Integrand size = 35, antiderivative size = 430 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=-\frac {\sqrt {2} A \sqrt {b^2-4 a c} \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} x^3 \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{a \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}}+\frac {2 \sqrt {2} B \sqrt {b^2-4 a c} \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} x^3 \operatorname {EllipticPi}\left (\frac {2 \sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}},\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \] Output:
-2^(1/2)*A*(-4*a*c+b^2)^(1/2)*(-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))/a/(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^ (1/2)+2*2^(1/2)*B*(-4*a*c+b^2)^(1/2)*(-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))/(b+(-4*a*c+b^2)^(1/2))/(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^( 1/2)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx \] 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.
\(\displaystyle \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 2260 |
\(\displaystyle \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}}dx\) |
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]
\[\int \frac {B \,x^{2}+A}{\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]
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)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {e x^{2} + d}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fr icas")
Output:
integral(sqrt(c*x^4 + b*x^2 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/(c*e*x^6 + (c *d + b*e)*x^4 + (b*d + a*e)*x^2 + a*d), x)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{\sqrt {d + e x^{2}} \sqrt {a + b x^{2} + c x^{4}}}\, dx \] 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)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {e x^{2} + d}} \,d x } \] 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(c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d)), x)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} \sqrt {e x^{2} + d}} \,d x } \] 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(c*x^4 + b*x^2 + a)*sqrt(e*x^2 + d)), x)
Timed out. \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {B\,x^2+A}{\sqrt {e\,x^2+d}\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \] 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)
\[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c e \,x^{6}+b e \,x^{4}+c d \,x^{4}+a e \,x^{2}+b d \,x^{2}+a d}d x \right ) b +\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c e \,x^{6}+b e \,x^{4}+c d \,x^{4}+a e \,x^{2}+b d \,x^{2}+a d}d x \right ) a \] 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*d + a*e*x**2 + b* d*x**2 + b*e*x**4 + c*d*x**4 + c*e*x**6),x)*b + int((sqrt(d + e*x**2)*sqrt (a + b*x**2 + c*x**4))/(a*d + a*e*x**2 + b*d*x**2 + b*e*x**4 + c*d*x**4 + c*e*x**6),x)*a