}} + \frac {\log \left (a x + 1\right )}{a^{3}} - \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right ) + \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 4\right )} a}{16 \, {\left (a^{6} x^{2} - a^{4}\right )}} \] Input:
integrate(x^2*arctanh(a*x)/(-a^2*x^2+1)^2,x, algorithm="maxima")
Output:
-1/4*(2*x/(a^4*x^2 - a^2) + log(a*x + 1)/a^3 - log(a*x - 1)/a^3)*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)*lo g(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 + 4)*a/(a^6*x^2 - a^4)
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate(x^2*arctanh(a*x)/(-a^2*x^2+1)^2,x, algorithm="giac")
Output:
integrate(x^2*arctanh(a*x)/(a^2*x^2 - 1)^2, x)
Time = 3.63 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.93 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{8\,a^3}+\frac {x}{2\,a^2\,\left (2\,a^2\,x^2-2\right )}\right )-\frac {{\ln \left (a\,x+1\right )}^2}{16\,a^3}-\frac {{\ln \left (1-a\,x\right )}^2}{16\,a^3}-\frac {1}{2\,a^2\,\left (2\,a-2\,a^3\,x^2\right )}-\frac {x\,\ln \left (a\,x+1\right )}{4\,a^3\,\left (a\,x^2-\frac {1}{a}\right )} \] Input:
int((x^2*atanh(a*x))/(a^2*x^2 - 1)^2,x)
Output:
log(1 - a*x)*(log(a*x + 1)/(8*a^3) + x/(2*a^2*(2*a^2*x^2 - 2))) - log(a*x + 1)^2/(16*a^3) - log(1 - a*x)^2/(16*a^3) - 1/(2*a^2*(2*a - 2*a^3*x^2)) - (x*log(a*x + 1))/(4*a^3*(a*x^2 - 1/a))
Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2} \, dx=\frac {-\mathit {atanh} \left (a x \right )^{2} a^{2} x^{2}+\mathit {atanh} \left (a x \right )^{2}-2 \mathit {atanh} \left (a x \right ) a x +a^{2} x^{2}}{4 a^{3} \left (a^{2} x^{2}-1\right )} \] Input:
int(x^2*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**3*(a**2*x**2 - 1))