3.4.51 \(\int \frac {x^6}{(1-a^2 x^2)^4 \text {arctanh}(a x)} \, dx\) [351]

3.4.51.1 Optimal result

Integrand size = 22, antiderivative size = 55 \[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {15 \text {Chi}(2 \text {arctanh}(a x))}{32 a^7}-\frac {3 \text {Chi}(4 \text {arctanh}(a x))}{16 a^7}+\frac {\text {Chi}(6 \text {arctanh}(a x))}{32 a^7}-\frac {5 \log (\text {arctanh}(a x))}{16 a^7} \]

output

15/32*Chi(2*arctanh(a*x))/a^7-3/16*Chi(4*arctanh(a*x))/a^7+1/32*Chi(6*arct 
anh(a*x))/a^7-5/16*ln(arctanh(a*x))/a^7
 

3.4.51.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\frac {15 \text {Chi}(2 \text {arctanh}(a x))-6 \text {Chi}(4 \text {arctanh}(a x))+\text {Chi}(6 \text {arctanh}(a x))-10 \log (\text {arctanh}(a x))}{32 a^7} \]

input

Integrate[x^6/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]
 

output

(15*CoshIntegral[2*ArcTanh[a*x]] - 6*CoshIntegral[4*ArcTanh[a*x]] + CoshIn 
tegral[6*ArcTanh[a*x]] - 10*Log[ArcTanh[a*x]])/(32*a^7)
 

3.4.51.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6596, 3042, 25, 3793, 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.

\(\displaystyle \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx\)

\(\Big \downarrow \) 6596

\(\displaystyle \frac {\int \frac {a^6 x^6}{\left (1-a^2 x^2\right )^3 \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^7}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int -\frac {\sin (i \text {arctanh}(a x))^6}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^7}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\sin (i \text {arctanh}(a x))^6}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^7}\)

\(\Big \downarrow \) 3793

\(\displaystyle -\frac {\int \left (-\frac {15 \cosh (2 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {3 \cosh (4 \text {arctanh}(a x))}{16 \text {arctanh}(a x)}-\frac {\cosh (6 \text {arctanh}(a x))}{32 \text {arctanh}(a x)}+\frac {5}{16 \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a^7}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {15}{32} \text {Chi}(2 \text {arctanh}(a x))-\frac {3}{16} \text {Chi}(4 \text {arctanh}(a x))+\frac {1}{32} \text {Chi}(6 \text {arctanh}(a x))-\frac {5}{16} \log (\text {arctanh}(a x))}{a^7}\)

input

Int[x^6/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]
 

output

((15*CoshIntegral[2*ArcTanh[a*x]])/32 - (3*CoshIntegral[4*ArcTanh[a*x]])/1 
6 + CoshIntegral[6*ArcTanh[a*x]]/32 - (5*Log[ArcTanh[a*x]])/16)/a^7
 

3.4.51.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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])
 

3.4.51.4 Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {-\frac {5 \ln \left (\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {15 \,\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}-\frac {3 \,\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{7}}\)	\(40\)
default	\(\frac {-\frac {5 \ln \left (\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {15 \,\operatorname {Chi}\left (2 \,\operatorname {arctanh}\left (a x \right )\right )}{32}-\frac {3 \,\operatorname {Chi}\left (4 \,\operatorname {arctanh}\left (a x \right )\right )}{16}+\frac {\operatorname {Chi}\left (6 \,\operatorname {arctanh}\left (a x \right )\right )}{32}}{a^{7}}\)	\(40\)

input

int(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x,method=_RETURNVERBOSE)
 

output

1/a^7*(-5/16*ln(arctanh(a*x))+15/32*Chi(2*arctanh(a*x))-3/16*Chi(4*arctanh 
(a*x))+1/32*Chi(6*arctanh(a*x)))
 

3.4.51.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (47) = 94\).

Time = 0.24 (sec) , antiderivative size = 220, normalized size of antiderivative = 4.00 \[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=-\frac {20 \, \log \left (\log \left (-\frac {a x + 1}{a x - 1}\right )\right ) - \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 6 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 6 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - 15 \, \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 15 \, \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{64 \, a^{7}} \]

input

integrate(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="fricas")
 

output

-1/64*(20*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)) + 6*log_int 
egral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + 6*log_integral((a^2*x 
^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) - 15*log_integral(-(a*x + 1)/(a*x - 
 1)) - 15*log_integral(-(a*x - 1)/(a*x + 1)))/a^7
 

3.4.51.6 Sympy [F]

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

input

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

output

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

3.4.51.7 Maxima [F]

\[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int { \frac {x^{6}}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )} \,d x } \]

input

integrate(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="maxima")
 

output

integrate(x^6/((a^2*x^2 - 1)^4*arctanh(a*x)), x)
 

3.4.51.8 Giac [F]

\[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int { \frac {x^{6}}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )} \,d x } \]

input

integrate(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="giac")
 

output

integrate(x^6/((a^2*x^2 - 1)^4*arctanh(a*x)), x)
 

3.4.51.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (1-a^2 x^2\right )^4 \text {arctanh}(a x)} \, dx=\int \frac {x^6}{\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^4} \,d x \]

input

int(x^6/(atanh(a*x)*(a^2*x^2 - 1)^4),x)
 

output

int(x^6/(atanh(a*x)*(a^2*x^2 - 1)^4), x)