Integrand size = 22, antiderivative size = 194 \[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^3}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a^2}+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {\arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{4 a^3}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3} \]
-1/12*(-a^2*x^2+1)^(3/2)/a^3-1/4*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arct anh(a*x)/a^3-1/8*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3+1/8*I*po lylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3+1/8*(-a^2*x^2+1)^(1/2)/a^3-1/8 *x*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^2+1/4*x^3*arctanh(a*x)*(-a^2*x^2+1)^( 1/2)
Time = 0.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.82 \[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\frac {\sqrt {1-a^2 x^2} \left (1+2 a^2 x^2+3 a x \text {arctanh}(a x)+6 a x \left (-1+a^2 x^2\right ) \text {arctanh}(a x)-\frac {3 i \text {arctanh}(a x) \left (\log \left (1-i e^{-\text {arctanh}(a x)}\right )-\log \left (1+i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-\frac {3 i \left (\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{24 a^3} \]
(Sqrt[1 - a^2*x^2]*(1 + 2*a^2*x^2 + 3*a*x*ArcTanh[a*x] + 6*a*x*(-1 + a^2*x ^2)*ArcTanh[a*x] - ((3*I)*ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2] - ((3*I)*(PolyLog[2, (-I)/E^ArcTan h[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(24*a^3)
Time = 0.60 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6572, 243, 53, 2009, 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.
\(\displaystyle \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6572 |
\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{4} a \int \frac {x^3}{\sqrt {1-a^2 x^2}}dx+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{8} a \int \frac {x^2}{\sqrt {1-a^2 x^2}}dx^2+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{8} a \int \left (\frac {1}{a^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a^2}\right )dx^2+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\) |
\(\Big \downarrow \) 6578 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}\right )+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )+\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )\) |
\(\Big \downarrow \) 6512 |
\(\displaystyle \frac {1}{4} x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-\frac {1}{8} a \left (\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}\right )+\frac {1}{4} \left (\frac {-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )\) |
-1/8*(a*((-2*Sqrt[1 - a^2*x^2])/a^4 + (2*(1 - a^2*x^2)^(3/2))/(3*a^4))) + (x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/4 + (-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])/Sqrt[1 + a*x]])/a)/(2*a^2))/4
3.5.28.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.) *(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c *x])/(f*(m + 2))), x] + (Simp[d/(m + 2) Int[(f*x)^m*((a + b*ArcTanh[c*x]) /Sqrt[d + e*x^2]), x], x] - Simp[b*c*(d/(f*(m + 2))) Int[(f*x)^(m + 1)/Sq rt[d + e*x^2], x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]
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 = 0.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (6 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )+2 a^{2} x^{2}-3 a x \,\operatorname {arctanh}\left (a x \right )+1\right )}{24 a^{3}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{8 a^{3}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \operatorname {arctanh}\left (a x \right )}{8 a^{3}}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3}}\) | \(175\) |
1/24/a^3*(-(a*x-1)*(a*x+1))^(1/2)*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2-3*a*x* arctanh(a*x)+1)-1/8*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3+ 1/8*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3-1/8*I*dilog(1+I* (a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/8*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2) )/a^3
\[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\int { \sqrt {-a^{2} x^{2} + 1} x^{2} \operatorname {artanh}\left (a x\right ) \,d x } \]
\[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\int x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}{\left (a x \right )}\, dx \]
\[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\int { \sqrt {-a^{2} x^{2} + 1} x^{2} \operatorname {artanh}\left (a x\right ) \,d x } \]
\[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\int { \sqrt {-a^{2} x^{2} + 1} x^{2} \operatorname {artanh}\left (a x\right ) \,d x } \]
Timed out. \[ \int x^2 \sqrt {1-a^2 x^2} \text {arctanh}(a x) \, dx=\int x^2\,\mathrm {atanh}\left (a\,x\right )\,\sqrt {1-a^2\,x^2} \,d x \]