\(\int \frac {(d+e x^2)^{3/2} (A+B x^2+C x^4)}{(a+c x^4)^2} \, dx\) [20]

Optimal result

Integrand size = 33, antiderivative size = 586 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=-\frac {B e x \sqrt {d+e x^2}}{4 a c}+\frac {x \left (A c-a C+B c x^2\right ) \left (d+e x^2\right )^{3/2}}{4 a c \left (a+c x^4\right )}+\frac {\sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (c d \left (B c d^2+3 A c d e+5 a C d e+2 a B e^2\right )+\left (4 a^2 C e^2-c d (3 A c d+a C d+a B e)\right ) \left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{8 \sqrt {2} a^{5/4} c^{5/2} d \sqrt {c d^2+a e^2}}+\frac {C e^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c^2}+\frac {\sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (c d \left (B c d^2+3 A c d e+5 a C d e+2 a B e^2\right )+\left (4 a^2 C e^2-c d (3 A c d+a C d+a B e)\right ) \left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{8 \sqrt {2} a^{5/4} c^{5/2} d \sqrt {c d^2+a e^2}} \] Output:

-1/4*B*e*x*(e*x^2+d)^(1/2)/a/c+1/4*x*(B*c*x^2+A*c-C*a)*(e*x^2+d)^(3/2)/a/c 
/(c*x^4+a)+1/16*(a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*(c*d*(3*A*c*d*e+2*B* 
a*e^2+B*c*d^2+5*C*a*d*e)+(4*a^2*C*e^2-c*d*(3*A*c*d+B*a*e+C*a*d))*(e-(a*e^2 
+c*d^2)^(1/2)/a^(1/2)))*arctan(2^(1/2)*a^(1/4)*c^(1/2)*(a^(1/2)*e+(a*e^2+c 
*d^2)^(1/2))^(1/2)*x*(e*x^2+d)^(1/2)/(a^(1/2)*(a^(1/2)*e+(a*e^2+c*d^2)^(1/ 
2))-c*d*x^2))*2^(1/2)/a^(5/4)/c^(5/2)/d/(a*e^2+c*d^2)^(1/2)+C*e^(3/2)*arct 
anh(e^(1/2)*x/(e*x^2+d)^(1/2))/c^2+1/16*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^( 
1/2)*(c*d*(3*A*c*d*e+2*B*a*e^2+B*c*d^2+5*C*a*d*e)+(4*a^2*C*e^2-c*d*(3*A*c* 
d+B*a*e+C*a*d))*(e+(a*e^2+c*d^2)^(1/2)/a^(1/2)))*arctanh(2^(1/2)*a^(1/4)*c 
^(1/2)*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*x*(e*x^2+d)^(1/2)/(a^(1/2)*( 
a^(1/2)*e-(a*e^2+c*d^2)^(1/2))-c*d*x^2))*2^(1/2)/a^(5/4)/c^(5/2)/d/(a*e^2+ 
c*d^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 2.05 (sec) , antiderivative size = 1422, normalized size of antiderivative = 2.43 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((2*c*x*Sqrt[d + e*x^2]*(B*c*d*x^2 + A*c*(d + e*x^2) - a*(C*d + B*e + C*e* 
x^2)))/(a*(a + c*x^4)) - 8*C*e^(3/2)*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]] + 
 4*e^(3/2)*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^2*#1^2 + 16*a*e^2*#1^2 - 4*c 
*d*#1^3 + c*#1^4 & , (2*c*C*d^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x 
^2] - #1] + 17*B*c*d^2*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - # 
1] + 32*A*c*d*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 32 
*a*C*d*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 16*a*B*e^ 
3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 6*B*c*d*e*Log[d + 
2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 4*A*c*e^2*Log[d + 2*e*x^2 
 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 8*a*C*e^2*Log[d + 2*e*x^2 - 2*Sq 
rt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 2*c*C*d*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*S 
qrt[d + e*x^2] - #1]*#1^2 + B*c*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e 
*x^2] - #1]*#1^2)/(c*d^3 - 3*c*d^2*#1 - 8*a*e^2*#1 + 3*c*d*#1^2 - c*#1^3) 
& ] + (Sqrt[e]*RootSum[c*d^4 - 4*c*d^3*#1 + 6*c*d^2*#1^2 + 16*a*e^2*#1^2 - 
 4*c*d*#1^3 + c*#1^4 & , (B*c^2*d^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + 
 e*x^2] - #1] + 3*A*c^2*d^3*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2 
] - #1] - 3*a*c*C*d^3*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1 
] - 66*a*B*c*d^2*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 
 128*a*A*c*d*e^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 128 
*a^2*C*d*e^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 64*a...
 

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.

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(A c-a C) \int \frac {\left (e x^2+d\right )^{3/2}}{\left (c x^4+a\right )^2}dx}{c}+B \int \frac {x^2 \left (e x^2+d\right )^{3/2}}{\left (c x^4+a\right )^2}dx-\frac {C \left (-2 \sqrt {-a} \sqrt {c} d e-a e^2+c d^2\right ) \arctan \left (\frac {x \sqrt {\sqrt {c} d-\sqrt {-a} e}}{\sqrt [4]{-a} \sqrt {d+e x^2}}\right )}{2 (-a)^{3/4} c^2 \sqrt {\sqrt {c} d-\sqrt {-a} e}}-\frac {C \left (2 \sqrt {-a} \sqrt {c} d e-a e^2+c d^2\right ) \text {arctanh}\left (\frac {x \sqrt {\sqrt {-a} e+\sqrt {c} d}}{\sqrt [4]{-a} \sqrt {d+e x^2}}\right )}{2 (-a)^{3/4} c^2 \sqrt {\sqrt {-a} e+\sqrt {c} d}}-\frac {C \sqrt {e} \left (3 \sqrt {c} d-2 \sqrt {-a} e\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {-a} c^2}+\frac {C \sqrt {e} \left (2 \sqrt {-a} e+3 \sqrt {c} d\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 \sqrt {-a} c^2}\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1110\) vs. \(2(492)=984\).

Time = 1.37 (sec) , antiderivative size = 1111, normalized size of antiderivative = 1.90

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1111\)
default	\(\text {Expression too large to display}\)	\(9082\)

Input:

int((e*x^2+d)^(3/2)*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/8*((((-C*a^(9/2)*e-1/4*c*((4*C*e*x^4+B*d)*a^(7/2)+B*a^(5/2)*c*d*x^4))*( 
a*e^2+c*d^2)^(1/2)+3/4*(-4/3*a^2*C*e^2+1/3*c*d*(B*e+C*d)*a+A*c^2*d^2)*(c*x 
^4+a)*a^2)*(a*(a*e^2+c*d^2))^(1/2)+((1/4*c*(4*C*e*x^4+B*d)*a^(9/2)+C*a^(11 
/2)*e+1/4*B*a^(7/2)*c^2*d*x^4)*(a*e^2+c*d^2)^(1/2)-3/4*(-4/3*a^2*C*e^2+1/3 
*c*d*(B*e+C*d)*a+A*c^2*d^2)*(c*x^4+a)*a^3)*e)*(4*(a*e^2+c*d^2)^(1/2)*a^(1/ 
2)-2*(a*(a*e^2+c*d^2))^(1/2)-2*a*e)^(1/2)*(ln((a^(1/2)*(e*x^2+d)-(e*x^2+d) 
^(1/2)*(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*x+x^2*(a*e^2+c*d^2)^(1/2))/ 
x^2)-ln((a^(1/2)*(e*x^2+d)+x^2*(a*e^2+c*d^2)^(1/2)+(e*x^2+d)^(1/2)*(2*(a*( 
a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*x)/x^2))*(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e 
)^(1/2)-8*c*d^2*((4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1/2)- 
2*a*e)^(1/2)*e^(3/2)*(a^(11/2)+a^(9/2)*c*x^4)*C*arctanh((e*x^2+d)^(1/2)/x/ 
e^(1/2))+1/2*(arctan(((2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e)^(1/2)*x-2*a^(1/2)* 
(e*x^2+d)^(1/2))/x/(4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1/2 
)-2*a*e)^(1/2))-arctan((2*a^(1/2)*(e*x^2+d)^(1/2)+(2*(a*(a*e^2+c*d^2))^(1/ 
2)+2*a*e)^(1/2)*x)/x/(4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2*(a*(a*e^2+c*d^2))^(1 
/2)-2*a*e)^(1/2)))*(1/4*c*(4*C*e*x^4+B*d)*a^(9/2)+C*a^(11/2)*e+1/4*B*a^(7/ 
2)*c^2*d*x^4)*(a*e^2+c*d^2)^(1/2)+1/4*c*x*(4*(a*e^2+c*d^2)^(1/2)*a^(1/2)-2 
*(a*(a*e^2+c*d^2))^(1/2)-2*a*e)^(1/2)*((-C*d-(C*x^2+B)*e)*a^(9/2)+c*a^(7/2 
)*(A*e*x^2+(B*x^2+A)*d))*(e*x^2+d)^(1/2)+3/8*(arctan(((2*(a*(a*e^2+c*d^2)) 
^(1/2)+2*a*e)^(1/2)*x-2*a^(1/2)*(e*x^2+d)^(1/2))/x/(4*(a*e^2+c*d^2)^(1/...
 

Fricas [F(-1)]

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

integrate((e*x^2+d)^(3/2)*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="fricas 
")
 

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((e*x^2+d)^(3/2)*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=-\frac {C e^{\frac {3}{2}} \log \left ({\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2}\right )}{2 \, c^{2}} - \frac {{\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{6} B c^{2} d^{2} \sqrt {e} - 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{6} C a c d e^{\frac {3}{2}} + 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{6} A c^{2} d e^{\frac {3}{2}} - 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{6} B a c e^{\frac {5}{2}} - 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} B c^{2} d^{3} \sqrt {e} + 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} C a c d^{2} e^{\frac {3}{2}} - 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} A c^{2} d^{2} e^{\frac {3}{2}} - 8 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} B a c d e^{\frac {5}{2}} + 8 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} C a^{2} e^{\frac {7}{2}} - 8 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} A a c e^{\frac {7}{2}} + 3 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} B c^{2} d^{4} \sqrt {e} - {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} C a c d^{3} e^{\frac {3}{2}} + {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} A c^{2} d^{3} e^{\frac {3}{2}} + 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} B a c d^{2} e^{\frac {5}{2}} - B c^{2} d^{5} \sqrt {e} + C a c d^{4} e^{\frac {3}{2}} - A c^{2} d^{4} e^{\frac {3}{2}}}{2 \, {\left ({\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{8} c - 4 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{6} c d + 6 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} c d^{2} + 16 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{4} a e^{2} - 4 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} c d^{3} + c d^{4}\right )} a c^{2}} \] Input:

integrate((e*x^2+d)^(3/2)*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="giac")
 

Output:

-1/2*C*e^(3/2)*log((sqrt(e)*x - sqrt(e*x^2 + d))^2)/c^2 - 1/2*((sqrt(e)*x 
- sqrt(e*x^2 + d))^6*B*c^2*d^2*sqrt(e) - 3*(sqrt(e)*x - sqrt(e*x^2 + d))^6 
*C*a*c*d*e^(3/2) + 3*(sqrt(e)*x - sqrt(e*x^2 + d))^6*A*c^2*d*e^(3/2) - 2*( 
sqrt(e)*x - sqrt(e*x^2 + d))^6*B*a*c*e^(5/2) - 3*(sqrt(e)*x - sqrt(e*x^2 + 
 d))^4*B*c^2*d^3*sqrt(e) + 3*(sqrt(e)*x - sqrt(e*x^2 + d))^4*C*a*c*d^2*e^( 
3/2) - 3*(sqrt(e)*x - sqrt(e*x^2 + d))^4*A*c^2*d^2*e^(3/2) - 8*(sqrt(e)*x 
- sqrt(e*x^2 + d))^4*B*a*c*d*e^(5/2) + 8*(sqrt(e)*x - sqrt(e*x^2 + d))^4*C 
*a^2*e^(7/2) - 8*(sqrt(e)*x - sqrt(e*x^2 + d))^4*A*a*c*e^(7/2) + 3*(sqrt(e 
)*x - sqrt(e*x^2 + d))^2*B*c^2*d^4*sqrt(e) - (sqrt(e)*x - sqrt(e*x^2 + d)) 
^2*C*a*c*d^3*e^(3/2) + (sqrt(e)*x - sqrt(e*x^2 + d))^2*A*c^2*d^3*e^(3/2) + 
 2*(sqrt(e)*x - sqrt(e*x^2 + d))^2*B*a*c*d^2*e^(5/2) - B*c^2*d^5*sqrt(e) + 
 C*a*c*d^4*e^(3/2) - A*c^2*d^4*e^(3/2))/(((sqrt(e)*x - sqrt(e*x^2 + d))^8* 
c - 4*(sqrt(e)*x - sqrt(e*x^2 + d))^6*c*d + 6*(sqrt(e)*x - sqrt(e*x^2 + d) 
)^4*c*d^2 + 16*(sqrt(e)*x - sqrt(e*x^2 + d))^4*a*e^2 - 4*(sqrt(e)*x - sqrt 
(e*x^2 + d))^2*c*d^3 + c*d^4)*a*c^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) a d +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{6}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) c e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) b e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) c d +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{2}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) a e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{2}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) b d \] Input:

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

Output:

int(sqrt(d + e*x**2)/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*a*d + int((sqrt(d 
+ e*x**2)*x**6)/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*c*e + int((sqrt(d + e*x 
**2)*x**4)/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*b*e + int((sqrt(d + e*x**2)* 
x**4)/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*c*d + int((sqrt(d + e*x**2)*x**2) 
/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*a*e + int((sqrt(d + e*x**2)*x**2)/(a** 
2 + 2*a*c*x**4 + c**2*x**8),x)*b*d