\(\int \frac {(d+e x^2)^{3/2} (A+B x^2+C x^4)}{a+c x^4} \, dx\) [14]

Optimal result

Integrand size = 33, antiderivative size = 615 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx=\frac {(3 C d+4 B e) x \sqrt {d+e x^2}}{8 c}+\frac {C x \left (d+e x^2\right )^{3/2}}{4 c}+\frac {\sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (c d \left (B c d^2+2 A c d e-2 a C d e-a B e^2\right )-\left (e-\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d+2 B e)\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/2} d \sqrt {c d^2+a e^2}}+\frac {\left (3 c C d^2+12 B c d e+8 A c e^2-8 a C e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 c^2 \sqrt {e}}+\frac {\sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (c d \left (B c d^2+2 A c d e-2 a C d e-a B e^2\right )-\left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d+2 B e)\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{2 \sqrt {2} \sqrt [4]{a} c^{5/2} d \sqrt {c d^2+a e^2}} \] Output:

1/8*(4*B*e+3*C*d)*x*(e*x^2+d)^(1/2)/c+1/4*C*x*(e*x^2+d)^(3/2)/c+1/4*(a^(1/ 
2)*e+(a*e^2+c*d^2)^(1/2))^(1/2)*(c*d*(2*A*c*d*e-B*a*e^2+B*c*d^2-2*C*a*d*e) 
-(e-(a*e^2+c*d^2)^(1/2)/a^(1/2))*(A*c*(-a*e^2+c*d^2)+a*(C*a*e^2-c*d*(2*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)+1/8*(8*A*c*e^2+12*B*c*d*e-8*C 
*a*e^2+3*C*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)*(c*d*(2*A*c*d*e-B*a*e^2+B*c*d^2-2*C*a*d* 
e)-(e+(a*e^2+c*d^2)^(1/2)/a^(1/2))*(A*c*(-a*e^2+c*d^2)+a*(C*a*e^2-c*d*(2*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 = 0.85 (sec) , antiderivative size = 705, normalized size of antiderivative = 1.15 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (A+B x^2+C x^4\right )}{a+c x^4} \, dx=\frac {c \sqrt {e} x \sqrt {d+e x^2} \left (5 C d+4 B e+2 C e x^2\right )+\left (-3 c C d^2+8 a C e^2-4 c e (3 B d+2 A e)\right ) \log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )+4 e \text {RootSum}\left [c d^4-4 c d^3 \text {$\#$1}+6 c d^2 \text {$\#$1}^2+16 a e^2 \text {$\#$1}^2-4 c d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {B c^2 d^4 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right )+2 A c^2 d^3 e \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right )-2 a c C d^3 e \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right )-a B c d^2 e^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right )-2 B c^2 d^3 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}-6 a B c d e^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 a A c e^3 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}+4 a^2 C e^3 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}+B c^2 d^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 A c^2 d e \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-2 a c C d e \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a B c e^2 \log \left (d+2 e x^2-2 \sqrt {e} x \sqrt {d+e x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{c d^3-3 c d^2 \text {$\#$1}-8 a e^2 \text {$\#$1}+3 c d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]}{8 c^2 \sqrt {e}} \] Input:

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

Output:

(c*Sqrt[e]*x*Sqrt[d + e*x^2]*(5*C*d + 4*B*e + 2*C*e*x^2) + (-3*c*C*d^2 + 8 
*a*C*e^2 - 4*c*e*(3*B*d + 2*A*e))*Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]] + 4* 
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] 
 + 2*A*c^2*d^3*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 2*a 
*c*C*d^3*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - a*B*c*d^2 
*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 2*B*c^2*d^3*Log 
[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 6*a*B*c*d*e^2*Log[d 
+ 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 4*a*A*c*e^3*Log[d + 2*e 
*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 4*a^2*C*e^3*Log[d + 2*e*x^2 
- 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + B*c^2*d^2*Log[d + 2*e*x^2 - 2*Sqr 
t[e]*x*Sqrt[d + e*x^2] - #1]*#1^2 + 2*A*c^2*d*e*Log[d + 2*e*x^2 - 2*Sqrt[e 
]*x*Sqrt[d + e*x^2] - #1]*#1^2 - 2*a*c*C*d*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x 
*Sqrt[d + e*x^2] - #1]*#1^2 - a*B*c*e^2*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) & ])/(8*c^2*Sqrt[e])
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 593, normalized size of antiderivative = 0.96, 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.

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(3*C*d*x*Sqrt[d + e*x^2])/(8*c) + ((Sqrt[-a]*B*Sqrt[c] + A*c - a*C)*e*x*Sq 
rt[d + e*x^2])/(4*Sqrt[-a]*c^(3/2)) + ((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*e* 
x*Sqrt[d + e*x^2])/(4*Sqrt[-a]*c^(3/2)) + (C*x*(d + e*x^2)^(3/2))/(4*c) + 
((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*(c*d^2 - 2*Sqrt[-a]*Sqrt[c]*d*e - a*e^2) 
*ArcTan[(Sqrt[Sqrt[c]*d - Sqrt[-a]*e]*x)/((-a)^(1/4)*Sqrt[d + e*x^2])])/(2 
*(-a)^(3/4)*c^2*Sqrt[Sqrt[c]*d - Sqrt[-a]*e]) + (3*C*d^2*ArcTanh[(Sqrt[e]* 
x)/Sqrt[d + e*x^2]])/(8*c*Sqrt[e]) + ((Sqrt[-a]*B*Sqrt[c] - A*c + a*C)*Sqr 
t[e]*(3*Sqrt[c]*d - 2*Sqrt[-a]*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(4 
*Sqrt[-a]*c^2) + ((Sqrt[-a]*B*Sqrt[c] + A*c - a*C)*Sqrt[e]*(3*Sqrt[c]*d + 
2*Sqrt[-a]*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(4*Sqrt[-a]*c^2) - ((S 
qrt[-a]*B*Sqrt[c] + A*c - a*C)*(c*d^2 + 2*Sqrt[-a]*Sqrt[c]*d*e - a*e^2)*Ar 
cTanh[(Sqrt[Sqrt[c]*d + Sqrt[-a]*e]*x)/((-a)^(1/4)*Sqrt[d + e*x^2])])/(2*( 
-a)^(3/4)*c^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. \(1058\) vs. \(2(519)=1038\).

Time = 0.95 (sec) , antiderivative size = 1059, normalized size of antiderivative = 1.72

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1059\)
risch	\(\text {Expression too large to display}\)	\(1156\)
default	\(\text {Expression too large to display}\)	\(3422\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

int((e*x^2+d)^(3/2)*(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 + int((sqrt(d + e*x**2)*x**6)/(a 
+ c*x**4),x)*c*e + int((sqrt(d + e*x**2)*x**4)/(a + c*x**4),x)*b*e + int(( 
sqrt(d + e*x**2)*x**4)/(a + c*x**4),x)*c*d + int((sqrt(d + e*x**2)*x**2)/( 
a + c*x**4),x)*a*e + int((sqrt(d + e*x**2)*x**2)/(a + c*x**4),x)*b*d