Integrand size = 35, antiderivative size = 470 \[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\frac {(B d-A e) \sqrt {a+b x^2+c x^4}}{\left (c d^2-b d e+a e^2\right ) x \sqrt {d+e x^2}}-\frac {\sqrt {b^2-4 a c} (B d-A e) \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} x \sqrt {d+e x^2} E\left (\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{\sqrt {2} d \left (c d^2-b d e+a e^2\right ) \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt {2} A \sqrt {b^2-4 a c} \sqrt {-\frac {a \left (c+\frac {a}{x^4}+\frac {b}{x^2}\right )}{b^2-4 a c}} \sqrt {-\frac {a \left (e+\frac {d}{x^2}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d-2 a e}} x^3 \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {b+\frac {2 a}{x^2}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {b^2-4 a c} d}{b d+\sqrt {b^2-4 a c} d-2 a e}\right )}{a d \sqrt {d+e x^2} \sqrt {a+b x^2+c x^4}} \] Output:
(-A*e+B*d)*(c*x^4+b*x^2+a)^(1/2)/(a*e^2-b*d*e+c*d^2)/x/(e*x^2+d)^(1/2)-1/2 *(-4*a*c+b^2)^(1/2)*(-A*e+B*d)*(-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)/d/(a*e^2-b*d*e+c*d^2)/(-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)-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/ d/(e*x^2+d)^(1/2)/(c*x^4+b*x^2+a)^(1/2)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx \] Input:
Integrate[(A + B*x^2)/((d + e*x^2)^(3/2)*Sqrt[a + b*x^2 + c*x^4]),x]
Output:
Integrate[(A + B*x^2)/((d + e*x^2)^(3/2)*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.
\(\displaystyle \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 2260 |
\(\displaystyle \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}}dx\) |
Input:
Int[(A + B*x^2)/((d + e*x^2)^(3/2)*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]
\[\int \frac {B \,x^{2}+A}{\left (e \,x^{2}+d \right )^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]
Input:
int((B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^(1/2),x)
Output:
int((B*x^2+A)/(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a)^(1/2),x)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x^2+d)^(3/2)/(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^2*x^8 + (2*c*d*e + b*e^2)*x^6 + (c*d^2 + 2*b*d*e + a*e^2)*x^4 + a*d^2 + (b*d^2 + 2 *a*d*e)*x^2), x)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{\left (d + e x^{2}\right )^{\frac {3}{2}} \sqrt {a + b x^{2} + c x^{4}}}\, dx \] Input:
integrate((B*x**2+A)/(e*x**2+d)**(3/2)/(c*x**4+b*x**2+a)**(1/2),x)
Output:
Integral((A + B*x**2)/((d + e*x**2)**(3/2)*sqrt(a + b*x**2 + c*x**4)), x)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x^2+d)^(3/2)/(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)*(e*x^2 + d)^(3/2)), x)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((B*x^2+A)/(e*x^2+d)^(3/2)/(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)*(e*x^2 + d)^(3/2)), x)
Timed out. \[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {B\,x^2+A}{{\left (e\,x^2+d\right )}^{3/2}\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:
int((A + B*x^2)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^(1/2)),x)
Output:
int((A + B*x^2)/((d + e*x^2)^(3/2)*(a + b*x^2 + c*x^4)^(1/2)), x)
\[ \int \frac {A+B x^2}{\left (d+e x^2\right )^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,e^{2} x^{8}+b \,e^{2} x^{6}+2 c d e \,x^{6}+a \,e^{2} x^{4}+2 b d e \,x^{4}+c \,d^{2} x^{4}+2 a d e \,x^{2}+b \,d^{2} x^{2}+a \,d^{2}}d x \right ) b +\left (\int \frac {\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,e^{2} x^{8}+b \,e^{2} x^{6}+2 c d e \,x^{6}+a \,e^{2} x^{4}+2 b d e \,x^{4}+c \,d^{2} x^{4}+2 a d e \,x^{2}+b \,d^{2} x^{2}+a \,d^{2}}d x \right ) a \] Input:
int((B*x^2+A)/(e*x^2+d)^(3/2)/(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**2 + 2*a*d*e*x* *2 + a*e**2*x**4 + b*d**2*x**2 + 2*b*d*e*x**4 + b*e**2*x**6 + c*d**2*x**4 + 2*c*d*e*x**6 + c*e**2*x**8),x)*b + int((sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4))/(a*d**2 + 2*a*d*e*x**2 + a*e**2*x**4 + b*d**2*x**2 + 2*b*d*e*x **4 + b*e**2*x**6 + c*d**2*x**4 + 2*c*d*e*x**6 + c*e**2*x**8),x)*a