Integrand size = 34, antiderivative size = 503 \[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\frac {e^2 (B d-A e) x \sqrt {a-c x^4}}{d^4 \sqrt {d+e x^2}}-\frac {A \sqrt {d+e x^2} \sqrt {a-c x^4}}{5 d^2 x^5}-\frac {(5 B d-9 A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{15 d^3 x^3}-\frac {e (B d-A e) \sqrt {d+e x^2} \sqrt {a-c x^4}}{d^4 x}-\frac {2 c \left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \left (20 a B d e+3 A \left (c d^2-8 a e^2\right )\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{15 a d^4 \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {2 \sqrt {c} \left (5 B c d^3-9 A c d^2 e-20 a B d e^2+24 a A e^3\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{15 \sqrt {a} d^4 \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
e^2*(-A*e+B*d)*x*(-c*x^4+a)^(1/2)/d^4/(e*x^2+d)^(1/2)-1/5*A*(e*x^2+d)^(1/2 )*(-c*x^4+a)^(1/2)/d^2/x^5-1/15*(-9*A*e+5*B*d)*(e*x^2+d)^(1/2)*(-c*x^4+a)^ (1/2)/d^3/x^3-e*(-A*e+B*d)*(e*x^2+d)^(1/2)*(-c*x^4+a)^(1/2)/d^4/x-2/15*c*( d+a^(1/2)*e/c^(1/2))*(20*a*B*d*e+3*A*(-8*a*e^2+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)*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^4/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-2/15*c^(1/2)*(24*A*a*e^3-9*A*c*d^2* e-20*B*a*d*e^2+5*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^4/(e*x^2+d)^(1/2)/( -c*x^4+a)^(1/2)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx \] Input:
Integrate[((A + B*x^2)*Sqrt[a - c*x^4])/(x^6*(d + e*x^2)^(3/2)),x]
Output:
Integrate[((A + B*x^2)*Sqrt[a - c*x^4])/(x^6*(d + e*x^2)^(3/2)), 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.
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right )}{x^6 \left (d+e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2251 |
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right )}{x^6 \left (d+e x^2\right )^{3/2}}dx\) |
Input:
Int[((A + B*x^2)*Sqrt[a - c*x^4])/(x^6*(d + e*x^2)^(3/2)),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]
\[\int \frac {\left (B \,x^{2}+A \right ) \sqrt {-c \,x^{4}+a}}{x^{6} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Input:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),x)
Output:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),x, algorithm="fri cas")
Output:
integral(sqrt(-c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/(e^2*x^10 + 2*d*e*x^ 8 + d^2*x^6), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {a - c x^{4}}}{x^{6} \left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((B*x**2+A)*(-c*x**4+a)**(1/2)/x**6/(e*x**2+d)**(3/2),x)
Output:
Integral((A + B*x**2)*sqrt(a - c*x**4)/(x**6*(d + e*x**2)**(3/2)), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),x, algorithm="max ima")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)/((e*x^2 + d)^(3/2)*x^6), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{6}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),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. \[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}}{x^6\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \] Input:
int(((A + B*x^2)*(a - c*x^4)^(1/2))/(x^6*(d + e*x^2)^(3/2)),x)
Output:
int(((A + B*x^2)*(a - c*x^4)^(1/2))/(x^6*(d + e*x^2)^(3/2)), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{x^6 \left (d+e x^2\right )^{3/2}} \, dx=\text {too large to display} \] Input:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/x^6/(e*x^2+d)^(3/2),x)
Output:
( - 6*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**3*d*e**2 - 3*sqrt(d + e*x**2)*s qrt(a - c*x**4)*a**2*c*d**2*e*x**2 - 6*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a **2*c*d*e**2*x**4 + 12*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a**2*c*e**3*x**6 + 5*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*b*c*d**3*x**2 + 10*sqrt(d + e*x**2 )*sqrt(a - c*x**4)*a*b*c*d**2*e*x**4 - 10*sqrt(d + e*x**2)*sqrt(a - c*x**4 )*a*b*c*d*e**2*x**6 + 3*sqrt(d + e*x**2)*sqrt(a - c*x**4)*a*c**2*d**2*e*x* *6 - 5*sqrt(d + e*x**2)*sqrt(a - c*x**4)*b*c**2*d**3*x**6 + 24*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**2*c**2*d*e**4*x**5 + 24*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**2*c**2*e**5*x**7 - 20*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**2*d**2*e* *3*x**5 - 20*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* *2*d*e**4*x**7 + 6*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**3*d**3*e**2*x**5 + 6*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**3*d**2*e**3*x**7 - 10*int((sqrt(d + e*x**2)*sqrt(a - c*x**4...