\(\int \frac {1}{(1-a^2 x^2)^3 \text {arctanh}(a x)^2} \, dx\) [334]

Optimal result

Integrand size = 19, antiderivative size = 49 \[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}+\frac {\text {Shi}(2 \text {arctanh}(a x))}{a}+\frac {\text {Shi}(4 \text {arctanh}(a x))}{2 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=\frac {-\frac {2}{\left (-1+a^2 x^2\right )^2 \text {arctanh}(a x)}+2 \text {Shi}(2 \text {arctanh}(a x))+\text {Shi}(4 \text {arctanh}(a x))}{2 a} \] Input:

Integrate[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.

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

\(\Big \downarrow \) 6528

\(\displaystyle 4 a \int \frac {x}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}dx-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 6596

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

\(\Big \downarrow \) 5971

\(\displaystyle \frac {4 \int \left (\frac {\sinh (2 \text {arctanh}(a x))}{4 \text {arctanh}(a x)}+\frac {\sinh (4 \text {arctanh}(a x))}{8 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 \left (\frac {1}{4} \text {Shi}(2 \text {arctanh}(a x))+\frac {1}{8} \text {Shi}(4 \text {arctanh}(a x))\right )}{a}-\frac {1}{a \left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}\)

Input:

Int[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^2),x]
 

Output:

-(1/(a*(1 - a^2*x^2)^2*ArcTanh[a*x])) + (4*(SinhIntegral[2*ArcTanh[a*x]]/4 
 + SinhIntegral[4*ArcTanh[a*x]]/8))/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 [A] (verified)

Time = 1.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {-\frac {3}{8 \,\operatorname {arctanh}\left (a x \right )}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{2 \,\operatorname {arctanh}\left (a x \right )}+\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )-\frac {\cosh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8 \,\operatorname {arctanh}\left (a x \right )}+\frac {\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{2}}{a}\)	\(60\)
default	\(\frac {-\frac {3}{8 \,\operatorname {arctanh}\left (a x \right )}-\frac {\cosh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{2 \,\operatorname {arctanh}\left (a x \right )}+\operatorname {Shi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )-\frac {\cosh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8 \,\operatorname {arctanh}\left (a x \right )}+\frac {\operatorname {Shi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{2}}{a}\)	\(60\)

Input:

int(1/(-a^2*x^2+1)^3/arctanh(a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/a*(-3/8/arctanh(a*x)-1/2/arctanh(a*x)*cosh(2*arctanh(a*x))+Shi(2*arctanh 
(a*x))-1/8/arctanh(a*x)*cosh(4*arctanh(a*x))+1/2*Shi(4*arctanh(a*x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (46) = 92\).

Time = 0.08 (sec) , antiderivative size = 222, normalized size of antiderivative = 4.53 \[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=\frac {{\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 8}{4 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \] Input:

integrate(1/(-a^2*x^2+1)^3/arctanh(a*x)^2,x, algorithm="fricas")
 

Output:

1/4*(((a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^ 
2 - 2*a*x + 1)) - (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x 
+ 1)/(a^2*x^2 + 2*a*x + 1)) + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a 
*x + 1)/(a*x - 1)) - 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/( 
a*x + 1)))*log(-(a*x + 1)/(a*x - 1)) - 8)/((a^5*x^4 - 2*a^3*x^2 + a)*log(- 
(a*x + 1)/(a*x - 1)))
 

Sympy [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=- \int \frac {1}{a^{6} x^{6} \operatorname {atanh}^{2}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{2}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{2}{\left (a x \right )} - \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \] Input:

integrate(1/(-a**2*x**2+1)**3/atanh(a*x)**2,x)
 

Output:

-Integral(1/(a**6*x**6*atanh(a*x)**2 - 3*a**4*x**4*atanh(a*x)**2 + 3*a**2* 
x**2*atanh(a*x)**2 - atanh(a*x)**2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=\int { -\frac {1}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \] Input:

integrate(1/(-a^2*x^2+1)^3/arctanh(a*x)^2,x, algorithm="maxima")
 

Output:

8*a*integrate(-x/((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x + 1) - (a^ 
6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(-a*x + 1)), x) - 2/((a^5*x^4 - 2*a^ 
3*x^2 + a)*log(a*x + 1) - (a^5*x^4 - 2*a^3*x^2 + a)*log(-a*x + 1))
 

Giac [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=\int { -\frac {1}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{2}} \,d x } \] Input:

integrate(1/(-a^2*x^2+1)^3/arctanh(a*x)^2,x, algorithm="giac")
 

Output:

integrate(-1/((a^2*x^2 - 1)^3*arctanh(a*x)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=-\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^3} \,d x \] Input:

int(-1/(atanh(a*x)^2*(a^2*x^2 - 1)^3),x)
 

Output:

-int(1/(atanh(a*x)^2*(a^2*x^2 - 1)^3), x)
 

Reduce [F]

\[ \int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)^2} \, dx=\frac {-4 \mathit {atanh} \left (a x \right ) \left (\int \frac {x}{\mathit {atanh} \left (a x \right ) a^{6} x^{6}-3 \mathit {atanh} \left (a x \right ) a^{4} x^{4}+3 \mathit {atanh} \left (a x \right ) a^{2} x^{2}-\mathit {atanh} \left (a x \right )}d x \right ) a^{6} x^{4}+8 \mathit {atanh} \left (a x \right ) \left (\int \frac {x}{\mathit {atanh} \left (a x \right ) a^{6} x^{6}-3 \mathit {atanh} \left (a x \right ) a^{4} x^{4}+3 \mathit {atanh} \left (a x \right ) a^{2} x^{2}-\mathit {atanh} \left (a x \right )}d x \right ) a^{4} x^{2}-4 \mathit {atanh} \left (a x \right ) \left (\int \frac {x}{\mathit {atanh} \left (a x \right ) a^{6} x^{6}-3 \mathit {atanh} \left (a x \right ) a^{4} x^{4}+3 \mathit {atanh} \left (a x \right ) a^{2} x^{2}-\mathit {atanh} \left (a x \right )}d x \right ) a^{2}-1}{\mathit {atanh} \left (a x \right ) a \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right )} \] Input:

int(1/(-a^2*x^2+1)^3/atanh(a*x)^2,x)
                                                                                    
                                                                                    
 

Output:

( - 4*atanh(a*x)*int(x/(atanh(a*x)*a**6*x**6 - 3*atanh(a*x)*a**4*x**4 + 3* 
atanh(a*x)*a**2*x**2 - atanh(a*x)),x)*a**6*x**4 + 8*atanh(a*x)*int(x/(atan 
h(a*x)*a**6*x**6 - 3*atanh(a*x)*a**4*x**4 + 3*atanh(a*x)*a**2*x**2 - atanh 
(a*x)),x)*a**4*x**2 - 4*atanh(a*x)*int(x/(atanh(a*x)*a**6*x**6 - 3*atanh(a 
*x)*a**4*x**4 + 3*atanh(a*x)*a**2*x**2 - atanh(a*x)),x)*a**2 - 1)/(atanh(a 
*x)*a*(a**4*x**4 - 2*a**2*x**2 + 1))