Integrand size = 21, antiderivative size = 66 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}+\frac {5 \text {Shi}(\text {arctanh}(a x))}{8 a}+\frac {15 \text {Shi}(3 \text {arctanh}(a x))}{16 a}+\frac {5 \text {Shi}(5 \text {arctanh}(a x))}{16 a} \]
-1/a/(-a^2*x^2+1)^(5/2)/arctanh(a*x)+5/8*Shi(arctanh(a*x))/a+15/16*Shi(3*a rctanh(a*x))/a+5/16*Shi(5*arctanh(a*x))/a
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\frac {-\frac {16}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}+5 (2 \text {Shi}(\text {arctanh}(a x))+3 \text {Shi}(3 \text {arctanh}(a x))+\text {Shi}(5 \text {arctanh}(a x)))}{16 a} \]
(-16/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + 5*(2*SinhIntegral[ArcTanh[a*x]] + 3*SinhIntegral[3*ArcTanh[a*x]] + SinhIntegral[5*ArcTanh[a*x]]))/(16*a)
Time = 0.48 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6528, 6596, 5971, 2009}
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.
\(\displaystyle \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx\) |
\(\Big \downarrow \) 6528 |
\(\displaystyle 5 a \int \frac {x}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}\) |
\(\Big \downarrow \) 6596 |
\(\displaystyle \frac {5 \int \frac {a x}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \frac {5 \int \left (\frac {a x}{8 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {3 \sinh (3 \text {arctanh}(a x))}{16 \text {arctanh}(a x)}+\frac {\sinh (5 \text {arctanh}(a x))}{16 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 \left (\frac {1}{8} \text {Shi}(\text {arctanh}(a x))+\frac {3}{16} \text {Shi}(3 \text {arctanh}(a x))+\frac {1}{16} \text {Shi}(5 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}\) |
-(1/(a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])) + (5*(SinhIntegral[ArcTanh[a*x]] /8 + (3*SinhIntegral[3*ArcTanh[a*x]])/16 + SinhIntegral[5*ArcTanh[a*x]]/16 ))/a
3.5.89.3.1 Defintions of rubi rules used
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_)^(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])
Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(58)=116\).
Time = 0.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.67
method | result | size |
default | \(\frac {10 \,\operatorname {arctanh}\left (a x \right ) \operatorname {Shi}\left (\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}+15 \,\operatorname {arctanh}\left (a x \right ) \operatorname {Shi}\left (3 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}+5 \,\operatorname {arctanh}\left (a x \right ) \operatorname {Shi}\left (5 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}-5 \cosh \left (3 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}-\cosh \left (5 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}-10 \,\operatorname {Shi}\left (\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )-15 \,\operatorname {Shi}\left (3 \,\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )-5 \,\operatorname {Shi}\left (5 \,\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cosh \left (3 \,\operatorname {arctanh}\left (a x \right )\right )+\cosh \left (5 \,\operatorname {arctanh}\left (a x \right )\right )}{16 a \left (a^{2} x^{2}-1\right ) \operatorname {arctanh}\left (a x \right )}\) | \(176\) |
1/16/a*(10*arctanh(a*x)*Shi(arctanh(a*x))*a^2*x^2+15*arctanh(a*x)*Shi(3*ar ctanh(a*x))*a^2*x^2+5*arctanh(a*x)*Shi(5*arctanh(a*x))*a^2*x^2-5*cosh(3*ar ctanh(a*x))*a^2*x^2-cosh(5*arctanh(a*x))*a^2*x^2-10*Shi(arctanh(a*x))*arct anh(a*x)-15*Shi(3*arctanh(a*x))*arctanh(a*x)-5*Shi(5*arctanh(a*x))*arctanh (a*x)+10*(-a^2*x^2+1)^(1/2)+5*cosh(3*arctanh(a*x))+cosh(5*arctanh(a*x)))/( a^2*x^2-1)/arctanh(a*x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \]
integral(sqrt(-a^2*x^2 + 1)/((a^8*x^8 - 4*a^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*arctanh(a*x)^2), x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \]
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \text {arctanh}(a x)^2} \, dx=\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{7/2}} \,d x \]