\(\int \frac {A+B x^2+C x^4}{(d+e x^2)^{3/2} (a+c x^4)^2} \, dx\) [23]

Optimal result

Integrand size = 33, antiderivative size = 719 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^2} \, dx=\frac {e \left (B c d^3-2 A c d^2 e+6 a C d^2 e-5 a B d e^2+4 a A e^3\right ) x}{4 a d \left (c d^2+a e^2\right )^2 \sqrt {d+e x^2}}+\frac {x \left (A c d-a C d+a B e+(B c d-A c e+a C e) x^2\right )}{4 a \left (c d^2+a e^2\right ) \sqrt {d+e x^2} \left (a+c x^4\right )}+\frac {\sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (d \left (B c d \left (c d^2+7 a e^2\right )-2 a e \left (2 c C d^2+3 A c e^2-a C e^2\right )\right )-\frac {\left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right ) \left (3 A c d \left (c d^2+3 a e^2\right )-a \left (a e^2 (5 C d-4 B e)-c d^2 (C d-2 B e)\right )\right )}{\sqrt {a}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{8 \sqrt {2} a^{5/4} \sqrt {c} d \left (c d^2+a e^2\right )^{5/2}}+\frac {\sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} \left (d \left (B c d \left (c d^2+7 a e^2\right )-2 a e \left (2 c C d^2+3 A c e^2-a C e^2\right )\right )-\left (e+\frac {\sqrt {c d^2+a e^2}}{\sqrt {a}}\right ) \left (3 A c d \left (c d^2+3 a e^2\right )-a \left (a e^2 (5 C d-4 B e)-c d^2 (C d-2 B e)\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c} \sqrt {-\sqrt {a} e+\sqrt {c d^2+a e^2}} x \sqrt {d+e x^2}}{\sqrt {a} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )-c d x^2}\right )}{8 \sqrt {2} a^{5/4} \sqrt {c} d \left (c d^2+a e^2\right )^{5/2}} \] Output:

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

Mathematica [C] (verified)

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

Time = 3.00 (sec) , antiderivative size = 1758, normalized size of antiderivative = 2.45 \[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((-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 & , (c^2*C*d^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e 
*x^2] - #1] - 17*B*c^2*d^3*e*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] 
 - #1] + 17*A*c^2*d^2*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - 
#1] - 16*a*c*C*d^2*e^2*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] 
 - 16*a*B*c*d*e^3*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] + 16 
*a*A*c*e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 16*a^2*C* 
e^4*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1] - 6*c^2*C*d^3*Log[ 
d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + 6*B*c^2*d^2*e*Log[d + 
 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 - 6*A*c^2*d*e^2*Log[d + 2* 
e*x^2 - 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1 + c^2*C*d^2*Log[d + 2*e*x^2 - 
 2*Sqrt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2 - B*c^2*d*e*Log[d + 2*e*x^2 - 2*Sq 
rt[e]*x*Sqrt[d + e*x^2] - #1]*#1^2 + A*c^2*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) & ])/c + ((2*x*(4*a^2*A*e^4 + B*c^2*d^3*x^2*(d + e*x^2) + a^2 
*d*e^2*(5*C*d - 4*B*e + C*e*x^2) - a*A*c*e^2*(d^2 + d*e*x^2 - 4*e^2*x^4) + 
 A*c^2*d^2*(d^2 - d*e*x^2 - 2*e^2*x^4) + a*c*d*(B*e*(2*d^2 + d*e*x^2 - 5*e 
^2*x^4) + C*d*(-d^2 + d*e*x^2 + 6*e^2*x^4))))/(d*Sqrt[d + e*x^2]*(a + c*x^ 
4)) + (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^3*d^5*Log[d + 2*e*x^2 - 2*Sqrt[e]*x*Sqrt[...
 

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.

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(A c-a C) \int \frac {1}{\left (e x^2+d\right )^{3/2} \left (c x^4+a\right )^2}dx}{c}+B \int \frac {x^2}{\left (e x^2+d\right )^{3/2} \left (c x^4+a\right )^2}dx-\frac {C \left (\sqrt {-a} e+\sqrt {c} d\right ) \arctan \left (\frac {x \sqrt {\sqrt {c} d-\sqrt {-a} e}}{\sqrt [4]{-a} \sqrt {d+e x^2}}\right )}{2 (-a)^{3/4} \sqrt {c} \sqrt {\sqrt {c} d-\sqrt {-a} e} \left (a e^2+c d^2\right )}-\frac {C \left (\sqrt {c} d-\sqrt {-a} e\right ) \text {arctanh}\left (\frac {x \sqrt {\sqrt {-a} e+\sqrt {c} d}}{\sqrt [4]{-a} \sqrt {d+e x^2}}\right )}{2 (-a)^{3/4} \sqrt {c} \sqrt {\sqrt {-a} e+\sqrt {c} d} \left (a e^2+c d^2\right )}+\frac {C e^2 x}{c d \sqrt {d+e x^2} \left (a e^2+c d^2\right )}\)

Input:

Int[(A + B*x^2 + C*x^4)/((d + e*x^2)^(3/2)*(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. \(1500\) vs. \(2(629)=1258\).

Time = 2.46 (sec) , antiderivative size = 1501, normalized size of antiderivative = 2.09

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(1501\)
default	\(\text {Expression too large to display}\)	\(5642\)

Input:

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

Output:

-3/32/a^(5/2)*((e*x^2+d)^(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))*((3*(1/3*(A*c^2+1/3*C*c*a)*d^3-2/9*B*a*c*d^ 
2*e+a*e^2*(A*c-5/9*C*a)*d+4/9*a^2*B*e^3)*(c*x^4+a)*(a*e^2+c*d^2)^(1/2)+c^2 
*d*(-1/3*B*d^3+e*(A+5/3*C*x^4)*d^2-3*B*e^2*d*x^4+5*A*e^3*x^4)*a^(3/2)+5*c* 
e*(1/3*C*d^3-3/5*B*e*d^2+e^2*(-7/15*C*x^4+A)*d+4/15*B*e^3*x^4)*a^(5/2)+1/3 
*(4*B*e^4-7*C*d*e^3)*a^(7/2)+a^(1/2)*c^3*d^3*x^4*(A*e-1/3*B*d))*(a*(a*e^2+ 
c*d^2))^(1/2)-e*(3*(1/3*(A*c^2+1/3*C*c*a)*d^3-2/9*B*a*c*d^2*e+a*e^2*(A*c-5 
/9*C*a)*d+4/9*a^2*B*e^3)*(c*x^4+a)*a*(a*e^2+c*d^2)^(1/2)+c^3*d^3*x^4*(A*e- 
1/3*B*d)*a^(3/2)+c^2*d*(-1/3*B*d^3+e*(A+5/3*C*x^4)*d^2-3*B*e^2*d*x^4+5*A*e 
^3*x^4)*a^(5/2)+5*e*(c*(1/3*C*d^3-3/5*B*e*d^2+e^2*(-7/15*C*x^4+A)*d+4/15*B 
*e^3*x^4)*a^(7/2)+4/15*(B*e-7/4*C*d)*a^(9/2)*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)*(2*(a*(a*e^2+c*d^2))^(1/2) 
+2*a*e)^(1/2)-8/3*c*d*((-9/2*d*(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))-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)))*(1/3*(A*c^2+1/3*C*c*a)*d^3-2/9*B*a*c* 
d^2*e+a*e^2*(A*c-5/9*C*a)*d+4/9*a^2*B*e^3)*(c*x^4+a)*a*(e*x^2+d)^(1/2)+...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2+C x^4}{\left (d+e x^2\right )^{3/2} \left (a+c x^4\right )^2} \, dx=\left (\int \frac {x^{4}}{\sqrt {e \,x^{2}+d}\, a^{2} d +\sqrt {e \,x^{2}+d}\, a^{2} e \,x^{2}+2 \sqrt {e \,x^{2}+d}\, a c d \,x^{4}+2 \sqrt {e \,x^{2}+d}\, a c e \,x^{6}+\sqrt {e \,x^{2}+d}\, c^{2} d \,x^{8}+\sqrt {e \,x^{2}+d}\, c^{2} e \,x^{10}}d x \right ) c +\left (\int \frac {x^{2}}{\sqrt {e \,x^{2}+d}\, a^{2} d +\sqrt {e \,x^{2}+d}\, a^{2} e \,x^{2}+2 \sqrt {e \,x^{2}+d}\, a c d \,x^{4}+2 \sqrt {e \,x^{2}+d}\, a c e \,x^{6}+\sqrt {e \,x^{2}+d}\, c^{2} d \,x^{8}+\sqrt {e \,x^{2}+d}\, c^{2} e \,x^{10}}d x \right ) b +\left (\int \frac {1}{\sqrt {e \,x^{2}+d}\, a^{2} d +\sqrt {e \,x^{2}+d}\, a^{2} e \,x^{2}+2 \sqrt {e \,x^{2}+d}\, a c d \,x^{4}+2 \sqrt {e \,x^{2}+d}\, a c e \,x^{6}+\sqrt {e \,x^{2}+d}\, c^{2} d \,x^{8}+\sqrt {e \,x^{2}+d}\, c^{2} e \,x^{10}}d x \right ) a \] Input:

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

Output:

int(x**4/(sqrt(d + e*x**2)*a**2*d + sqrt(d + e*x**2)*a**2*e*x**2 + 2*sqrt( 
d + e*x**2)*a*c*d*x**4 + 2*sqrt(d + e*x**2)*a*c*e*x**6 + sqrt(d + e*x**2)* 
c**2*d*x**8 + sqrt(d + e*x**2)*c**2*e*x**10),x)*c + int(x**2/(sqrt(d + e*x 
**2)*a**2*d + sqrt(d + e*x**2)*a**2*e*x**2 + 2*sqrt(d + e*x**2)*a*c*d*x**4 
 + 2*sqrt(d + e*x**2)*a*c*e*x**6 + sqrt(d + e*x**2)*c**2*d*x**8 + sqrt(d + 
 e*x**2)*c**2*e*x**10),x)*b + int(1/(sqrt(d + e*x**2)*a**2*d + sqrt(d + e* 
x**2)*a**2*e*x**2 + 2*sqrt(d + e*x**2)*a*c*d*x**4 + 2*sqrt(d + e*x**2)*a*c 
*e*x**6 + sqrt(d + e*x**2)*c**2*d*x**8 + sqrt(d + e*x**2)*c**2*e*x**10),x) 
*a