Integrand size = 35, antiderivative size = 690 \[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {B \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}}{2 c x}-\frac {B \sqrt {b^2-4 a c} \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} x \sqrt {d+e x^2} E\left (\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{2 \sqrt {2} c \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt {b^2-4 a c} (2 A c d-a B e) \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} x^3 \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{\sqrt {2} a c \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} (B c d-b B e+2 A c e) \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} x^3 \operatorname {EllipticPi}\left (\frac {2 \sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}},\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{c \left (b+\sqrt {b^2-4 a c}\right ) \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \] Output:
1/2*B*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/c/x-1/4*B*(-4*a*c+b^2)^(1/2)*( -a*(c+a/x^4+b/x^2)/(-4*a*c+b^2))^(1/2)*x*(e*x^2+d)^(1/2)*EllipticE(1/2*(1+ (b+2*a/x^2)/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2)*((-4*a*c+b^2)^(1/2)* d/(b*d+(-4*a*c+b^2)^(1/2)*d-2*a*e))^(1/2))*2^(1/2)/c/(-a*(e+d/x^2)/((b+(-4 *a*c+b^2)^(1/2))*d-2*a*e))^(1/2)/(c*x^4+b*x^2+a)^(1/2)-1/2*(-4*a*c+b^2)^(1 /2)*(2*A*c*d-B*a*e)*(-a*(c+a/x^4+b/x^2)/(-4*a*c+b^2))^(1/2)*(-a*(e+d/x^2)/ ((b+(-4*a*c+b^2)^(1/2))*d-2*a*e))^(1/2)*x^3*EllipticF(1/2*(1+(b+2*a/x^2)/( -4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2)*((-4*a*c+b^2)^(1/2)*d/(b*d+(-4*a* c+b^2)^(1/2)*d-2*a*e))^(1/2))*2^(1/2)/a/c/(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^ (1/2)+2^(1/2)*(-4*a*c+b^2)^(1/2)*(2*A*c*e-B*b*e+B*c*d)*(-a*(c+a/x^4+b/x^2) /(-4*a*c+b^2))^(1/2)*(-a*(e+d/x^2)/((b+(-4*a*c+b^2)^(1/2))*d-2*a*e))^(1/2) *x^3*EllipticPi(1/2*(1+(b+2*a/x^2)/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2*(-4 *a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2)),2^(1/2)*((-4*a*c+b^2)^(1/2)*d/(b*d+ (-4*a*c+b^2)^(1/2)*d-2*a*e))^(1/2))/c/(b+(-4*a*c+b^2)^(1/2))/(e*x^2+d)^(1/ 2)/(c*x^4+b*x^2+a)^(1/2)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx \] Input:
Integrate[((A + B*x^2)*Sqrt[d + e*x^2])/Sqrt[a + b*x^2 + c*x^4],x]
Output:
Integrate[((A + B*x^2)*Sqrt[d + e*x^2])/Sqrt[a + b*x^2 + c*x^4], 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 2260 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}}dx\) |
Input:
Int[((A + B*x^2)*Sqrt[d + e*x^2])/Sqrt[a + b*x^2 + c*x^4],x]
Output:
$Aborted
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_.), x_Symbol] :> Unintegrable[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x] /; FreeQ[{a, b, c, d, e, p, q}, x] && PolyQ[Px, x]
\[\int \frac {\left (B \,x^{2}+A \right ) \sqrt {e \,x^{2}+d}}{\sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]
Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x)
Output:
int((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\text {Timed out} \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fr icas")
Output:
Timed out
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {\left (A + B x^{2}\right ) \sqrt {d + e x^{2}}}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \] Input:
integrate((B*x**2+A)*(e*x**2+d)**(1/2)/(c*x**4+b*x**2+a)**(1/2),x)
Output:
Integral((A + B*x**2)*sqrt(d + e*x**2)/sqrt(a + b*x**2 + c*x**4), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {e x^{2} + d}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="ma xima")
Output:
integrate((B*x^2 + A)*sqrt(e*x^2 + d)/sqrt(c*x^4 + b*x^2 + a), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \sqrt {e x^{2} + d}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="gi ac")
Output:
integrate((B*x^2 + A)*sqrt(e*x^2 + d)/sqrt(c*x^4 + b*x^2 + a), x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {e\,x^2+d}}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:
int(((A + B*x^2)*(d + e*x^2)^(1/2))/(a + b*x^2 + c*x^4)^(1/2),x)
Output:
int(((A + B*x^2)*(d + e*x^2)^(1/2))/(a + b*x^2 + c*x^4)^(1/2), x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {d+e x^2}}{\sqrt {a+b x^2+c x^4}} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b +\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a \] Input:
int((B*x^2+A)*(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x)
Output:
int((sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4 ),x)*b + int((sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4))/(a + b*x**2 + c* x**4),x)*a