Integrand size = 22, antiderivative size = 146 \[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{a^3}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a^3}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a^3} \]
-arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^3-1/2*I*polylog(2,-I* (-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3+1/2*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1) ^(1/2))/a^3-1/2*(-a^2*x^2+1)^(1/2)/a^3-1/2*x*arctanh(a*x)*(-a^2*x^2+1)^(1/ 2)/a^2
Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}+a x \sqrt {1-a^2 x^2} \text {arctanh}(a x)+i \text {arctanh}(a x) \log \left (1-i e^{-\text {arctanh}(a x)}\right )-i \text {arctanh}(a x) \log \left (1+i e^{-\text {arctanh}(a x)}\right )+i \operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-i \operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )}{2 a^3} \]
-1/2*(Sqrt[1 - a^2*x^2] + a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + I*ArcTanh[a *x]*Log[1 - I/E^ArcTanh[a*x]] - I*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] + I*PolyLog[2, (-I)/E^ArcTanh[a*x]] - I*PolyLog[2, I/E^ArcTanh[a*x]])/a^3
Time = 0.39 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6578, 241, 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.
\(\displaystyle \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6578 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\) |
\(\Big \downarrow \) 6512 |
\(\displaystyle \frac {-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\) |
-1/2*Sqrt[1 - a^2*x^2]/a^3 - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*a^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)/(2*a^2)
3.4.67.3.1 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_.))/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_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-f)*(f*x)^(m - 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])^p/(c^2*d*m)), x] + (Simp[b*f*(p/(c*m)) Int[(f*x)^(m - 1 )*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Simp[f^2*((m - 1 )/(c^2*m)) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/Sqrt[d + e*x^2]), x] , x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05
method | result | size |
default | \(-\frac {\left (a x \,\operatorname {arctanh}\left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{3}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{2 a^{3}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{2 a^{3}}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3}}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3}}\) | \(154\) |
-1/2*(a*x*arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3-1/2*I*ln(1+I*(a*x+1 )/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3+1/2*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^( 1/2))*arctanh(a*x)/a^3-1/2*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/2 *I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3
\[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x^{2} \operatorname {atanh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{2} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x^2\,\mathrm {atanh}\left (a\,x\right )}{\sqrt {1-a^2\,x^2}} \,d x \]