\(\int \frac {x^4}{(1-a^2 x^2)^4 \text {arctanh}(a x)} \, dx\) [353]

Optimal result

Integrand size = 22, antiderivative size = 55 \[ \int \frac {x^4}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=-\frac {\text {Chi}(2 \text {arctanh}(a x))}{32 a^5}-\frac {\text {Chi}(4 \text {arctanh}(a x))}{16 a^5}+\frac {\text {Chi}(6 \text {arctanh}(a x))}{32 a^5}+\frac {\log (\text {arctanh}(a x))}{16 a^5} \] Output:

-1/32*Chi(2*arctanh(a*x))/a^5-1/16*Chi(4*arctanh(a*x))/a^5+1/32*Chi(6*arct 
anh(a*x))/a^5+1/16*ln(arctanh(a*x))/a^5
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {x^4}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {-\text {Chi}(2 \text {arctanh}(a x))-2 \text {Chi}(4 \text {arctanh}(a x))+\text {Chi}(6 \text {arctanh}(a x))+2 \log (\text {arctanh}(a x))}{32 a^5} \] Input:

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

Output:

(-CoshIntegral[2*ArcTanh[a*x]] - 2*CoshIntegral[4*ArcTanh[a*x]] + CoshInte 
gral[6*ArcTanh[a*x]] + 2*Log[ArcTanh[a*x]])/(32*a^5)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {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.

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

\(\Big \downarrow \) 6596

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

\(\Big \downarrow \) 5971

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/32*CoshIntegral[2*ArcTanh[a*x]] - CoshIntegral[4*ArcTanh[a*x]]/16 + Co 
shIntegral[6*ArcTanh[a*x]]/32 + Log[ArcTanh[a*x]]/16)/a^5
 

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 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 = 0.88 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}-\frac {\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{5}}\)	\(40\)
default	\(\frac {\frac {\ln \left (\operatorname {arctanh}\left (a x \right )\right )}{16}-\frac {\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}-\frac {\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{5}}\)	\(40\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (47) = 94\).

Time = 0.08 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.93 \[ \int \frac {x^4}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {4 \, \log \left (\log \left (-\frac {a x + 1}{a x - 1}\right )\right ) + \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) - 2 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{64 \, a^{5}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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