Integrand size = 31, antiderivative size = 584 \[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {\left (\frac {A}{d}-\frac {B}{e}\right ) x \sqrt {a-c x^4}}{3 \left (d+e x^2\right )^{3/2}}-\frac {\left (3 B c d^3-a B d e^2-2 a A e^3\right ) \sqrt {a-c x^4}}{3 d e^2 \left (c d^2-a e^2\right ) x \sqrt {d+e x^2}}-\frac {\sqrt {c} \left (3 B c d^3-a B d e^2-2 a A e^3\right ) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} E\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right )|\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{3 d^2 e^2 \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {d+e x^2} \sqrt {a-c x^4}}+\frac {\sqrt {a} \sqrt {c} (B d+2 A e) \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{3 d^2 e \sqrt {d+e x^2} \sqrt {a-c x^4}}-\frac {B c \sqrt {1-\frac {a}{c x^4}} x^3 \sqrt {\frac {\sqrt {a} \left (d+e x^2\right )}{\left (\sqrt {c} d+\sqrt {a} e\right ) x^2}} \operatorname {EllipticPi}\left (2,\arcsin \left (\frac {\sqrt {1-\frac {\sqrt {a}}{\sqrt {c} x^2}}}{\sqrt {2}}\right ),\frac {2 d}{d+\frac {\sqrt {a} e}{\sqrt {c}}}\right )}{e^2 \sqrt {d+e x^2} \sqrt {a-c x^4}} \] Output:
1/3*(A/d-B/e)*x*(-c*x^4+a)^(1/2)/(e*x^2+d)^(3/2)-1/3*(-2*A*a*e^3-B*a*d*e^2 +3*B*c*d^3)*(-c*x^4+a)^(1/2)/d/e^2/(-a*e^2+c*d^2)/x/(e*x^2+d)^(1/2)-1/3*c^ (1/2)*(-2*A*a*e^3-B*a*d*e^2+3*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))/d^2/e^2/(c^(1/2) *d-a^(1/2)*e)/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)+1/3*a^(1/2)*c^(1/2)*(2*A*e+ 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)*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))/d^2/e/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)-B*c*(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)*Ellipti cPi(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^2/(e*x^2+d)^(1/2)/(-c*x^4+a)^(1/2)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx \] Input:
Integrate[((A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(5/2),x]
Output:
Integrate[((A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(5/2), 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.
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right )}{\left (d+e x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2261 |
\(\displaystyle \int \frac {\sqrt {a-c x^4} \left (A+B x^2\right )}{\left (d+e x^2\right )^{5/2}}dx\) |
Input:
Int[((A + B*x^2)*Sqrt[a - c*x^4])/(d + e*x^2)^(5/2),x]
Output:
$Aborted
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Unintegrable[Px*(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x] && PolyQ[Px, x]
\[\int \frac {\left (B \,x^{2}+A \right ) \sqrt {-c \,x^{4}+a}}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Input:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x)
Output:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x, algorithm="fricas" )
Output:
integral(sqrt(-c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/(e^3*x^6 + 3*d*e^2*x ^4 + 3*d^2*e*x^2 + d^3), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {a - c x^{4}}}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x**2+A)*(-c*x**4+a)**(1/2)/(e*x**2+d)**(5/2),x)
Output:
Integral((A + B*x**2)*sqrt(a - c*x**4)/(d + e*x**2)**(5/2), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x, algorithm="maxima" )
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)/(e*x^2 + d)^(5/2), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {\sqrt {-c x^{4} + a} {\left (B x^{2} + A\right )}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x, algorithm="giac")
Output:
integrate(sqrt(-c*x^4 + a)*(B*x^2 + A)/(e*x^2 + d)^(5/2), x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {a-c\,x^4}}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \] Input:
int(((A + B*x^2)*(a - c*x^4)^(1/2))/(d + e*x^2)^(5/2),x)
Output:
int(((A + B*x^2)*(a - c*x^4)^(1/2))/(d + e*x^2)^(5/2), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {a-c x^4}}{\left (d+e x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
int((B*x^2+A)*(-c*x^4+a)^(1/2)/(e*x^2+d)^(5/2),x)
Output:
( - sqrt(d + e*x**2)*sqrt(a - c*x**4)*b*x - 2*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a*d**3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x**4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2*x**8 - c*e**3*x**10),x)*b*c* d**2*e - 4*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**6)/(a*d**3 + 3*a*d**2 *e*x**2 + 3*a*d*e**2*x**4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2*x**8 - c*e**3*x**10),x)*b*c*d*e**2*x**2 - 2*int((sqrt(d + e*x** 2)*sqrt(a - c*x**4)*x**6)/(a*d**3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x**4 + a* e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2*x**8 - c*e**3*x**10 ),x)*b*c*e**3*x**4 - 2*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(a*d** 3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x**4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d* *2*e*x**6 - 3*c*d*e**2*x**8 - c*e**3*x**10),x)*a*c*d**2*e - 4*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(a*d**3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x* *4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2*x**8 - c*e** 3*x**10),x)*a*c*d*e**2*x**2 - 2*int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x** 4)/(a*d**3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x**4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2*x**8 - c*e**3*x**10),x)*a*c*e**3*x**4 - 3* int((sqrt(d + e*x**2)*sqrt(a - c*x**4)*x**4)/(a*d**3 + 3*a*d**2*e*x**2 + 3 *a*d*e**2*x**4 + a*e**3*x**6 - c*d**3*x**4 - 3*c*d**2*e*x**6 - 3*c*d*e**2* x**8 - c*e**3*x**10),x)*b*c*d**3 - 6*int((sqrt(d + e*x**2)*sqrt(a - c*x**4 )*x**4)/(a*d**3 + 3*a*d**2*e*x**2 + 3*a*d*e**2*x**4 + a*e**3*x**6 - c*d...