\(\int \frac {(A+B x^2) (d+e x^2)^{5/2}}{(a+b x^2+c x^4)^{3/2}} \, dx\) [243]

Optimal result

Integrand size = 35, antiderivative size = 1091 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx =\text {Too large to display} \] Output:

x*(A*b^2-B*a*b-2*A*a*c+(A*b-2*B*a)*c*x^2)*(e*x^2+d)^(5/2)/a/(-4*a*c+b^2)/( 
c*x^4+b*x^2+a)^(1/2)-1/2*(2*A*c*(a*b*e^2-4*a*c*d*e+b*c*d^2)-a*B*(4*c^2*d^2 
+3*b^2*e^2-4*c*e*(2*a*e+b*d)))*(e*x^2+d)^(1/2)*(c*x^4+b*x^2+a)^(1/2)/a/c^2 
/(-4*a*c+b^2)/x-e*(2*A*c*(-a*e+b*d)-a*B*(-b*e+4*c*d))*x*(e*x^2+d)^(1/2)*(c 
*x^4+b*x^2+a)^(1/2)/a/c/(-4*a*c+b^2)-(A*b-2*B*a)*e^2*x^3*(e*x^2+d)^(1/2)*( 
c*x^4+b*x^2+a)^(1/2)/a/(-4*a*c+b^2)+1/4*(2*A*c*(a*b*e^2-4*a*c*d*e+b*c*d^2) 
-a*B*(4*c^2*d^2+3*b^2*e^2-4*c*e*(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)/a/c^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)/(c*x^4+b*x^2+a)^(1/2)+1/2*(3*a*b^2*B*e^3-2*b* 
c*(A*a*e^3+3*A*c*d^2*e+3*B*a*d*e^2+B*c*d^3)+4*c*(a*B*e*(-2*a*e^2+3*c*d^2)+ 
A*c*d*(3*a*e^2+c*d^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))*2^(1/2)/a/c^2/(-4*a*c+b^2)^(1/2)/(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)*e^2*(2*A*c*e-3*B 
*b*e+5*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))...
 

Mathematica [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx \] Input:

Integrate[((A + B*x^2)*(d + e*x^2)^(5/2))/(a + b*x^2 + c*x^4)^(3/2),x]
 

Output:

Integrate[((A + B*x^2)*(d + e*x^2)^(5/2))/(a + b*x^2 + c*x^4)^(3/2), x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2260

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}}dx\)

Input:

Int[((A + B*x^2)*(d + e*x^2)^(5/2))/(a + b*x^2 + c*x^4)^(3/2),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2260

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]
 

Maple [F]

\[\int \frac {\left (B \,x^{2}+A \right ) \left (e \,x^{2}+d \right )^{\frac {5}{2}}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]

Input:

int((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x)
 

Output:

int((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fr 
icas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\left (A + B x^{2}\right ) \left (d + e x^{2}\right )^{\frac {5}{2}}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x**2+A)*(e*x**2+d)**(5/2)/(c*x**4+b*x**2+a)**(3/2),x)
 

Output:

Integral((A + B*x**2)*(d + e*x**2)**(5/2)/(a + b*x**2 + c*x**4)**(3/2), x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="ma 
xima")
 

Output:

integrate((B*x^2 + A)*(e*x^2 + d)^(5/2)/(c*x^4 + b*x^2 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="gi 
ac")
 

Output:

integrate((B*x^2 + A)*(e*x^2 + d)^(5/2)/(c*x^4 + b*x^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x^2+d\right )}^{5/2}}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(((A + B*x^2)*(d + e*x^2)^(5/2))/(a + b*x^2 + c*x^4)^(3/2),x)
 

Output:

int(((A + B*x^2)*(d + e*x^2)^(5/2))/(a + b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^{5/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\text {too large to display} \] Input:

int((B*x^2+A)*(e*x^2+d)^(5/2)/(c*x^4+b*x^2+a)^(3/2),x)
 

Output:

( - 4*sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4)*a**2*e**4*x - 5*sqrt(d + 
e*x**2)*sqrt(a + b*x**2 + c*x**4)*a*b*d*e**3*x + sqrt(d + e*x**2)*sqrt(a + 
 b*x**2 + c*x**4)*a*b*e**4*x**3 + 3*sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x 
**4)*a*c*d**2*e**2*x - sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4)*a*c*d*e* 
*3*x**3 + 3*sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4)*b**2*d**2*e**2*x + 
3*sqrt(d + e*x**2)*sqrt(a + b*x**2 + c*x**4)*b**2*d*e**3*x**3 + 3*sqrt(d + 
 e*x**2)*sqrt(a + b*x**2 + c*x**4)*b*c*d**3*e*x - 3*sqrt(d + e*x**2)*sqrt( 
a + b*x**2 + c*x**4)*b*c*d**2*e**2*x**3 - 6*int((sqrt(d + e*x**2)*sqrt(a + 
 b*x**2 + c*x**4)*x**8)/(3*a**2*b**2*d*e**2 + 3*a**2*b**2*e**3*x**2 - 2*a* 
*2*b*c*d**2*e - 2*a**2*b*c*d*e**2*x**2 - a**2*c**2*d**3 - a**2*c**2*d**2*e 
*x**2 + 6*a*b**3*d*e**2*x**2 + 6*a*b**3*e**3*x**4 - 4*a*b**2*c*d**2*e*x**2 
 + 2*a*b**2*c*d*e**2*x**4 + 6*a*b**2*c*e**3*x**6 - 2*a*b*c**2*d**3*x**2 - 
6*a*b*c**2*d**2*e*x**4 - 4*a*b*c**2*d*e**2*x**6 - 2*a*c**3*d**3*x**4 - 2*a 
*c**3*d**2*e*x**6 + 3*b**4*d*e**2*x**4 + 3*b**4*e**3*x**6 - 2*b**3*c*d**2* 
e*x**4 + 4*b**3*c*d*e**2*x**6 + 6*b**3*c*e**3*x**8 - b**2*c**2*d**3*x**4 - 
 5*b**2*c**2*d**2*e*x**6 - b**2*c**2*d*e**2*x**8 + 3*b**2*c**2*e**3*x**10 
- 2*b*c**3*d**3*x**6 - 4*b*c**3*d**2*e*x**8 - 2*b*c**3*d*e**2*x**10 - c**4 
*d**3*x**8 - c**4*d**2*e*x**10),x)*a**2*b**3*c*e**7 + 10*int((sqrt(d + e*x 
**2)*sqrt(a + b*x**2 + c*x**4)*x**8)/(3*a**2*b**2*d*e**2 + 3*a**2*b**2*e** 
3*x**2 - 2*a**2*b*c*d**2*e - 2*a**2*b*c*d*e**2*x**2 - a**2*c**2*d**3 - ...