\(\int \frac {x^2 (A+B x^2) \sqrt {a-c x^4}}{(d+e x^2)^{3/2}} \, dx\) [61]

Optimal result

Integrand size = 34, antiderivative size = 580 \[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {(B d-A e) x \sqrt {a-c x^4}}{e^2 \sqrt {d+e x^2}}-\frac {3 (5 B d-4 A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{8 e^3 x}+\frac {B x \sqrt {d+e x^2} \sqrt {a-c x^4}}{4 e^2}-\frac {3 \sqrt {c} \left (\sqrt {c} d+\sqrt {a} e\right ) (5 B d-4 A e) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{8 e^3 \sqrt {d+e x^2} \sqrt {a-c x^4}}+\frac {\sqrt {a} \sqrt {c} (5 B d-4 A e) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{8 e^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {\left (15 B c d^2-12 A c d e-4 a B e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{8 e^3 \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:

(-A*e+B*d)*x*(-c*x^4+a)^(1/2)/e^2/(e*x^2+d)^(1/2)-3/8*(-4*A*e+5*B*d)*(e*x^ 
2+d)^(1/2)*(-c*x^4+a)^(1/2)/e^3/x+1/4*B*x*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2) 
/e^2-3/8*c^(1/2)*(c^(1/2)*d+a^(1/2)*e)*(-4*A*e+5*B*d)*(1-a/c/x^4)^(1/2)*x^ 
3*(a^(1/2)*(e*x^2+d)/(c^(1/2)*d+a^(1/2)*e)/x^2)^(1/2)*EllipticE(1/2*(1-a^( 
1/2)/c^(1/2)/x^2)^(1/2)*2^(1/2),2^(1/2)*(d/(d+a^(1/2)*e/c^(1/2)))^(1/2))/e 
^3/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)+1/8*a^(1/2)*c^(1/2)*(-4*A*e+5*B*d)*(1- 
a/c/x^4)^(1/2)*x^3*(a^(1/2)*(e*x^2+d)/(c^(1/2)*d+a^(1/2)*e)/x^2)^(1/2)*Ell 
ipticF(1/2*(1-a^(1/2)/c^(1/2)/x^2)^(1/2)*2^(1/2),2^(1/2)*(d/(d+a^(1/2)*e/c 
^(1/2)))^(1/2))/e^2/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-1/8*(-12*A*c*d*e-4*B* 
a*e^2+15*B*c*d^2)*(1-a/c/x^4)^(1/2)*x^3*(a^(1/2)*(e*x^2+d)/(c^(1/2)*d+a^(1 
/2)*e)/x^2)^(1/2)*EllipticPi(1/2*(1-a^(1/2)/c^(1/2)/x^2)^(1/2)*2^(1/2),2,2 
^(1/2)*(d/(d+a^(1/2)*e/c^(1/2)))^(1/2))/e^3/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/ 
2)
 

Mathematica [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx \] Input:

Integrate[(x^2*(A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(3/2),x]
 

Output:

Integrate[(x^2*(A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(3/2), x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {x^2 \sqrt {a-c x^4} \left (A+B x^2\right )}{\left (d+e x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2251

\(\displaystyle \int \frac {x^2 \sqrt {a-c x^4} \left (A+B x^2\right )}{\left (d+e x^2\right )^{3/2}}dx\)

Input:

Int[(x^2*(A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(3/2),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2251

Int[(Px_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Unintegrable[Px*(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p 
, x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && PolyQ[Px, x]
 

Maple [F]

\[\int \frac {x^{2} \left (B \,x^{2}+A \right ) \sqrt {-c \,x^{4}+a}}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]

Input:

int(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x)
 

Output:

int(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x)
 

Fricas [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x, algorithm="fri 
cas")
 

Output:

integral((B*x^4 + A*x^2)*sqrt(-c*x^4 + a)*sqrt(e*x^2 + d)/(e^2*x^4 + 2*d*e 
*x^2 + d^2), x)
 

Sympy [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (A + B x^{2}\right ) \sqrt {a - c x^{4}}}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**2*(B*x**2+A)*(-c*x**4+a)**(1/2)/(e*x**2+d)**(3/2),x)
 

Output:

Integral(x**2*(A + B*x**2)*sqrt(a - c*x**4)/(d + e*x**2)**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x, algorithm="max 
ima")
 

Output:

integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*x^2/(e*x^2 + d)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x, algorithm="gia 
c")
 

Output:

integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*x^2/(e*x^2 + d)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \] Input:

int((x^2*(A + B*x^2)*(a - c*x^4)^(1/2))/(d + e*x^2)^(3/2),x)
 

Output:

int((x^2*(A + B*x^2)*(a - c*x^4)^(1/2))/(d + e*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(x^2*(B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2),x)
 

Output:

( - 2*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*b*e*x + 3*sqrt(d + e*x**2)*sqrt( 
a - c*x**4)*b*c*d*x**3 - 4*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a 
*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e**2*x 
**8),x)*a*b*c*d*e**2 - 4*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a*d 
**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e**2*x** 
8),x)*a*b*c*e**3*x**2 - 12*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a 
*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e**2*x 
**8),x)*a*c**2*d**2*e - 12*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a 
*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e**2*x 
**8),x)*a*c**2*d*e**2*x**2 + 15*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x** 
6)/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e 
**2*x**8),x)*b*c**2*d**3 + 15*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6) 
/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - c*e** 
2*x**8),x)*b*c**2*d**2*e*x**2 + 12*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)* 
x**2)/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - 
c*e**2*x**8),x)*a**2*c*d**2*e + 12*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)* 
x**2)/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x**6 - 
c*e**2*x**8),x)*a**2*c*d*e**2*x**2 - 9*int((sqrt(d + e*x**2)*sqrt(a - c*x* 
*4)*x**2)/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 - c*d**2*x**4 - 2*c*d*e*x** 
6 - c*e**2*x**8),x)*a*b*c*d**3 - 9*int((sqrt(d + e*x**2)*sqrt(a - c*x**...