Integrand size = 34, antiderivative size = 592 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=-\frac {A d \sqrt {d+e x^2} \sqrt {a-c x^4}}{5 x^5}-\frac {(5 B d+6 A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{15 x^3}+\frac {B e \sqrt {d+e x^2} \sqrt {a-c x^4}}{2 x}-\frac {c \left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \left (12 A c d^2-55 a B d e-6 a A e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{30 a d \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {\sqrt {c} \left (20 B c d^3+24 A c d^2 e+25 a B d e^2+6 a A e^3\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{30 \sqrt {a} d \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {c e (3 B d+2 A e) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{2 \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
-1/5*A*d*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^5-1/15*(6*A*e+5*B*d)*(e*x^2+d) ^(1/2)*(-c*x^4+a)^(1/2)/x^3+1/2*B*e*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x-1/3 0*c*(d+a^(1/2)*e/c^(1/2))*(-6*A*a*e^2+12*A*c*d^2-55*B*a*d*e)*(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))/a/d/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-1/30*c^(1/2)*(6*A*a*e^3+24*A*c* d^2*e+25*B*a*d*e^2+20*B*c*d^3)*(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)*EllipticF(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))/a^(1/2)/d/(e*x^2+d)^(1/2 )/(-c*x^4+a)^(1/2)-1/2*c*e*(2*A*e+3*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)*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*x^2+d)^( 1/2)/(-c*x^4+a)^(1/2)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx \] Input:
Integrate[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^6,x]
Output:
Integrate[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^6, x]
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.
| \(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \left (d+e x^2\right )^{3/2}}{x^6} \, dx\) |
| \(\Big \downarrow \) 2251 |
| \(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \left (d+e x^2\right )^{3/2}}{x^6}dx\) |
Input:
Int[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^6,x]
Output:
$Aborted
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]
\[\int \frac {\left (B \,x^{2}+A \right ) \left (e \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {-c \,x^{4}+a}}{x^{6}}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x)
Output:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="fri cas")
Output:
integral((B*e*x^4 + (B*d + A*e)*x^2 + A*d)*sqrt(-c*x^4 + a)*sqrt(e*x^2 + d )/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {a - c x^{4}} \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \] Input:
integrate((B*x**2+A)*(e*x**2+d)**(3/2)*(-c*x**4+a)**(1/2)/x**6,x)
Output:
Integral((A + B*x**2)*sqrt(a - c*x**4)*(d + e*x**2)**(3/2)/x**6, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="max ima")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*(e*x^2 + d)^(3/2)/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="gia c")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*(e*x^2 + d)^(3/2)/x^6, x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}\,{\left (e\,x^2+d\right )}^{3/2}}{x^6} \,d x \] Input:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(3/2))/x^6,x)
Output:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(3/2))/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^6} \, dx=\text {too large to display} \] Input:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^6,x)
Output:
( - sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*d**2*e - 5*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*d*e**2*x**2 + 5*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*e**3*x**4 - 10*sqrt(d + e*x**2)*sqrt(a - c*x**4)*b*d**2*e*x**2 + 10*sqrt(d + e*x**2 )*sqrt(a - c*x**4)*b*d*e**2*x**4 + sqrt(d + e*x**2)*sqrt(a - c*x**4)*c*d** 3*x**2 + 40*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a**2*d*e**2 + 4*a**2*e**3*x**2 + 3*a*c*d**3 + 3*a*c*d**2*e*x**2 - 4*a*c*d*e**2*x**4 - 4* a*c*e**3*x**6 - 3*c**2*d**3*x**4 - 3*c**2*d**2*e*x**6),x)*a**2*c*e**6*x**5 + 60*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a**2*d*e**2 + 4*a**2 *e**3*x**2 + 3*a*c*d**3 + 3*a*c*d**2*e*x**2 - 4*a*c*d*e**2*x**4 - 4*a*c*e* *3*x**6 - 3*c**2*d**3*x**4 - 3*c**2*d**2*e*x**6),x)*a*b*c*d*e**5*x**5 + 30 *int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a**2*d*e**2 + 4*a**2*e**3 *x**2 + 3*a*c*d**3 + 3*a*c*d**2*e*x**2 - 4*a*c*d*e**2*x**4 - 4*a*c*e**3*x* *6 - 3*c**2*d**3*x**4 - 3*c**2*d**2*e*x**6),x)*a*c**2*d**2*e**4*x**5 + 45* int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(4*a**2*d*e**2 + 4*a**2*e**3* x**2 + 3*a*c*d**3 + 3*a*c*d**2*e*x**2 - 4*a*c*d*e**2*x**4 - 4*a*c*e**3*x** 6 - 3*c**2*d**3*x**4 - 3*c**2*d**2*e*x**6),x)*b*c**2*d**3*e**3*x**5 - 36*i nt((sqrt(d + e*x**2)*sqrt(a - c*x**4))/(4*a**2*d*e**2*x**4 + 4*a**2*e**3*x **6 + 3*a*c*d**3*x**4 + 3*a*c*d**2*e*x**6 - 4*a*c*d*e**2*x**8 - 4*a*c*e**3 *x**10 - 3*c**2*d**3*x**8 - 3*c**2*d**2*e*x**10),x)*a**3*d**2*e**4*x**5 - 100*int((sqrt(d + e*x**2)*sqrt(a - c*x**4))/(4*a**2*d*e**2*x**4 + 4*a**...