\(\int e^{-\text {arctanh}(a x)} x^3 (1-a^2 x^2)^p \, dx\) [1238]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^4 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^4 (3+2 p)}-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \] Output:

-(-a^2*x^2+1)^(1/2+p)/a^4/(1+2*p)+(-a^2*x^2+1)^(3/2+p)/a^4/(3+2*p)-1/5*a*x 
^5*hypergeom([5/2, 1/2-p],[7/2],a^2*x^2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p} \left (2+a^2 (1+2 p) x^2\right )}{a^4 \left (3+8 p+4 p^2\right )}-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \] Input:

Integrate[(x^3*(1 - a^2*x^2)^p)/E^ArcTanh[a*x],x]
 

Output:

-(((1 - a^2*x^2)^(1/2 + p)*(2 + a^2*(1 + 2*p)*x^2))/(a^4*(3 + 8*p + 4*p^2) 
)) - (a*x^5*Hypergeometric2F1[5/2, 1/2 - p, 7/2, a^2*x^2])/5
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6699, 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.

\(\displaystyle \int x^3 e^{-\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 6699

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

\(\Big \downarrow \) 542

\(\displaystyle \int x^3 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx-a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\)

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 53

\(\displaystyle \frac {1}{2} \int \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2-a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {1}{2} \int \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}-\frac {2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}\right )-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right )\)

Input:

Int[(x^3*(1 - a^2*x^2)^p)/E^ArcTanh[a*x],x]
 

Output:

((-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 - (a*x^5*Hypergeometric2F1[5/2, 1/2 - p, 7/2, a^2*x^2 
])/5
 

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 6699

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]) && ILtQ[(n - 1)/2, 0] &&  !IntegerQ[p - n/2]
 

Maple [F]

\[\int \frac {x^{3} \left (-a^{2} x^{2}+1\right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^3*(-a^2*x^2+1)^p/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fric 
as")
 

Output:

integral(sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p*x^3/(a*x + 1), x)
 

Sympy [F]

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

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

Output:

Integral(x**3*sqrt(-(a*x - 1)*(a*x + 1))*(-(a*x - 1)*(a*x + 1))**p/(a*x + 
1), x)
 

Maxima [F]

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

integrate(x^3*(-a^2*x^2+1)^p/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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