\(\int e^{n \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\) [1351]

Optimal result

Integrand size = 27, antiderivative size = 244 \[ \int e^{n \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {n (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{12 a^3 \sqrt {1-a^2 x^2}}-\frac {x (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{4 a^2 \sqrt {1-a^2 x^2}}-\frac {2^{\frac {1}{2} (-1+n)} \left (3+n^2\right ) (1-a x)^{\frac {3-n}{2}} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{3 a^3 (3-n) \sqrt {1-a^2 x^2}} \] Output:

-1/12*n*(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(-a^2*c*x^2+c)^(1/2)/a^3/ 
(-a^2*x^2+1)^(1/2)-1/4*x*(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(-a^2*c* 
x^2+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)-1/3*2^(-1/2+1/2*n)*(n^2+3)*(-a*x+1)^(3 
/2-1/2*n)*(-a^2*c*x^2+c)^(1/2)*hypergeom([3/2-1/2*n, -1/2-1/2*n],[5/2-1/2* 
n],-1/2*a*x+1/2)/a^3/(3-n)/(-a^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.55 \[ \int e^{n \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {(1-a x)^{\frac {3}{2}-\frac {n}{2}} \sqrt {c-a^2 c x^2} \left (-\left ((-3+n) (1+a x)^{\frac {3+n}{2}} (n+3 a x)\right )+2^{\frac {3+n}{2}} \left (3+n^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {5}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )\right )}{12 a^3 (-3+n) \sqrt {1-a^2 x^2}} \] Input:

Integrate[E^(n*ArcTanh[a*x])*x^2*Sqrt[c - a^2*c*x^2],x]
 

Output:

((1 - a*x)^(3/2 - n/2)*Sqrt[c - a^2*c*x^2]*(-((-3 + n)*(1 + a*x)^((3 + n)/ 
2)*(n + 3*a*x)) + 2^((3 + n)/2)*(3 + n^2)*Hypergeometric2F1[-1/2 - n/2, 3/ 
2 - n/2, 5/2 - n/2, 1/2 - (a*x)/2]))/(12*a^3*(-3 + n)*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.80, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6703, 6700, 101, 25, 90, 79}

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.

\(\displaystyle \int x^2 \sqrt {c-a^2 c x^2} e^{n \text {arctanh}(a x)} \, dx\)

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{n \text {arctanh}(a x)} x^2 \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 79

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

Input:

Int[E^(n*ArcTanh[a*x])*x^2*Sqrt[c - a^2*c*x^2],x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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])
 

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 [F]

\[\int {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{2} \sqrt {-a^{2} c \,x^{2}+c}d x\]

Input:

int(exp(n*arctanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x)
 

Output:

int(exp(n*arctanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x)
 

Fricas [F]

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

integrate(exp(n*arctanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

\[ \int e^{n \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int x^{2} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} e^{n \operatorname {atanh}{\left (a x \right )}}\, dx \] Input:

integrate(exp(n*atanh(a*x))*x**2*(-a**2*c*x**2+c)**(1/2),x)
 

Output:

Integral(x**2*sqrt(-c*(a*x - 1)*(a*x + 1))*exp(n*atanh(a*x)), x)
 

Maxima [F]

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

integrate(exp(n*arctanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm="maxim 
a")
 

Output:

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

Giac [F]

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

integrate(exp(n*arctanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm="giac" 
)
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int e^{n \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int x^2\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,\sqrt {c-a^2\,c\,x^2} \,d x \] Input:

int(x^2*exp(n*atanh(a*x))*(c - a^2*c*x^2)^(1/2),x)
 

Output:

int(x^2*exp(n*atanh(a*x))*(c - a^2*c*x^2)^(1/2), x)
 

Reduce [F]

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

int(exp(n*atanh(a*x))*x^2*(-a^2*c*x^2+c)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

sqrt(c)*int(e**(atanh(a*x)*n)*sqrt( - a**2*x**2 + 1)*x**2,x)