Integrand size = 34, antiderivative size = 551 \[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=-\frac {A \sqrt {d+e x^2} \sqrt {a-c x^4}}{5 x^5}-\frac {(5 B d+A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{15 d x^3}-\frac {c \left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \left (6 A c d^2-5 a B d e+2 a A e^2\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{15 a d^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {\sqrt {c} \left (10 B c d^3+2 A c d^2 e+5 a B d e^2-2 a A e^3\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{15 \sqrt {a} d^2 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {B c e \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{\sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
-1/5*A*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^5-1/15*(A*e+5*B*d)*(e*x^2+d)^(1/ 2)*(-c*x^4+a)^(1/2)/d/x^3-1/15*c*(d+a^(1/2)*e/c^(1/2))*(2*A*a*e^2+6*A*c*d^ 2-5*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^2/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-1/1 5*c^(1/2)*(-2*A*a*e^3+2*A*c*d^2*e+5*B*a*d*e^2+10*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^2/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-B*c*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)*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)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx \] Input:
Integrate[((A + B*x^2)*Sqrt[d + e*x^2]*Sqrt[a - c*x^4])/x^6,x]
Output:
Integrate[((A + B*x^2)*Sqrt[d + e*x^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.
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \sqrt {d+e x^2}}{x^6} \, dx\) |
\(\Big \downarrow \) 2251 |
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right ) \sqrt {d+e x^2}}{x^6}dx\) |
Input:
Int[((A + B*x^2)*Sqrt[d + e*x^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]
\[\int \frac {\left (B \,x^{2}+A \right ) \sqrt {e \,x^{2}+d}\, \sqrt {-c \,x^{4}+a}}{x^{6}}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x)
Output:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="fri cas")
Output:
integral(sqrt(-c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {a - c x^{4}} \sqrt {d + e x^{2}}}{x^{6}}\, dx \] Input:
integrate((B*x**2+A)*(e*x**2+d)**(1/2)*(-c*x**4+a)**(1/2)/x**6,x)
Output:
Integral((A + B*x**2)*sqrt(a - c*x**4)*sqrt(d + e*x**2)/x**6, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="max ima")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )} \sqrt {e x^{2} + d}}{x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x, algorithm="gia c")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/x^6, x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}\,\sqrt {e\,x^2+d}}{x^6} \,d x \] Input:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(1/2))/x^6,x)
Output:
int(((A + B*x^2)*(a - c*x^4)^(1/2)*(d + e*x^2)^(1/2))/x^6, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}{x^6} \, dx =\text {Too large to display} \] Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/x^6,x)
Output:
( - 2*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*e - 5*sqrt(d + e*x**2)*sqrt(a - c*x**4)*b*e*x**2 + 2*sqrt(d + e*x**2)*sqrt(a - c*x**4)*c*d*x**2 - 40*int(( sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**2)/(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*e**4*x**5 - 30*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**2)/(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**2*e**2*x**5 + 8*int((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*e**4*x**5 - 20*int((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**2*b*d*e**3*x**5 + 30*int((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**2*c*d**2*e**2*x**5 - 15*int((s qrt(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**1 0 - 3*c**2*d**3*x**8 - 3*c**2*d**2*e*x**10),x)*a*b*c*d**3*e*x**5 + 18*i...