\(\int \frac {1}{(1-a^2 x^2)^{5/2} \text {arctanh}(a x)^2} \, dx\) [488]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=-\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}+\frac {3 \text {Shi}(\text {arctanh}(a x))}{4 a}+\frac {3 \text {Shi}(3 \text {arctanh}(a x))}{4 a} \] Output:

-1/a/(-a^2*x^2+1)^(3/2)/arctanh(a*x)+3/4*Shi(arctanh(a*x))/a+3/4*Shi(3*arc 
tanh(a*x))/a
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\frac {-\frac {4}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}+3 \text {Shi}(\text {arctanh}(a x))+3 \text {Shi}(3 \text {arctanh}(a x))}{4 a} \] Input:

Integrate[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2),x]
 

Output:

(-4/((1 - a^2*x^2)^(3/2)*ArcTanh[a*x]) + 3*SinhIntegral[ArcTanh[a*x]] + 3* 
SinhIntegral[3*ArcTanh[a*x]])/(4*a)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, 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.

\(\displaystyle \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx\)

\(\Big \downarrow \) 6528

\(\displaystyle 3 a \int \frac {x}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}\)

\(\Big \downarrow \) 6596

\(\displaystyle \frac {3 \int \frac {a x}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}\)

\(\Big \downarrow \) 5971

\(\displaystyle \frac {3 \int \left (\frac {a x}{4 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {\sinh (3 \text {arctanh}(a x))}{4 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (\frac {1}{4} \text {Shi}(\text {arctanh}(a x))+\frac {1}{4} \text {Shi}(3 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}\)

Input:

Int[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2),x]
 

Output:

-(1/(a*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x])) + (3*(SinhIntegral[ArcTanh[a*x]] 
/4 + SinhIntegral[3*ArcTanh[a*x]]/4))/a
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5971

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]
 

rule 6528

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]
 

rule 6596

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(46)=92\).

Time = 1.01 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.31

method	result	size
default	\(\frac {3 \,\operatorname {arctanh}\left (a x \right ) \operatorname {Shi}\left (\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}+3 \,\operatorname {arctanh}\left (a x \right ) \operatorname {Shi}\left (3 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}-\cosh \left (3 \,\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}-3 \,\operatorname {Shi}\left (\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )-3 \,\operatorname {Shi}\left (3 \,\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )+3 \sqrt {-a^{2} x^{2}+1}+\cosh \left (3 \,\operatorname {arctanh}\left (a x \right )\right )}{4 a \,\operatorname {arctanh}\left (a x \right ) \left (a^{2} x^{2}-1\right )}\)	\(120\)

Input:

int(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4/a*(3*arctanh(a*x)*Shi(arctanh(a*x))*a^2*x^2+3*arctanh(a*x)*Shi(3*arcta 
nh(a*x))*a^2*x^2-cosh(3*arctanh(a*x))*a^2*x^2-3*Shi(arctanh(a*x))*arctanh( 
a*x)-3*Shi(3*arctanh(a*x))*arctanh(a*x)+3*(-a^2*x^2+1)^(1/2)+cosh(3*arctan 
h(a*x)))/arctanh(a*x)/(a^2*x^2-1)
 

Fricas [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \] Input:

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="fricas")
 

Output:

integral(-sqrt(-a^2*x^2 + 1)/((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*arctan 
h(a*x)^2), x)
 

Sympy [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \] Input:

integrate(1/(-a**2*x**2+1)**(5/2)/atanh(a*x)**2,x)
 

Output:

Integral(1/((-(a*x - 1)*(a*x + 1))**(5/2)*atanh(a*x)**2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \] Input:

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="maxima")
 

Output:

integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)^2), x)
 

Giac [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \] Input:

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="giac")
 

Output:

integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(5/2)),x)
 

Output:

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)^2} \, dx=\int \frac {1}{\sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )^{2} a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )^{2} a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )^{2}}d x \] Input:

int(1/(-a^2*x^2+1)^(5/2)/atanh(a*x)^2,x)
 

Output:

int(1/(sqrt( - a**2*x**2 + 1)*atanh(a*x)**2*a**4*x**4 - 2*sqrt( - a**2*x** 
2 + 1)*atanh(a*x)**2*a**2*x**2 + sqrt( - a**2*x**2 + 1)*atanh(a*x)**2),x)