\(\int \frac {(d+e x^2)^{5/2} (A+B x^2+C x^4)}{a+c x^4} \, dx\) [13]

Optimal result

Integrand size = 33, antiderivative size = 717 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx=\frac {\left (5 c C d^2+14 B c d e+8 A c e^2-8 a C e^2\right ) x \sqrt {d+e x^2}}{16 c^2}+\frac {(5 C d+6 B e) x \left (d+e x^2\right )^{3/2}}{24 c}+\frac {C x \left (d+e x^2\right )^{5/2}}{6 c}+\frac {\sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (d \left (B c d \left (c d^2-3 a e^2\right )+(A c-a C) e \left (3 c d^2-a e^2\right )\right )-\left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (3 C d+B e)-c d^2 (C d+3 B e)\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/2} d \sqrt {c d^2+a e^2}}-\frac {\left (8 a e^2 (5 C d+2 B e)-5 c d \left (C d^2+6 B d e+8 A e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{16 c^2 \sqrt {e}}+\frac {\sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (d \left (B c d \left (c d^2-3 a e^2\right )+(A c-a C) e \left (3 c d^2-a e^2\right )\right )-\left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (3 C d+B e)-c d^2 (C d+3 B e)\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/2} d \sqrt {c d^2+a e^2}} \] Output:

1/16*(8*A*c*e^2+14*B*c*d*e-8*C*a*e^2+5*C*c*d^2)*x*(e*x^2+d)^(1/2)/c^2+1/24 
*(6*B*e+5*C*d)*x*(e*x^2+d)^(3/2)/c+1/6*C*x*(e*x^2+d)^(5/2)/c+1/4*(a^(1/2)* 
e+(a*e^2+c*d^2)^(1/2))^(1/2)*(d*(B*c*d*(-3*a*e^2+c*d^2)+(A*c-C*a)*e*(-a*e^ 
2+3*c*d^2))-(e-(a*e^2+c*d^2)^(1/2)/a^(1/2))*(A*c*d*(-3*a*e^2+c*d^2)+a*(a*e 
^2*(B*e+3*C*d)-c*d^2*(3*B*e+C*d))))*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^(1/4)/c^(5/2)/d/(a*e^2+c*d^2)^(1/2)-1/ 
16*(8*a*e^2*(2*B*e+5*C*d)-5*c*d*(8*A*e^2+6*B*d*e+C*d^2))*arctanh(e^(1/2)*x 
/(e*x^2+d)^(1/2))/c^2/e^(1/2)+1/4*(-a^(1/2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*( 
d*(B*c*d*(-3*a*e^2+c*d^2)+(A*c-C*a)*e*(-a*e^2+3*c*d^2))-(e+(a*e^2+c*d^2)^( 
1/2)/a^(1/2))*(A*c*d*(-3*a*e^2+c*d^2)+a*(a*e^2*(B*e+3*C*d)-c*d^2*(3*B*e+C* 
d))))*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^(1/4)/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 = 1.33 (sec) , antiderivative size = 1063, normalized size of antiderivative = 1.48 \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-24*a*C*e^2 + 6*c*e*(9*B*d + 4*A*e + 2*B*e*x^2 
) + c*C*(33*d^2 + 26*d*e*x^2 + 8*e^2*x^4)) - 3*(-8*a*e^2*(5*C*d + 2*B*e) + 
 5*c*d*(C*d^2 + 6*B*d*e + 8*A*e^2))*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]] + 
24*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^5*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - 
#1] + 3*A*c^2*d^4*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 
3*a*c*C*d^4*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 3*a*B* 
c*d^3*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - a*A*c*d^2* 
e^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + a^2*C*d^2*e^3*Lo 
g[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 2*B*c^2*d^4*Log[d + 2* 
e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 2*A*c^2*d^3*e*Log[d + 2*e*x 
^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 2*a*c*C*d^3*e*Log[d + 2*e*x^2 
- 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 6*a*B*c*d^2*e^2*Log[d + 2*e*x^2 - 
 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 10*a*A*c*d*e^3*Log[d + 2*e*x^2 - 2 
*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 10*a^2*C*d*e^3*Log[d + 2*e*x^2 - 2*S 
qrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 4*a^2*B*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e 
]*x*Sqrt[d + e*x^2] - #1]*#1 + B*c^2*d^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqr 
t[d + e*x^2] - #1]*#1^2 + 3*A*c^2*d^2*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt 
[d + e*x^2] - #1]*#1^2 - 3*a*c*C*d^2*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[ 
d + e*x^2] - #1]*#1^2 - 3*a*B*c*d*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqr...
 

Rubi [A] (verified)

Time = 2.05 (sec) , antiderivative size = 819, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2257, 2009}

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.

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 C \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^2+d}}\right ) d^3}{16 c \sqrt {e}}+\frac {5 C x \sqrt {e x^2+d} d^2}{16 c}+\frac {5 C x \left (e x^2+d\right )^{3/2} d}{24 c}+\frac {C x \left (e x^2+d\right )^{5/2}}{6 c}+\frac {\left (\sqrt {-a} \sqrt {c} B+A c-a C\right ) e x \left (e x^2+d\right )^{3/2}}{8 \sqrt {-a} c^{3/2}}+\frac {\left (\sqrt {-a} \sqrt {c} B-A c+a C\right ) e x \left (e x^2+d\right )^{3/2}}{8 \sqrt {-a} c^{3/2}}+\frac {\left (\sqrt {-a} \sqrt {c} B-A c+a C\right ) \left (c^{3/2} d^3-3 \sqrt {-a} c e d^2-3 a \sqrt {c} e^2 d+\sqrt {-a} a e^3\right ) \arctan \left (\frac {\sqrt {\sqrt {c} d-\sqrt {-a} e} x}{\sqrt [4]{-a} \sqrt {e x^2+d}}\right )}{2 (-a)^{3/4} c^{5/2} \sqrt {\sqrt {c} d-\sqrt {-a} e}}+\frac {\left (\sqrt {-a} \sqrt {c} B-A c+a C\right ) \sqrt {e} \left (15 c d^2-20 \sqrt {-a} \sqrt {c} e d-8 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^2+d}}\right )}{16 \sqrt {-a} c^{5/2}}+\frac {\left (\sqrt {-a} \sqrt {c} B+A c-a C\right ) \sqrt {e} \left (15 c d^2+20 \sqrt {-a} \sqrt {c} e d-8 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^2+d}}\right )}{16 \sqrt {-a} c^{5/2}}-\frac {\left (\sqrt {-a} \sqrt {c} B+A c-a C\right ) \left (c^{3/2} d^3+3 \sqrt {-a} c e d^2-3 a \sqrt {c} e^2 d+(-a)^{3/2} e^3\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {c} d+\sqrt {-a} e} x}{\sqrt [4]{-a} \sqrt {e x^2+d}}\right )}{2 (-a)^{3/4} c^{5/2} \sqrt {\sqrt {c} d+\sqrt {-a} e}}+\frac {\left (\sqrt {-a} \sqrt {c} B-A c+a C\right ) e \left (7 \sqrt {c} d-4 \sqrt {-a} e\right ) x \sqrt {e x^2+d}}{16 \sqrt {-a} c^2}+\frac {\left (\sqrt {-a} \sqrt {c} B+A c-a C\right ) e \left (7 \sqrt {c} d+4 \sqrt {-a} e\right ) x \sqrt {e x^2+d}}{16 \sqrt {-a} c^2}\)

Input:

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

Output:

(5*C*d^2*x*Sqrt[d + e*x^2])/(16*c) + ((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*e*( 
7*Sqrt[c]*d - 4*Sqrt[-a]*e)*x*Sqrt[d + e*x^2])/(16*Sqrt[-a]*c^2) + ((Sqrt[ 
-a]*B*Sqrt[c] + A*c - a*C)*e*(7*Sqrt[c]*d + 4*Sqrt[-a]*e)*x*Sqrt[d + e*x^2 
])/(16*Sqrt[-a]*c^2) + (5*C*d*x*(d + e*x^2)^(3/2))/(24*c) + ((Sqrt[-a]*B*S 
qrt[c] + A*c - a*C)*e*x*(d + e*x^2)^(3/2))/(8*Sqrt[-a]*c^(3/2)) + ((Sqrt[- 
a]*B*Sqrt[c] - A*c + a*C)*e*x*(d + e*x^2)^(3/2))/(8*Sqrt[-a]*c^(3/2)) + (C 
*x*(d + e*x^2)^(5/2))/(6*c) + ((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*(c^(3/2)*d 
^3 - 3*Sqrt[-a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 + Sqrt[-a]*a*e^3)*ArcTan[(Sqrt 
[Sqrt[c]*d - Sqrt[-a]*e]*x)/((-a)^(1/4)*Sqrt[d + e*x^2])])/(2*(-a)^(3/4)*c 
^(5/2)*Sqrt[Sqrt[c]*d - Sqrt[-a]*e]) + (5*C*d^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d 
 + e*x^2]])/(16*c*Sqrt[e]) + ((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*Sqrt[e]*(15 
*c*d^2 - 20*Sqrt[-a]*Sqrt[c]*d*e - 8*a*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e 
*x^2]])/(16*Sqrt[-a]*c^(5/2)) + ((Sqrt[-a]*B*Sqrt[c] + A*c - a*C)*Sqrt[e]* 
(15*c*d^2 + 20*Sqrt[-a]*Sqrt[c]*d*e - 8*a*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d 
+ e*x^2]])/(16*Sqrt[-a]*c^(5/2)) - ((Sqrt[-a]*B*Sqrt[c] + A*c - a*C)*(c^(3 
/2)*d^3 + 3*Sqrt[-a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 + (-a)^(3/2)*e^3)*ArcTanh 
[(Sqrt[Sqrt[c]*d + Sqrt[-a]*e]*x)/((-a)^(1/4)*Sqrt[d + e*x^2])])/(2*(-a)^( 
3/4)*c^(5/2)*Sqrt[Sqrt[c]*d + Sqrt[-a]*e])
 

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. \(1274\) vs. \(2(617)=1234\).

Time = 3.50 (sec) , antiderivative size = 1275, normalized size of antiderivative = 1.78

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1275\)
risch	\(\text {Expression too large to display}\)	\(1374\)
default	\(\text {Expression too large to display}\)	\(5484\)

Input:

int((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

-(-1/4*(((d*(A*c*a^(1/2)-C*a^(3/2))*e^(3/2)+1/2*B*(c*d^2*a^(1/2)*e^(1/2)-a 
^(3/2)*e^(5/2)))*(a*e^2+c*d^2)^(1/2)+3/2*a*d*(A*c-C*a)*e^(5/2)+3/2*B*a*c*d 
^2*e^(3/2)-1/2*B*a^2*e^(7/2)-1/2*e^(1/2)*c*d^3*(A*c-C*a))*(a*(a*e^2+c*d^2) 
)^(1/2)+(-1/2*B*a^(3/2)*c*d^2*e^(3/2)-A*a^(3/2)*c*d*e^(5/2)+a^(5/2)*(C*d*e 
^(5/2)+1/2*B*e^(7/2)))*(a*e^2+c*d^2)^(1/2)+1/2*(c*d^3*(A*c-C*a)*e^(3/2)-3* 
(e^(5/2)*B*c*d^2+d*(A*c-C*a)*e^(7/2)-1/3*e^(9/2)*B*a)*a)*a)*(2*(a*(a*e^2+c 
*d^2))^(1/2)+2*a*e)^(1/2)*(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)*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)+1/4*(((d*(A*c*a^ 
(1/2)-C*a^(3/2))*e^(3/2)+1/2*B*(c*d^2*a^(1/2)*e^(1/2)-a^(3/2)*e^(5/2)))*(a 
*e^2+c*d^2)^(1/2)+3/2*a*d*(A*c-C*a)*e^(5/2)+3/2*B*a*c*d^2*e^(3/2)-1/2*B*a^ 
2*e^(7/2)-1/2*e^(1/2)*c*d^3*(A*c-C*a))*(a*(a*e^2+c*d^2))^(1/2)+(-1/2*B*a^( 
3/2)*c*d^2*e^(3/2)-A*a^(3/2)*c*d*e^(5/2)+a^(5/2)*(C*d*e^(5/2)+1/2*B*e^(7/2 
)))*(a*e^2+c*d^2)^(1/2)+1/2*(c*d^3*(A*c-C*a)*e^(3/2)-3*(e^(5/2)*B*c*d^2+d* 
(A*c-C*a)*e^(7/2)-1/3*e^(9/2)*B*a)*a)*a)*(2*(a*(a*e^2+c*d^2))^(1/2)+2*a*e) 
^(1/2)*(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)*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)+c*d^2*(((-5/2*c*(A*e^2+3/4*B*d*e+1/ 
8*C*d^2)*d*a^(3/2)+a^(5/2)*e^2*(B*e+5/2*C*d))*arctanh((e*x^2+d)^(1/2)/x/e^ 
(1/2))-1/2*(e*x^2+d)^(1/2)*x*(9/4*c*d*a^(3/2)*(13/27*C*x^2+B)*e^(3/2)+c...
 

Fricas [F(-1)]

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

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+a),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x^2+d)^(5/2)*(C*x^4+B*x^2+A)/(c*x^4+a),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^{5/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx=\left (\int \frac {\sqrt {e \,x^{2}+d}}{c \,x^{4}+a}d x \right ) a \,d^{2}+\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{8}}{c \,x^{4}+a}d x \right ) c \,e^{2}+\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{6}}{c \,x^{4}+a}d x \right ) b \,e^{2}+2 \left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{6}}{c \,x^{4}+a}d x \right ) c d e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c \,x^{4}+a}d x \right ) a \,e^{2}+2 \left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c \,x^{4}+a}d x \right ) b d e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{4}}{c \,x^{4}+a}d x \right ) c \,d^{2}+2 \left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{2}}{c \,x^{4}+a}d x \right ) a d e +\left (\int \frac {\sqrt {e \,x^{2}+d}\, x^{2}}{c \,x^{4}+a}d x \right ) b \,d^{2} \] Input:

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

Output:

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