Integrand size = 21, antiderivative size = 65 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}-\frac {x}{2 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {\text {Chi}(\text {arctanh}(a x))}{2 a} \]
1/2*Chi(arctanh(a*x))/a-1/2/a/arctanh(a*x)^2/(-a^2*x^2+1)^(1/2)-1/2*x/arct anh(a*x)/(-a^2*x^2+1)^(1/2)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\frac {-\frac {1+a x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}+\text {Chi}(\text {arctanh}(a x))}{2 a} \]
(-((1 + a*x*ArcTanh[a*x])/(Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)) + CoshIntegr al[ArcTanh[a*x]])/(2*a)
Time = 0.53 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6528, 6568, 6530, 3042, 3782}
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.
\(\displaystyle \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx\) |
\(\Big \downarrow \) 6528 |
\(\displaystyle \frac {1}{2} a \int \frac {x}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}dx-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}\) |
\(\Big \downarrow \) 6568 |
\(\displaystyle \frac {1}{2} a \left (\frac {\int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}dx}{a}-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}\right )-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}\) |
\(\Big \downarrow \) 6530 |
\(\displaystyle \frac {1}{2} a \left (\frac {\int \frac {1}{\sqrt {1-a^2 x^2} \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}\right )-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}+\frac {1}{2} a \left (-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {\int \frac {\sin \left (i \text {arctanh}(a x)+\frac {\pi }{2}\right )}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}\right )\) |
\(\Big \downarrow \) 3782 |
\(\displaystyle \frac {1}{2} a \left (\frac {\text {Chi}(\text {arctanh}(a x))}{a^2}-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}\right )-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}\) |
-1/2*1/(a*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2) + (a*(-(x/(a*Sqrt[1 - a^2*x^2] *ArcTanh[a*x])) + CoshIntegral[ArcTanh[a*x]]/a^2))/2
3.5.24.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[CoshIntegral[c*f*(fz/d) + f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz }, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_ Symbol] :> Simp[(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1))), x] + Simp[2*c*((q + 1)/(b*(p + 1))) Int[x*(d + e*x^2)^q*(a + b*A rcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && LtQ[p, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x _Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && I LtQ[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*Arc Tanh[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[f*(m/(b*c*(p + 1))) Int[(f *x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d + e, 0] && EqQ[m + 2*q + 2, 0] && Lt Q[p, -1]
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32
method | result | size |
default | \(\frac {\operatorname {arctanh}\left (a x \right )^{2} \operatorname {Chi}\left (\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}+\operatorname {arctanh}\left (a x \right ) \sqrt {-a^{2} x^{2}+1}\, a x -\operatorname {Chi}\left (\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )^{2}+\sqrt {-a^{2} x^{2}+1}}{2 a \left (a^{2} x^{2}-1\right ) \operatorname {arctanh}\left (a x \right )^{2}}\) | \(86\) |
1/2/a*(arctanh(a*x)^2*Chi(arctanh(a*x))*a^2*x^2+arctanh(a*x)*(-a^2*x^2+1)^ (1/2)*a*x-Chi(arctanh(a*x))*arctanh(a*x)^2+(-a^2*x^2+1)^(1/2))/(a^2*x^2-1) /arctanh(a*x)^2
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )^{3}} \,d x } \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )^{3}} \,d x } \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]