Integrand size = 22, antiderivative size = 128 \[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\frac {6 \arctan \left (e^{\text {arctanh}(a x)}\right ) \text {arctanh}(a x)^2}{a^2}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}-\frac {6 i \text {arctanh}(a x) \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )}{a^2}+\frac {6 i \text {arctanh}(a x) \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )}{a^2}+\frac {6 i \operatorname {PolyLog}\left (3,-i e^{\text {arctanh}(a x)}\right )}{a^2}-\frac {6 i \operatorname {PolyLog}\left (3,i e^{\text {arctanh}(a x)}\right )}{a^2} \]
6*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a^2-6*I*arctanh(a*x)*p olylog(2,-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^2+6*I*arctanh(a*x)*polylog(2,I*( a*x+1)/(-a^2*x^2+1)^(1/2))/a^2+6*I*polylog(3,-I*(a*x+1)/(-a^2*x^2+1)^(1/2) )/a^2-6*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^2-arctanh(a*x)^3*(-a^2 *x^2+1)^(1/2)/a^2
Time = 0.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.23 \[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3+3 i \text {arctanh}(a x)^2 \log \left (1-i e^{-\text {arctanh}(a x)}\right )-3 i \text {arctanh}(a x)^2 \log \left (1+i e^{-\text {arctanh}(a x)}\right )+6 i \text {arctanh}(a x) \operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-6 i \text {arctanh}(a x) \operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )+6 i \operatorname {PolyLog}\left (3,-i e^{-\text {arctanh}(a x)}\right )-6 i \operatorname {PolyLog}\left (3,i e^{-\text {arctanh}(a x)}\right )}{a^2} \]
-((Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3 + (3*I)*ArcTanh[a*x]^2*Log[1 - I/E^Arc Tanh[a*x]] - (3*I)*ArcTanh[a*x]^2*Log[1 + I/E^ArcTanh[a*x]] + (6*I)*ArcTan h[a*x]*PolyLog[2, (-I)/E^ArcTanh[a*x]] - (6*I)*ArcTanh[a*x]*PolyLog[2, I/E ^ArcTanh[a*x]] + (6*I)*PolyLog[3, (-I)/E^ArcTanh[a*x]] - (6*I)*PolyLog[3, I/E^ArcTanh[a*x]])/a^2)
Time = 0.63 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6556, 6514, 3042, 4668, 3011, 2720, 7143}
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.
\(\displaystyle \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6556 |
\(\displaystyle \frac {3 \int \frac {\text {arctanh}(a x)^2}{\sqrt {1-a^2 x^2}}dx}{a}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}\) |
\(\Big \downarrow \) 6514 |
\(\displaystyle \frac {3 \int \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2d\text {arctanh}(a x)}{a^2}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}+\frac {3 \int \text {arctanh}(a x)^2 \csc \left (i \text {arctanh}(a x)+\frac {\pi }{2}\right )d\text {arctanh}(a x)}{a^2}\) |
\(\Big \downarrow \) 4668 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}+\frac {3 \left (-2 i \int \text {arctanh}(a x) \log \left (1-i e^{\text {arctanh}(a x)}\right )d\text {arctanh}(a x)+2 i \int \text {arctanh}(a x) \log \left (1+i e^{\text {arctanh}(a x)}\right )d\text {arctanh}(a x)+2 \text {arctanh}(a x)^2 \arctan \left (e^{\text {arctanh}(a x)}\right )\right )}{a^2}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}+\frac {3 \left (2 i \left (\int \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )d\text {arctanh}(a x)-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )\right )-2 i \left (\int \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )d\text {arctanh}(a x)-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )\right )+2 \text {arctanh}(a x)^2 \arctan \left (e^{\text {arctanh}(a x)}\right )\right )}{a^2}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}+\frac {3 \left (2 i \left (\int e^{-\text {arctanh}(a x)} \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )de^{\text {arctanh}(a x)}-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )\right )-2 i \left (\int e^{-\text {arctanh}(a x)} \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )de^{\text {arctanh}(a x)}-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )\right )+2 \text {arctanh}(a x)^2 \arctan \left (e^{\text {arctanh}(a x)}\right )\right )}{a^2}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^3}{a^2}+\frac {3 \left (2 \text {arctanh}(a x)^2 \arctan \left (e^{\text {arctanh}(a x)}\right )+2 i \left (\operatorname {PolyLog}\left (3,-i e^{\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,-i e^{\text {arctanh}(a x)}\right )\right )-2 i \left (\operatorname {PolyLog}\left (3,i e^{\text {arctanh}(a x)}\right )-\text {arctanh}(a x) \operatorname {PolyLog}\left (2,i e^{\text {arctanh}(a x)}\right )\right )\right )}{a^2}\) |
-((Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3)/a^2) + (3*(2*ArcTan[E^ArcTanh[a*x]]*A rcTanh[a*x]^2 + (2*I)*(-(ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]]) + P olyLog[3, (-I)*E^ArcTanh[a*x]]) - (2*I)*(-(ArcTanh[a*x]*PolyLog[2, I*E^Arc Tanh[a*x]]) + PolyLog[3, I*E^ArcTanh[a*x]])))/a^2
3.4.82.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ ))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[ 1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I)) Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c , d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[1/(c*Sqrt[d]) Subst[Int[(a + b*x)^p*Sech[x], x], x, ArcTa nh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0 ] && GtQ[d, 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]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {x \operatorname {arctanh}\left (a x \right )^{3}}{\sqrt {-a^{2} x^{2}+1}}d x\]
\[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x \operatorname {atanh}^{3}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
\[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x\,{\mathrm {atanh}\left (a\,x\right )}^3}{\sqrt {1-a^2\,x^2}} \,d x \]