\(\int e^{\text {arctanh}(a x)} x^2 (1-a^2 x^2)^p \, dx\) [1043]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int e^{\text {arctanh}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^3 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^3 (3+2 p)}+\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right ) \] Output:

-(-a^2*x^2+1)^(1/2+p)/a^3/(1+2*p)+(-a^2*x^2+1)^(3/2+p)/a^3/(3+2*p)+1/3*x^3 
*hypergeom([3/2, 1/2-p],[5/2],a^2*x^2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int e^{\text {arctanh}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p} \left (2+a^2 (1+2 p) x^2\right )}{a^3 \left (3+8 p+4 p^2\right )}+\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right ) \] Input:

Integrate[E^ArcTanh[a*x]*x^2*(1 - a^2*x^2)^p,x]
 

Output:

-(((1 - a^2*x^2)^(1/2 + p)*(2 + a^2*(1 + 2*p)*x^2))/(a^3*(3 + 8*p + 4*p^2) 
)) + (x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6698, 542, 243, 53, 278, 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.

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

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 542

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {1}{2} a \int \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2+\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right )+\frac {1}{2} a \left (\frac {2 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}-\frac {2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}\right )\)

Input:

Int[E^ArcTanh[a*x]*x^2*(1 - a^2*x^2)^p,x]
 

Output:

(a*((-2*(1 - a^2*x^2)^(1/2 + p))/(a^4*(1 + 2*p)) + (2*(1 - a^2*x^2)^(3/2 + 
 p))/(a^4*(3 + 2*p))))/2 + (x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x 
^2])/3
 

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 542

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[c   Int[x^m*(a + b*x^2)^p, x], x] + Simp[d   Int[x^(m + 1)*(a + b*x^2 
)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && IntegerQ[m] &&  !IntegerQ[2*p]
 

rule 2009

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

rule 6698

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.56

method	result	size
meijerg	\(\frac {a \,x^{4} \operatorname {hypergeom}\left (\left [2, \frac {1}{2}-p \right ], \left [3\right ], a^{2} x^{2}\right )}{4}+\frac {x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {1}{2}-p \right ], \left [\frac {5}{2}\right ], a^{2} x^{2}\right )}{3}\)	\(47\)

Input:

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

Output:

1/4*a*x^4*hypergeom([2,1/2-p],[3],a^2*x^2)+1/3*x^3*hypergeom([3/2,1/2-p],[ 
5/2],a^2*x^2)
 

Fricas [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2*(-a^2*x^2+1)^p,x, algorithm="fric 
as")
 

Output:

integral(-sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p*x^2/(a*x - 1), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.95 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.06 \[ \int e^{\text {arctanh}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx=- \frac {a^{2 p} x^{2 p + 3} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {3}{2} \\ \frac {1}{2}, - p - \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p + 3} x^{2 p + 3} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {3}{2} \\ p + 1, p + \frac {5}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } a^{3} \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} - \frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} - p - 1, 1 & -1 \\- p - \frac {3}{2}, - p - 1 & 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 \pi a^{3}} - \frac {{G_{3, 3}^{1, 3}\left (\begin {matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a^{3} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \] Input:

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

Output:

-a**(2*p)*x**(2*p + 3)*exp(I*pi*p)*gamma(-p - 3/2)*gamma(p + 1/2)*hyper((1 
, -p, -p - 3/2), (1/2, -p - 1/2), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(-p - 1/ 
2)*gamma(p + 1)) - a**(2*p + 3)*x**(2*p + 3)*exp(I*pi*p)*gamma(-p - 3/2)*g 
amma(p + 1/2)*hyper((1/2, 1, p + 3/2), (p + 1, p + 5/2), a**2*x**2*exp_pol 
ar(2*I*pi))/(2*sqrt(pi)*a**3*gamma(-p - 1/2)*gamma(p + 1)) - meijerg(((-p 
- 1, 1), (-1,)), ((-p - 3/2, -p - 1), (0,)), exp_polar(-I*pi)/(a**2*x**2)) 
*gamma(p + 1/2)/(2*pi*a**3) - meijerg(((-1, -p - 2, 1), ()), ((-p - 2,), ( 
-p - 3/2, 0)), exp_polar(-I*pi)/(a**2*x**2))*gamma(p + 1/2)/(2*a**3*gamma( 
-p)*gamma(p + 1))
 

Maxima [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2*(-a^2*x^2+1)^p,x, algorithm="maxi 
ma")
 

Output:

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

Giac [F]

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

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

Output:

integrate((a*x + 1)*(-a^2*x^2 + 1)^p*x^2/sqrt(-a^2*x^2 + 1), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int e^{\text {arctanh}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx=\frac {-2 \left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}} a^{2} p \,x^{2}-\left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}} a^{2} x^{2}-2 \left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}}+4 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{2}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{3} p^{2}+8 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{2}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{3} p +3 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{2}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{3}}{a^{3} \left (4 p^{2}+8 p +3\right )} \] Input:

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

Output:

( - 2*( - a**2*x**2 + 1)**((2*p + 1)/2)*a**2*p*x**2 - ( - a**2*x**2 + 1)** 
((2*p + 1)/2)*a**2*x**2 - 2*( - a**2*x**2 + 1)**((2*p + 1)/2) + 4*int((( - 
 a**2*x**2 + 1)**p*x**2)/sqrt( - a**2*x**2 + 1),x)*a**3*p**2 + 8*int((( - 
a**2*x**2 + 1)**p*x**2)/sqrt( - a**2*x**2 + 1),x)*a**3*p + 3*int((( - a**2 
*x**2 + 1)**p*x**2)/sqrt( - a**2*x**2 + 1),x)*a**3)/(a**3*(4*p**2 + 8*p + 
3))