Integrand size = 34, antiderivative size = 637 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=-\frac {A d \sqrt {d+e x^2} \sqrt {a-c x^4}}{7 x^7}-\frac {(7 B d+8 A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{35 x^5}+\frac {\left (10 A c d^2-42 a B d e-3 a A e^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{105 a d x^3}-\frac {c \left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \left (42 B c d^3+58 A c d^2 e-21 a B d e^2+6 a A e^3\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{105 a d^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {\sqrt {c} \left (21 a B d e \left (4 c d^2+a e^2\right )+2 A \left (5 c^2 d^4-2 a c d^2 e^2-3 a^2 e^4\right )\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{105 a^{3/2} d^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {B c e^2 \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{\sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
-1/7*A*d*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^7-1/35*(8*A*e+7*B*d)*(e*x^2+d) ^(1/2)*(-c*x^4+a)^(1/2)/x^5+1/105*(-3*A*a*e^2+10*A*c*d^2-42*B*a*d*e)*(e*x^ 2+d)^(1/2)*(-c*x^4+a)^(1/2)/a/d/x^3-1/105*c*(d+a^(1/2)*e/c^(1/2))*(6*A*a*e ^3+58*A*c*d^2*e-21*B*a*d*e^2+42*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)*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^2/(e*x^2+d )^(1/2)/(-c*x^4+a)^(1/2)-1/105*c^(1/2)*(21*a*B*d*e*(a*e^2+4*c*d^2)+2*A*(-3 *a^2*e^4-2*a*c*d^2*e^2+5*c^2*d^4))*(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^(3/2)/d^2/(e*x^2+d )^(1/2)/(-c*x^4+a)^(1/2)-B*c*e^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*x^2+d)^(1/2)/(-c *x^4+a)^(1/2)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx \] Input:
Integrate[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^8,x]
Output:
Integrate[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^8, 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.
| \(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \left (d+e x^2\right )^{3/2}}{x^8} \, dx\) |
| \(\Big \downarrow \) 2251 |
| \(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \left (d+e x^2\right )^{3/2}}{x^8}dx\) |
Input:
Int[((A + B*x^2)*(d + e*x^2)^(3/2)*Sqrt[a - c*x^4])/x^8,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]
\[\int \frac {\left (B \,x^{2}+A \right ) \left (e \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {-c \,x^{4}+a}}{x^{8}}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,x)
Output:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,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^8, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {a - c x^{4}} \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{8}}\, dx \] Input:
integrate((B*x**2+A)*(e*x**2+d)**(3/2)*(-c*x**4+a)**(1/2)/x**8,x)
Output:
Integral((A + B*x**2)*sqrt(a - c*x**4)*(d + e*x**2)**(3/2)/x**8, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,x, algorithm="max ima")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*(e*x^2 + d)^(3/2)/x^8, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,x, algorithm="gia c")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*(e*x^2 + d)^(3/2)/x^8, x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}\,{\left (e\,x^2+d\right )}^{3/2}}{x^8} \,d x \] Input:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(3/2))/x^8,x)
Output:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(3/2))/x^8, x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{3/2} \sqrt {a-c x^4}}{x^8} \, dx=\text {too large to display} \] Input:
int((B*x^2+A)*(e*x^2+d)^(3/2)*(-c*x^4+a)^(1/2)/x^8,x)
Output:
(336*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**4*e**6*x**2 + 840*sqrt(d + e*x** 2)*sqrt(a - c*x**4)*a**3*b*d*e**5*x**2 - 120*sqrt(d + e*x**2)*sqrt(a - c*x **4)*a**3*c*d**3*e**3 - 532*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**3*c*d**2* e**4*x**2 + 1344*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**3*c*d*e**5*x**4 - 16 80*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**3*c*e**6*x**6 - 1078*sqrt(d + e*x* *2)*sqrt(a - c*x**4)*a**2*b*c*d**3*e**3*x**2 + 2940*sqrt(d + e*x**2)*sqrt( a - c*x**4)*a**2*b*c*d**2*e**4*x**4 - 4200*sqrt(d + e*x**2)*sqrt(a - c*x** 4)*a**2*b*c*d*e**5*x**6 - 150*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**2*c**2* d**5*e - 918*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**2*c**2*d**4*e**2*x**2 + 1752*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**2*c**2*d**3*e**3*x**4 - 3900*sqr t(d + e*x**2)*sqrt(a - c*x**4)*a**2*c**2*d**2*e**4*x**6 - 1960*sqrt(d + e* x**2)*sqrt(a - c*x**4)*a*b*c**2*d**5*e*x**2 + 3003*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*b*c**2*d**4*e**2*x**4 - 9450*sqrt(d + e*x**2)*sqrt(a - c*x**4 )*a*b*c**2*d**3*e**3*x**6 - 150*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*c**3*d **6*x**2 + 678*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*c**3*d**5*e*x**4 - 2610 *sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*c**3*d**4*e**2*x**6 + 210*sqrt(d + e* x**2)*sqrt(a - c*x**4)*b*c**3*d**6*x**4 - 5250*sqrt(d + e*x**2)*sqrt(a - c *x**4)*b*c**3*d**5*e*x**6 - 450*sqrt(d + e*x**2)*sqrt(a - c*x**4)*c**4*d** 6*x**6 - 13440*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 + 5*a*c*d**3 + 5*a*c*d**2*e*x**2 - 4*a*c*d*e**2*x**...