\(\int \frac {e^{\text {arctanh}(a x)} x^m}{(1-a^2 x^2)^{5/2}} \, dx\) [1019]

Optimal result

Integrand size = 24, antiderivative size = 70 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{2+m} \] Output:

x^(1+m)*hypergeom([3, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+a*x^(2+m)*hype 
rgeom([3, 1+1/2*m],[2+1/2*m],a^2*x^2)/(2+m)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=x^{1+m} \left (\frac {a x \operatorname {Hypergeometric2F1}\left (3,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )}{2+m}+\frac {\operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}\right ) \] Input:

Integrate[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(5/2),x]
 

Output:

x^(1 + m)*((a*x*Hypergeometric2F1[3, 1 + m/2, 2 + m/2, a^2*x^2])/(2 + m) + 
 Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2]/(1 + m))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6700, 92, 82, 278}

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.

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 92

\(\displaystyle a \int \frac {x^{m+1}}{(1-a x)^3 (a x+1)^3}dx+\int \frac {x^m}{(1-a x)^3 (a x+1)^3}dx\)

\(\Big \downarrow \) 82

\(\displaystyle a \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^3}dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^3}dx\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (3,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\)

Input:

Int[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(5/2),x]
 

Output:

(x^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + 
(a*x^(2 + m)*Hypergeometric2F1[3, (2 + m)/2, (4 + m)/2, a^2*x^2])/(2 + m)
 

Defintions of rubi rules used

rule 82

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, 
 e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
 

rule 92

Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), 
x_] :> Simp[a   Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f   In 
t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, 
 n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] && 
!IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

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])
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5.

Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 3.20

method	result	size
meijerg	\(\frac {\left (-a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {x^{1+m} \left (-a^{2}\right )^{\frac {m}{2}+\frac {1}{2}} \left (a^{2} m^{2} x^{2}-2 a^{2} m \,x^{2}-3 a^{2} x^{2}-m^{2}+4 m +5\right )}{2 \left (1+m \right ) \left (-a^{2} x^{2}+1\right )^{2}}+\frac {4 x^{1+m} \left (-a^{2}\right )^{\frac {m}{2}+\frac {1}{2}} \left (\frac {1}{16} m^{3}-\frac {3}{16} m^{2}-\frac {1}{16} m +\frac {3}{16}\right ) \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}+\frac {1}{2}\right )}{1+m}\right )}{4}-\frac {\left (-a^{2}\right )^{-\frac {m}{2}} \left (-\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (a^{2} m \,x^{2}-m +2\right )}{2 \left (-a^{2} x^{2}+1\right )^{2}}-\frac {x^{m} \left (-a^{2}\right )^{\frac {m}{2}} \left (-2+m \right ) m \operatorname {LerchPhi}\left (a^{2} x^{2}, 1, \frac {m}{2}\right )}{4}\right )}{4 a}\)	\(224\)

Input:

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

Output:

1/4*(-a^2)^(-1/2-1/2*m)*(1/2/(1+m)*x^(1+m)*(-a^2)^(1/2*m+1/2)*(a^2*m^2*x^2 
-2*a^2*m*x^2-3*a^2*x^2-m^2+4*m+5)/(-a^2*x^2+1)^2+4/(1+m)*x^(1+m)*(-a^2)^(1 
/2*m+1/2)*(1/16*m^3-3/16*m^2-1/16*m+3/16)*LerchPhi(a^2*x^2,1,1/2*m+1/2))-1 
/4/a*(-a^2)^(-1/2*m)*(-1/2*x^m*(-a^2)^(1/2*m)*(a^2*m*x^2-m+2)/(-a^2*x^2+1) 
^2-1/4*x^m*(-a^2)^(1/2*m)*(-2+m)*m*LerchPhi(a^2*x^2,1,1/2*m))
 

Fricas [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{3}} \,d x } \] Input:

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

Output:

integral(-x^m/(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.35 (sec) , antiderivative size = 2173, normalized size of antiderivative = 31.04 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

a**4*m**3*x**4*x**(m + 1)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1 
/2)*gamma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 3/2) - 64*a**2*x**2*gamma(m 
/2 + 3/2) + 32*gamma(m/2 + 3/2)) - 3*a**4*m**2*x**4*x**(m + 1)*lerchphi(a* 
*2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**4*x**4*ga 
mma(m/2 + 3/2) - 64*a**2*x**2*gamma(m/2 + 3/2) + 32*gamma(m/2 + 3/2)) - a* 
*4*m*x**4*x**(m + 1)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*g 
amma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 3/2) - 64*a**2*x**2*gamma(m/2 + 
3/2) + 32*gamma(m/2 + 3/2)) + 3*a**4*x**4*x**(m + 1)*lerchphi(a**2*x**2*ex 
p_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 
3/2) - 64*a**2*x**2*gamma(m/2 + 3/2) + 32*gamma(m/2 + 3/2)) - 2*a**2*m**3* 
x**2*x**(m + 1)*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma( 
m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 3/2) - 64*a**2*x**2*gamma(m/2 + 3/2) 
+ 32*gamma(m/2 + 3/2)) + 6*a**2*m**2*x**2*x**(m + 1)*lerchphi(a**2*x**2*ex 
p_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 
3/2) - 64*a**2*x**2*gamma(m/2 + 3/2) + 32*gamma(m/2 + 3/2)) + 2*a**2*m**2* 
x**2*x**(m + 1)*gamma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/2 + 3/2) - 64*a**2* 
x**2*gamma(m/2 + 3/2) + 32*gamma(m/2 + 3/2)) + 2*a**2*m*x**2*x**(m + 1)*le 
rchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(32*a** 
4*x**4*gamma(m/2 + 3/2) - 64*a**2*x**2*gamma(m/2 + 3/2) + 32*gamma(m/2 + 3 
/2)) - 4*a**2*m*x**2*x**(m + 1)*gamma(m/2 + 1/2)/(32*a**4*x**4*gamma(m/...
 

Maxima [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{3}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (a x + 1\right )} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{3}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int -\frac {x^m\,\left (a\,x+1\right )}{{\left (a^2\,x^2-1\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx=-\left (\int \frac {x^{m}}{a^{5} x^{5}-a^{4} x^{4}-2 a^{3} x^{3}+2 a^{2} x^{2}+a x -1}d x \right ) \] Input:

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

Output:

 - int(x**m/(a**5*x**5 - a**4*x**4 - 2*a**3*x**3 + 2*a**2*x**2 + a*x - 1), 
x)