\(\int \frac {e^{\text {arctanh}(a x)} x^4}{(c-a^2 c x^2)^{5/2}} \, dx\) [1000]

Optimal result

Integrand size = 25, antiderivative size = 232 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{4 a^5 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^5 c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (1+a x)}{16 a^5 c^2 \sqrt {c-a^2 c x^2}} \] Output:

1/8*(-a^2*x^2+1)^(1/2)/a^5/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)-3/4*(-a^2*x 
^2+1)^(1/2)/a^5/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a 
^5/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)-11/16*(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/a^ 
5/c^2/(-a^2*c*x^2+c)^(1/2)-5/16*(-a^2*x^2+1)^(1/2)*ln(a*x+1)/a^5/c^2/(-a^2 
*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.38 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (\frac {2 \left (-6+3 a x+5 a^2 x^2\right )}{(-1+a x)^2 (1+a x)}-11 \log (1-a x)-5 \log (1+a x)\right )}{16 a^5 c^2 \sqrt {c-a^2 c x^2}} \] Input:

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]
 

Output:

(Sqrt[1 - a^2*x^2]*((2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) 
- 11*Log[1 - a*x] - 5*Log[1 + a*x]))/(16*a^5*c^2*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.45, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 99, 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.

\(\displaystyle \int \frac {x^4 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^4}{(1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {5}{16 a^4 (a x+1)}+\frac {1}{8 a^4 (a x+1)^2}-\frac {11}{16 a^4 (a x-1)}-\frac {3}{4 a^4 (a x-1)^2}-\frac {1}{4 a^4 (a x-1)^3}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (a x+1)}+\frac {1}{8 a^5 (1-a x)^2}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (a x+1)}{16 a^5}\right )}{c^2 \sqrt {c-a^2 c x^2}}\)

Input:

Int[(E^ArcTanh[a*x]*x^4)/(c - a^2*c*x^2)^(5/2),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5 
*(1 + a*x)) - (11*Log[1 - a*x])/(16*a^5) - (5*Log[1 + a*x])/(16*a^5)))/(c^ 
2*Sqrt[c - a^2*c*x^2])
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

rule 6700

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], 
 x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || 
 GtQ[c, 0])
 

rule 6703

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar 
t[p])   Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, 
d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !I 
ntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {\left (5 \ln \left (a x +1\right ) x^{3} a^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-5 \ln \left (a x +1\right ) x^{2} a^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}-10 a^{2} x^{2}-5 \ln \left (a x +1\right ) x a -11 a \ln \left (a x -1\right ) x -6 a x +5 \ln \left (a x +1\right )+11 \ln \left (a x -1\right )+12\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (a x -1\right )^{2} c^{3} a^{5}}\)	\(155\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/16*(5*ln(a*x+1)*x^3*a^3+11*a^3*ln(a*x-1)*x^3-5*ln(a*x+1)*x^2*a^2-11*a^2 
*ln(a*x-1)*x^2-10*a^2*x^2-5*ln(a*x+1)*x*a-11*a*ln(a*x-1)*x-6*a*x+5*ln(a*x+ 
1)+11*ln(a*x-1)+12)/(-a^2*x^2+1)^(1/2)/(a*x+1)/(a*x-1)^2*(-c*(a^2*x^2-1))^ 
(1/2)/c^3/a^5
 

Fricas [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="fricas")
 

Output:

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^4/(a^7*c^3*x^7 - a^6*c^ 
3*x^6 - 3*a^5*c^3*x^5 + 3*a^4*c^3*x^4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a* 
c^3*x + c^3), x)
 

Sympy [F]

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

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4/(-a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral(x**4*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 
))**(5/2)), x)
 

Maxima [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="maxima")
 

Output:

integrate((a*x + 1)*x^4/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)
 

Giac [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="giac")
 

Output:

integrate((a*x + 1)*x^4/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)
 

Mupad [F(-1)]

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

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)
 

Output:

int((x^4*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-11 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+11 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+11 \,\mathrm {log}\left (a x -1\right ) a x -11 \,\mathrm {log}\left (a x -1\right )-5 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+5 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+5 \,\mathrm {log}\left (a x +1\right ) a x -5 \,\mathrm {log}\left (a x +1\right )+10 a^{3} x^{3}-4 a x -2\right )}{16 a^{5} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4/(-a^2*c*x^2+c)^(5/2),x)
 

Output:

(sqrt(c)*( - 11*log(a*x - 1)*a**3*x**3 + 11*log(a*x - 1)*a**2*x**2 + 11*lo 
g(a*x - 1)*a*x - 11*log(a*x - 1) - 5*log(a*x + 1)*a**3*x**3 + 5*log(a*x + 
1)*a**2*x**2 + 5*log(a*x + 1)*a*x - 5*log(a*x + 1) + 10*a**3*x**3 - 4*a*x 
- 2))/(16*a**5*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))