Integrand size = 22, antiderivative size = 137 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {1}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \text {arctanh}(a x)}{a^2 \sqrt {1-a^2 x^2}}+\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{a^3}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3} \]
2*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^3+I*polylog(2,-I*(-a *x+1)^(1/2)/(a*x+1)^(1/2))/a^3-I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2)) /a^3-1/a^3/(-a^2*x^2+1)^(1/2)+x*arctanh(a*x)/a^2/(-a^2*x^2+1)^(1/2)
Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {i \left (\frac {i}{\sqrt {1-a^2 x^2}}-\frac {i a x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}+\text {arctanh}(a x) \log \left (1-i e^{-\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \log \left (1+i e^{-\text {arctanh}(a x)}\right )+\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{a^3} \]
(I*(I/Sqrt[1 - a^2*x^2] - (I*a*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] + ArcTanh [a*x]*Log[1 - I/E^ArcTanh[a*x]] - ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] + PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/a^3
Time = 0.39 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6560, 6512}
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.
\(\displaystyle \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6560 |
\(\displaystyle -\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{a^2}+\frac {x \text {arctanh}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {1}{a^3 \sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6512 |
\(\displaystyle -\frac {-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}}{a^2}+\frac {x \text {arctanh}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {1}{a^3 \sqrt {1-a^2 x^2}}\) |
-(1/(a^3*Sqrt[1 - a^2*x^2])) + (x*ArcTanh[a*x])/(a^2*Sqrt[1 - a^2*x^2]) - ((-2*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/a - (I*PolyLog[2, ( (-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + (I*PolyLog[2, (I*Sqrt[1 - a*x])/Sq rt[1 + a*x]])/a)/a^2
3.4.89.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol ] :> Simp[-2*(a + b*ArcTanh[c*x])*(ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]]/(c*S qrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/( c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/(c* Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]
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 = 0.24 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.39
method | result | size |
default | \(-\frac {\left (\operatorname {arctanh}\left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x -1\right )}-\frac {\left (1+\operatorname {arctanh}\left (a x \right )\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3} \left (a x +1\right )}+\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{a^{3}}-\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{a^{3}}+\frac {i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}-\frac {i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}\) | \(190\) |
-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(a*x-1)-1/2*(1+arctanh( a*x))*(-(a*x-1)*(a*x+1))^(1/2)/a^3/(a*x+1)+I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^( 1/2))*arctanh(a*x)/a^3-I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a ^3+I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3-I*dilog(1-I*(a*x+1)/(-a^2*x ^2+1)^(1/2))/a^3
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]