\(\int \frac {x}{(1-a^2 x^2)^4 \text {arctanh}(a x)^2} \, dx\) [360]

Optimal result

Integrand size = 20, antiderivative size = 67 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2} \, dx=-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}+\frac {5 \text {Chi}(2 \text {arctanh}(a x))}{16 a^2}+\frac {\text {Chi}(4 \text {arctanh}(a x))}{2 a^2}+\frac {3 \text {Chi}(6 \text {arctanh}(a x))}{16 a^2} \] Output:

-x/a/(-a^2*x^2+1)^3/arctanh(a*x)+5/16*Chi(2*arctanh(a*x))/a^2+1/2*Chi(4*ar 
ctanh(a*x))/a^2+3/16*Chi(6*arctanh(a*x))/a^2
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2} \, dx=\frac {\frac {16 a x}{\left (-1+a^2 x^2\right )^3 \text {arctanh}(a x)}+5 \text {Chi}(2 \text {arctanh}(a x))+8 \text {Chi}(4 \text {arctanh}(a x))+3 \text {Chi}(6 \text {arctanh}(a x))}{16 a^2} \] Input:

Integrate[x/((1 - a^2*x^2)^4*ArcTanh[a*x]^2),x]
 

Output:

((16*a*x)/((-1 + a^2*x^2)^3*ArcTanh[a*x]) + 5*CoshIntegral[2*ArcTanh[a*x]] 
 + 8*CoshIntegral[4*ArcTanh[a*x]] + 3*CoshIntegral[6*ArcTanh[a*x]])/(16*a^ 
2)
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.79, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6594, 6530, 3042, 3793, 2009, 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.

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

\(\Big \downarrow \) 6594

\(\displaystyle \frac {\int \frac {1}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx}{a}+5 a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 6530

\(\displaystyle 5 a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx+\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle 5 a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx+\frac {\int \frac {\sin \left (i \text {arctanh}(a x)+\frac {\pi }{2}\right )^6}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 3793

\(\displaystyle 5 a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx+\frac {\int \left (\frac {15 \cosh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {3 \cosh (4 \text {arctanh}(a x))}{16 \text {arctanh}(a x)}+\frac {\cosh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {5}{16 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 2009

\(\displaystyle 5 a \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)}dx+\frac {\frac {15}{32} \text {Chi}(2 \text {arctanh}(a x))+\frac {3}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))+\frac {5}{16} \log (\text {arctanh}(a x))}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 6596

\(\displaystyle \frac {5 \int \frac {a^2 x^2}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}+\frac {\frac {15}{32} \text {Chi}(2 \text {arctanh}(a x))+\frac {3}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))+\frac {5}{16} \log (\text {arctanh}(a x))}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 5971

\(\displaystyle \frac {5 \int \left (-\frac {\cosh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {\cosh (4 \text {arctanh}(a x))}{16 \text {arctanh}(a x)}+\frac {\cosh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}-\frac {1}{16 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a^2}+\frac {\frac {15}{32} \text {Chi}(2 \text {arctanh}(a x))+\frac {3}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))+\frac {5}{16} \log (\text {arctanh}(a x))}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 \left (-\frac {1}{32} \text {Chi}(2 \text {arctanh}(a x))+\frac {1}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))-\frac {1}{16} \log (\text {arctanh}(a x))\right )}{a^2}+\frac {\frac {15}{32} \text {Chi}(2 \text {arctanh}(a x))+\frac {3}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))+\frac {5}{16} \log (\text {arctanh}(a x))}{a^2}-\frac {x}{a \left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}\)

Input:

Int[x/((1 - a^2*x^2)^4*ArcTanh[a*x]^2),x]
 

Output:

-(x/(a*(1 - a^2*x^2)^3*ArcTanh[a*x])) + (5*(-1/32*CoshIntegral[2*ArcTanh[a 
*x]] + CoshIntegral[4*ArcTanh[a*x]]/16 + CoshIntegral[6*ArcTanh[a*x]]/32 - 
 Log[ArcTanh[a*x]]/16))/a^2 + ((15*CoshIntegral[2*ArcTanh[a*x]])/32 + (3*C 
oshIntegral[4*ArcTanh[a*x]])/16 + CoshIntegral[6*ArcTanh[a*x]]/32 + (5*Log 
[ArcTanh[a*x]])/16)/a^2
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

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 6530

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])
 

rule 6594

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) 
^2)^(q_), x_Symbol] :> Simp[x^m*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^( 
p + 1)/(b*c*d*(p + 1))), x] + (Simp[c*((m + 2*q + 2)/(b*(p + 1)))   Int[x^( 
m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] - Simp[m/(b*c*(p 
+ 1))   Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x]) / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && LtQ[q, - 
1] && LtQ[p, -1] && NeQ[m + 2*q + 2, 0]
 

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 = 1.56 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {\sinh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8 \,\operatorname {arctanh}\left (a x \right )}+\frac {\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{2}-\frac {\sinh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {3 \,\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {5 \sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {5 \,\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{16}}{a^{2}}\)	\(78\)
default	\(\frac {-\frac {\sinh \left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{8 \,\operatorname {arctanh}\left (a x \right )}+\frac {\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{2}-\frac {\sinh \left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {3 \,\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {5 \sinh \left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32 \,\operatorname {arctanh}\left (a x \right )}+\frac {5 \,\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{16}}{a^{2}}\)	\(78\)

Input:

int(x/(-a^2*x^2+1)^4/arctanh(a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/a^2*(-1/8*sinh(4*arctanh(a*x))/arctanh(a*x)+1/2*Chi(4*arctanh(a*x))-1/32 
/arctanh(a*x)*sinh(6*arctanh(a*x))+3/16*Chi(6*arctanh(a*x))-5/32/arctanh(a 
*x)*sinh(2*arctanh(a*x))+5/16*Chi(2*arctanh(a*x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (59) = 118\).

Time = 0.10 (sec) , antiderivative size = 418, normalized size of antiderivative = 6.24 \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2} \, dx=\frac {64 \, a x + {\left (3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{32 \, {\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \] Input:

integrate(x/(-a^2*x^2+1)^4/arctanh(a*x)^2,x, algorithm="fricas")
 

Output:

1/32*(64*a*x + (3*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3 
*x^3 + 3*a^2*x^2 + 3*a*x + 1)/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) + 3*(a^6* 
x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3*x^3 - 3*a^2*x^2 + 3*a* 
x - 1)/(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)) + 8*(a^6*x^6 - 3*a^4*x^4 + 3*a^2 
*x^2 - 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + 8*(a 
^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^ 
2*x^2 + 2*a*x + 1)) + 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral 
(-(a*x + 1)/(a*x - 1)) + 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integ 
ral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1)))/((a^8*x^6 - 3*a^6*x^ 
4 + 3*a^4*x^2 - a^2)*log(-(a*x + 1)/(a*x - 1)))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x/(-a^2*x^2+1)^4/arctanh(a*x)^2,x, algorithm="maxima")
 

Output:

2*x/((a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(a*x + 1) - (a^7*x^6 - 3*a^5 
*x^4 + 3*a^3*x^2 - a)*log(-a*x + 1)) - integrate(-2*(5*a^2*x^2 + 1)/((a^9* 
x^8 - 4*a^7*x^6 + 6*a^5*x^4 - 4*a^3*x^2 + a)*log(a*x + 1) - (a^9*x^8 - 4*a 
^7*x^6 + 6*a^5*x^4 - 4*a^3*x^2 + a)*log(-a*x + 1)), x)
 

Giac [F]

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

integrate(x/(-a^2*x^2+1)^4/arctanh(a*x)^2,x, algorithm="giac")
 

Output:

integrate(x/((a^2*x^2 - 1)^4*arctanh(a*x)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)^2} \, dx=\int \frac {x}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^4} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int(x/(atanh(a*x)**2*a**8*x**8 - 4*atanh(a*x)**2*a**6*x**6 + 6*atanh(a*x)* 
*2*a**4*x**4 - 4*atanh(a*x)**2*a**2*x**2 + atanh(a*x)**2),x)