Integrand size = 19, antiderivative size = 154 \[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}-\frac {x}{10 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {1+a^2 x^2}{30 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {1+a^2 x^2}{15 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)}+\frac {2 \text {Shi}(2 \text {arctanh}(a x))}{15 a} \]
-1/5/a/(-a^2*x^2+1)/arctanh(a*x)^5-1/10*x/(-a^2*x^2+1)/arctanh(a*x)^4+1/30 *(-a^2*x^2-1)/a/(-a^2*x^2+1)/arctanh(a*x)^3-1/15*x/(-a^2*x^2+1)/arctanh(a* x)^2+1/15*(-a^2*x^2-1)/a/(-a^2*x^2+1)/arctanh(a*x)+2/15*Shi(2*arctanh(a*x) )/a
Time = 0.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\frac {6+3 a x \text {arctanh}(a x)+\left (1+a^2 x^2\right ) \text {arctanh}(a x)^2+2 a x \text {arctanh}(a x)^3+2 \left (1+a^2 x^2\right ) \text {arctanh}(a x)^4+4 \left (-1+a^2 x^2\right ) \text {arctanh}(a x)^5 \text {Shi}(2 \text {arctanh}(a x))}{30 a \left (-1+a^2 x^2\right ) \text {arctanh}(a x)^5} \]
(6 + 3*a*x*ArcTanh[a*x] + (1 + a^2*x^2)*ArcTanh[a*x]^2 + 2*a*x*ArcTanh[a*x ]^3 + 2*(1 + a^2*x^2)*ArcTanh[a*x]^4 + 4*(-1 + a^2*x^2)*ArcTanh[a*x]^5*Sin hIntegral[2*ArcTanh[a*x]])/(30*a*(-1 + a^2*x^2)*ArcTanh[a*x]^5)
Time = 0.89 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6528, 6558, 6558, 6596, 5971, 27, 3042, 26, 3779}
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.
| \(\displaystyle \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx\) |
| \(\Big \downarrow \) 6528 |
| \(\displaystyle \frac {2}{5} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^5}dx-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 6558 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \int \frac {x}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^3}dx-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 6558 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \left (2 \int \frac {x}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}dx-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 6596 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \left (\frac {2 \int \frac {a x}{\left (1-a^2 x^2\right ) \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \left (\frac {2 \int \frac {\sinh (2 \text {arctanh}(a x))}{2 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \left (\frac {\int \frac {\sinh (2 \text {arctanh}(a x))}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}+\frac {2}{5} a \left (\frac {1}{3} \left (\frac {\int -\frac {i \sin (2 i \text {arctanh}(a x))}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}+\frac {2}{5} a \left (\frac {1}{3} \left (-\frac {i \int \frac {\sin (2 i \text {arctanh}(a x))}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )\) |
| \(\Big \downarrow \) 3779 |
| \(\displaystyle \frac {2}{5} a \left (\frac {1}{3} \left (\frac {\text {Shi}(2 \text {arctanh}(a x))}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^2}-\frac {a^2 x^2+1}{2 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)}\right )-\frac {x}{4 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^4}-\frac {a^2 x^2+1}{12 a^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x)^3}\right )-\frac {1}{5 a \left (1-a^2 x^2\right ) \text {arctanh}(a x)^5}\) |
-1/5*1/(a*(1 - a^2*x^2)*ArcTanh[a*x]^5) + (2*a*(-1/4*x/(a*(1 - a^2*x^2)*Ar cTanh[a*x]^4) - (1 + a^2*x^2)/(12*a^2*(1 - a^2*x^2)*ArcTanh[a*x]^3) + (-1/ 2*x/(a*(1 - a^2*x^2)*ArcTanh[a*x]^2) - (1 + a^2*x^2)/(2*a^2*(1 - a^2*x^2)* ArcTanh[a*x]) + SinhIntegral[2*ArcTanh[a*x]]/a^2)/3))/5
3.3.99.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[I*(SinhIntegral[c*f*(fz/d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f , fz}, x] && EqQ[d*e - c*f*fz*I, 0]
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 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_)*(x_))/((d_) + (e_.)*(x_)^2)^2 , x_Symbol] :> Simp[x*((a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x ^2))), x] + (Simp[(1 + c^2*x^2)*((a + b*ArcTanh[c*x])^(p + 2)/(b^2*e*(p + 1 )*(p + 2)*(d + e*x^2))), x] + Simp[4/(b^2*(p + 1)*(p + 2)) Int[x*((a + b* ArcTanh[c*x])^(p + 2)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[p, -1] && NeQ[p, -2]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) ^2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1) Subst[Int[(a + b*x)^p*(Sinh[x]^ m/Cosh[x]^(m + 2*(q + 1))), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (In tegerQ[q] || GtQ[d, 0])
Time = 0.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{10 \operatorname {arctanh}\left (a x \right )^{5}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{10 \operatorname {arctanh}\left (a x \right )^{5}}-\frac {\sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{20 \operatorname {arctanh}\left (a x \right )^{4}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{30 \operatorname {arctanh}\left (a x \right )^{3}}-\frac {\sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{30 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{15 \,\operatorname {arctanh}\left (a x \right )}+\frac {2 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{15}}{a}\) | \(98\) |
| default | \(\frac {-\frac {1}{10 \operatorname {arctanh}\left (a x \right )^{5}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{10 \operatorname {arctanh}\left (a x \right )^{5}}-\frac {\sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{20 \operatorname {arctanh}\left (a x \right )^{4}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{30 \operatorname {arctanh}\left (a x \right )^{3}}-\frac {\sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{30 \operatorname {arctanh}\left (a x \right )^{2}}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{15 \,\operatorname {arctanh}\left (a x \right )}+\frac {2 \,\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{15}}{a}\) | \(98\) |
1/a*(-1/10/arctanh(a*x)^5-1/10/arctanh(a*x)^5*cosh(2*arctanh(a*x))-1/20/ar ctanh(a*x)^4*sinh(2*arctanh(a*x))-1/30/arctanh(a*x)^3*cosh(2*arctanh(a*x)) -1/30*sinh(2*arctanh(a*x))/arctanh(a*x)^2-1/15/arctanh(a*x)*cosh(2*arctanh (a*x))+2/15*Shi(2*arctanh(a*x)))
Time = 0.25 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\frac {{\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} + 4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 96}{15 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5}} \]
1/15*(((a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) - (a^2*x^2 - 1)*lo g_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1))^5 + 4*a*x*log( -(a*x + 1)/(a*x - 1))^3 + 2*(a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^4 + 24 *a*x*log(-(a*x + 1)/(a*x - 1)) + 4*(a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)) ^2 + 96)/((a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1))^5)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\int \frac {1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{6}{\left (a x \right )}}\, dx \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\int { \frac {1}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{6}} \,d x } \]
-8*a*integrate(-1/15*x/((a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1) - (a^4*x^4 - 2*a^2*x^2 + 1)*log(-a*x + 1)), x) + 2/15*(2*a*x*log(a*x + 1)^3 + (a^2*x^ 2 + 1)*log(a*x + 1)^4 + (a^2*x^2 + 1)*log(-a*x + 1)^4 - 2*(a*x + 2*(a^2*x^ 2 + 1)*log(a*x + 1))*log(-a*x + 1)^3 + 12*a*x*log(a*x + 1) + 2*(a^2*x^2 + 1)*log(a*x + 1)^2 + 2*(a^2*x^2 + 3*a*x*log(a*x + 1) + 3*(a^2*x^2 + 1)*log( a*x + 1)^2 + 1)*log(-a*x + 1)^2 - 2*(3*a*x*log(a*x + 1)^2 + 2*(a^2*x^2 + 1 )*log(a*x + 1)^3 + 6*a*x + 2*(a^2*x^2 + 1)*log(a*x + 1))*log(-a*x + 1) + 4 8)/((a^3*x^2 - a)*log(a*x + 1)^5 - 5*(a^3*x^2 - a)*log(a*x + 1)^4*log(-a*x + 1) + 10*(a^3*x^2 - a)*log(a*x + 1)^3*log(-a*x + 1)^2 - 10*(a^3*x^2 - a) *log(a*x + 1)^2*log(-a*x + 1)^3 + 5*(a^3*x^2 - a)*log(a*x + 1)*log(-a*x + 1)^4 - (a^3*x^2 - a)*log(-a*x + 1)^5)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\int { \frac {1}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{6}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)^6} \, dx=\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^6\,{\left (a^2\,x^2-1\right )}^2} \,d x \]