\(\int \frac {1}{(d+e x^2)^2 (a+c x^4)^2} \, dx\) [323]

Optimal result

Integrand size = 19, antiderivative size = 492 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=-\frac {e^2 \left (c d^2-a e^2\right ) x}{2 a d \left (c d^2+a e^2\right )^2 \left (d+e x^2\right )}+\frac {c x \left (d-e x^2\right )}{4 a \left (c d^2+a e^2\right ) \left (d+e x^2\right ) \left (a+c x^4\right )}+\frac {e^{7/2} \left (9 c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} \left (c d^2+a e^2\right )^3}-\frac {c^{3/4} \left (3 c^2 d^4-2 \sqrt {a} c^{3/2} d^3 e+12 a c d^2 e^2-18 a^{3/2} \sqrt {c} d e^3-7 a^2 e^4\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^3}+\frac {c^{3/4} \left (3 c^2 d^4-2 \sqrt {a} c^{3/2} d^3 e+12 a c d^2 e^2-18 a^{3/2} \sqrt {c} d e^3-7 a^2 e^4\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^3}+\frac {c^{3/4} \left (3 c^2 d^4+12 a c d^2 e^2-7 a^2 e^4+2 \sqrt {a} \sqrt {c} d e \left (c d^2+9 a e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^3} \] Output:

-1/2*e^2*(-a*e^2+c*d^2)*x/a/d/(a*e^2+c*d^2)^2/(e*x^2+d)+1/4*c*x*(-e*x^2+d) 
/a/(a*e^2+c*d^2)/(e*x^2+d)/(c*x^4+a)+1/2*e^(7/2)*(a*e^2+9*c*d^2)*arctan(e^ 
(1/2)*x/d^(1/2))/d^(3/2)/(a*e^2+c*d^2)^3+1/16*c^(3/4)*(3*c^2*d^4-2*a^(1/2) 
*c^(3/2)*d^3*e+12*a*c*d^2*e^2-18*a^(3/2)*c^(1/2)*d*e^3-7*a^2*e^4)*arctan(- 
1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/(a*e^2+c*d^2)^3+1/16*c^(3/4)* 
(3*c^2*d^4-2*a^(1/2)*c^(3/2)*d^3*e+12*a*c*d^2*e^2-18*a^(3/2)*c^(1/2)*d*e^3 
-7*a^2*e^4)*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/(a*e^2+c*d 
^2)^3+1/16*c^(3/4)*(3*c^2*d^4+12*a*c*d^2*e^2-7*a^2*e^4+2*a^(1/2)*c^(1/2)*d 
*e*(9*a*e^2+c*d^2))*arctanh(2^(1/2)*a^(1/4)*c^(1/4)*x/(a^(1/2)+c^(1/2)*x^2 
))*2^(1/2)/a^(7/4)/(a*e^2+c*d^2)^3
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 540, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=\frac {\frac {16 e^4 \left (c d^2+a e^2\right ) x}{d \left (d+e x^2\right )}+\frac {8 c \left (c d^2+a e^2\right ) x \left (-a e^2+c d \left (d-2 e x^2\right )\right )}{a \left (a+c x^4\right )}+\frac {16 e^{7/2} \left (9 c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{3/2}}+\frac {2 \sqrt {2} c^{3/4} \left (-3 c^2 d^4+2 \sqrt {a} c^{3/2} d^3 e-12 a c d^2 e^2+18 a^{3/2} \sqrt {c} d e^3+7 a^2 e^4\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{7/4}}-\frac {2 \sqrt {2} c^{3/4} \left (-3 c^2 d^4+2 \sqrt {a} c^{3/2} d^3 e-12 a c d^2 e^2+18 a^{3/2} \sqrt {c} d e^3+7 a^2 e^4\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{7/4}}-\frac {\sqrt {2} c^{3/4} \left (3 c^2 d^4+2 \sqrt {a} c^{3/2} d^3 e+12 a c d^2 e^2+18 a^{3/2} \sqrt {c} d e^3-7 a^2 e^4\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{7/4}}+\frac {\sqrt {2} c^{3/4} \left (3 c^2 d^4+2 \sqrt {a} c^{3/2} d^3 e+12 a c d^2 e^2+18 a^{3/2} \sqrt {c} d e^3-7 a^2 e^4\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{a^{7/4}}}{32 \left (c d^2+a e^2\right )^3} \] Input:

Integrate[1/((d + e*x^2)^2*(a + c*x^4)^2),x]
 

Output:

((16*e^4*(c*d^2 + a*e^2)*x)/(d*(d + e*x^2)) + (8*c*(c*d^2 + a*e^2)*x*(-(a* 
e^2) + c*d*(d - 2*e*x^2)))/(a*(a + c*x^4)) + (16*e^(7/2)*(9*c*d^2 + a*e^2) 
*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/d^(3/2) + (2*Sqrt[2]*c^(3/4)*(-3*c^2*d^4 + 2 
*Sqrt[a]*c^(3/2)*d^3*e - 12*a*c*d^2*e^2 + 18*a^(3/2)*Sqrt[c]*d*e^3 + 7*a^2 
*e^4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(7/4) - (2*Sqrt[2]*c^(3/4 
)*(-3*c^2*d^4 + 2*Sqrt[a]*c^(3/2)*d^3*e - 12*a*c*d^2*e^2 + 18*a^(3/2)*Sqrt 
[c]*d*e^3 + 7*a^2*e^4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/a^(7/4) - 
(Sqrt[2]*c^(3/4)*(3*c^2*d^4 + 2*Sqrt[a]*c^(3/2)*d^3*e + 12*a*c*d^2*e^2 + 1 
8*a^(3/2)*Sqrt[c]*d*e^3 - 7*a^2*e^4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4) 
*x + Sqrt[c]*x^2])/a^(7/4) + (Sqrt[2]*c^(3/4)*(3*c^2*d^4 + 2*Sqrt[a]*c^(3/ 
2)*d^3*e + 12*a*c*d^2*e^2 + 18*a^(3/2)*Sqrt[c]*d*e^3 - 7*a^2*e^4)*Log[Sqrt 
[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/a^(7/4))/(32*(c*d^2 + a*e^ 
2)^3)
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 864, normalized size of antiderivative = 1.76, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1568, 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.

\(\displaystyle \int \frac {1}{\left (a+c x^4\right )^2 \left (d+e x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1568

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x e^4}{2 d \left (c d^2+a e^2\right )^2 \left (e x^2+d\right )}+\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) e^{7/2}}{2 d^{3/2} \left (c d^2+a e^2\right )^2}+\frac {4 c \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) e^{7/2}}{\left (c d^2+a e^2\right )^3}-\frac {c^{3/4} \left (3 c d^2-4 \sqrt {a} \sqrt {c} e d-a e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right ) e^2}{2 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )^3}+\frac {c^{3/4} \left (3 c d^2-4 \sqrt {a} \sqrt {c} e d-a e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right ) e^2}{2 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )^3}-\frac {c^{3/4} \left (3 c d^2+4 \sqrt {a} \sqrt {c} e d-a e^2\right ) \log \left (\sqrt {c} x^2-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}\right ) e^2}{4 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )^3}+\frac {c^{3/4} \left (3 c d^2+4 \sqrt {a} \sqrt {c} e d-a e^2\right ) \log \left (\sqrt {c} x^2+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}\right ) e^2}{4 \sqrt {2} a^{3/4} \left (c d^2+a e^2\right )^3}-\frac {c^{3/4} \left (3 c d^2-2 \sqrt {a} \sqrt {c} e d-3 a e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}+\frac {c^{3/4} \left (3 c d^2-2 \sqrt {a} \sqrt {c} e d-3 a e^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}-\frac {c^{3/4} \left (3 c d^2+2 \sqrt {a} \sqrt {c} e d-3 a e^2\right ) \log \left (\sqrt {c} x^2-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}\right )}{16 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}+\frac {c^{3/4} \left (3 c d^2+2 \sqrt {a} \sqrt {c} e d-3 a e^2\right ) \log \left (\sqrt {c} x^2+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}\right )}{16 \sqrt {2} a^{7/4} \left (c d^2+a e^2\right )^2}+\frac {c x \left (c d^2-2 c e x^2 d-a e^2\right )}{4 a \left (c d^2+a e^2\right )^2 \left (c x^4+a\right )}\)

Input:

Int[1/((d + e*x^2)^2*(a + c*x^4)^2),x]
 

Output:

(e^4*x)/(2*d*(c*d^2 + a*e^2)^2*(d + e*x^2)) + (c*x*(c*d^2 - a*e^2 - 2*c*d* 
e*x^2))/(4*a*(c*d^2 + a*e^2)^2*(a + c*x^4)) + (4*c*Sqrt[d]*e^(7/2)*ArcTan[ 
(Sqrt[e]*x)/Sqrt[d]])/(c*d^2 + a*e^2)^3 + (e^(7/2)*ArcTan[(Sqrt[e]*x)/Sqrt 
[d]])/(2*d^(3/2)*(c*d^2 + a*e^2)^2) - (c^(3/4)*e^2*(3*c*d^2 - 4*Sqrt[a]*Sq 
rt[c]*d*e - a*e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^( 
3/4)*(c*d^2 + a*e^2)^3) - (c^(3/4)*(3*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - 3*a* 
e^2)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(c*d^2 + 
a*e^2)^2) + (c^(3/4)*e^2*(3*c*d^2 - 4*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTan[ 
1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)^3) + 
(c^(3/4)*(3*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - 3*a*e^2)*ArcTan[1 + (Sqrt[2]*c 
^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)^2) - (c^(3/4)*e^2*( 
3*c*d^2 + 4*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^( 
1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)^3) - (c^(3/4)*(3 
*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - 3*a*e^2)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^ 
(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)^2) + (c^(3/4)* 
e^2*(3*c*d^2 + 4*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/4 
)*c^(1/4)*x + Sqrt[c]*x^2])/(4*Sqrt[2]*a^(3/4)*(c*d^2 + a*e^2)^3) + (c^(3/ 
4)*(3*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - 3*a*e^2)*Log[Sqrt[a] + Sqrt[2]*a^(1/ 
4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*(c*d^2 + a*e^2)^2)
 

Defintions of rubi rules used

rule 1568

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int 
[ExpandIntegrand[(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, 
p, q}, x] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0])
 

rule 2009

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 402, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {e^{4} \left (\frac {\left (a \,e^{2}+c \,d^{2}\right ) x}{2 d \left (e \,x^{2}+d \right )}+\frac {\left (a \,e^{2}+9 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 d \sqrt {d e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {c \left (\frac {\frac {e c d \left (a \,e^{2}+c \,d^{2}\right ) x^{3}}{2 a}+\frac {\left (a^{2} e^{4}-c^{2} d^{4}\right ) x}{4 a}}{c \,x^{4}+a}+\frac {\frac {\left (7 a^{2} e^{4}-12 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {\left (18 a c d \,e^{3}+2 c^{2} d^{3} e \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}}{4 a}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}\)	\(402\)
risch	\(\text {Expression too large to display}\)	\(3814\)

Input:

int(1/(e*x^2+d)^2/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

e^4/(a*e^2+c*d^2)^3*(1/2*(a*e^2+c*d^2)/d*x/(e*x^2+d)+1/2*(a*e^2+9*c*d^2)/d 
/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))-c/(a*e^2+c*d^2)^3*((1/2*e*c*d*(a*e^2 
+c*d^2)/a*x^3+1/4*(a^2*e^4-c^2*d^4)/a*x)/(c*x^4+a)+1/4/a*(1/8*(7*a^2*e^4-1 
2*a*c*d^2*e^2-3*c^2*d^4)*(1/c*a)^(1/4)/a*2^(1/2)*(ln((x^2+(1/c*a)^(1/4)*x* 
2^(1/2)+(1/c*a)^(1/2))/(x^2-(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))+2*arct 
an(2^(1/2)/(1/c*a)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1))+1/8*(18 
*a*c*d*e^3+2*c^2*d^3*e)/c/(1/c*a)^(1/4)*2^(1/2)*(ln((x^2-(1/c*a)^(1/4)*x*2 
^(1/2)+(1/c*a)^(1/2))/(x^2+(1/c*a)^(1/4)*x*2^(1/2)+(1/c*a)^(1/2)))+2*arcta 
n(2^(1/2)/(1/c*a)^(1/4)*x+1)+2*arctan(2^(1/2)/(1/c*a)^(1/4)*x-1))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7634 vs. \(2 (412) = 824\).

Time = 100.17 (sec) , antiderivative size = 15292, normalized size of antiderivative = 31.08 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x^2+d)^2/(c*x^4+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x**2+d)**2/(c*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(e*x^2+d)^2/(c*x^4+a)^2,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 879 vs. \(2 (412) = 824\).

Time = 0.16 (sec) , antiderivative size = 879, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(1/(e*x^2+d)^2/(c*x^4+a)^2,x, algorithm="giac")
 

Output:

1/8*(3*(a*c^3)^(1/4)*c^3*d^4 + 12*(a*c^3)^(1/4)*a*c^2*d^2*e^2 - 7*(a*c^3)^ 
(1/4)*a^2*c*e^4 - 2*(a*c^3)^(3/4)*c*d^3*e - 18*(a*c^3)^(3/4)*a*d*e^3)*arct 
an(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a^2*c^4*d 
^6 + 3*sqrt(2)*a^3*c^3*d^4*e^2 + 3*sqrt(2)*a^4*c^2*d^2*e^4 + sqrt(2)*a^5*c 
*e^6) + 1/8*(3*(a*c^3)^(1/4)*c^3*d^4 + 12*(a*c^3)^(1/4)*a*c^2*d^2*e^2 - 7* 
(a*c^3)^(1/4)*a^2*c*e^4 - 2*(a*c^3)^(3/4)*c*d^3*e - 18*(a*c^3)^(3/4)*a*d*e 
^3)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(sqrt(2)*a 
^2*c^4*d^6 + 3*sqrt(2)*a^3*c^3*d^4*e^2 + 3*sqrt(2)*a^4*c^2*d^2*e^4 + sqrt( 
2)*a^5*c*e^6) + 1/16*(3*(a*c^3)^(1/4)*c^3*d^4 + 12*(a*c^3)^(1/4)*a*c^2*d^2 
*e^2 - 7*(a*c^3)^(1/4)*a^2*c*e^4 + 2*(a*c^3)^(3/4)*c*d^3*e + 18*(a*c^3)^(3 
/4)*a*d*e^3)*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^4 
*d^6 + 3*sqrt(2)*a^3*c^3*d^4*e^2 + 3*sqrt(2)*a^4*c^2*d^2*e^4 + sqrt(2)*a^5 
*c*e^6) - 1/16*(3*(a*c^3)^(1/4)*c^3*d^4 + 12*(a*c^3)^(1/4)*a*c^2*d^2*e^2 - 
 7*(a*c^3)^(1/4)*a^2*c*e^4 + 2*(a*c^3)^(3/4)*c*d^3*e + 18*(a*c^3)^(3/4)*a* 
d*e^3)*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(sqrt(2)*a^2*c^4*d^6 + 
 3*sqrt(2)*a^3*c^3*d^4*e^2 + 3*sqrt(2)*a^4*c^2*d^2*e^4 + sqrt(2)*a^5*c*e^6 
) + 1/2*(9*c*d^2*e^4 + a*e^6)*arctan(e*x/sqrt(d*e))/((c^3*d^7 + 3*a*c^2*d^ 
5*e^2 + 3*a^2*c*d^3*e^4 + a^3*d*e^6)*sqrt(d*e)) - 1/4*(2*c^2*d^2*e^2*x^5 - 
 2*a*c*e^4*x^5 + c^2*d^3*e*x^3 + a*c*d*e^3*x^3 - c^2*d^4*x + a*c*d^2*e^2*x 
 - 2*a^2*e^4*x)/((a*c^2*d^5 + 2*a^2*c*d^3*e^2 + a^3*d*e^4)*(c*e*x^6 + c...
 

Mupad [B] (verification not implemented)

Time = 21.29 (sec) , antiderivative size = 28923, normalized size of antiderivative = 58.79 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/((a + c*x^4)^2*(d + e*x^2)^2),x)
 

Output:

((x*(2*a^2*e^4 + c^2*d^4 - a*c*d^2*e^2))/(4*a*d*(a^2*e^4 + c^2*d^4 + 2*a*c 
*d^2*e^2)) - (c*e*x^3)/(4*a*(a*e^2 + c*d^2)) + (c*e^2*x^5*(a*e^2 - c*d^2)) 
/(2*a*d*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))/(a*d + a*e*x^2 + c*d*x^4 + c 
*e*x^6) + atan(((((3584*a^10*c^5*e^21 + 1152*a*c^14*d^18*e^3 + 13184*a^2*c 
^13*d^16*e^5 + 54912*a^3*c^12*d^14*e^7 + 296832*a^4*c^11*d^12*e^9 + 128243 
2*a^5*c^10*d^10*e^11 + 769152*a^6*c^9*d^8*e^13 - 1421440*a^7*c^8*d^6*e^15 
- 1254784*a^8*c^7*d^4*e^17 - 89088*a^9*c^6*d^2*e^19)/(512*(a^4*c^8*d^18 + 
a^12*d^2*e^16 + 8*a^11*c*d^4*e^14 + 8*a^5*c^7*d^16*e^2 + 28*a^6*c^6*d^14*e 
^4 + 56*a^7*c^5*d^12*e^6 + 70*a^8*c^4*d^10*e^8 + 56*a^9*c^3*d^8*e^10 + 28* 
a^10*c^2*d^6*e^12)) - (((65536*a^15*c^4*d*e^24 - 24576*a^4*c^15*d^23*e^2 - 
 212992*a^5*c^14*d^21*e^4 - 352256*a^6*c^13*d^19*e^6 + 1966080*a^7*c^12*d^ 
17*e^8 + 10960896*a^8*c^11*d^15*e^10 + 25460736*a^9*c^10*d^13*e^12 + 34750 
464*a^10*c^9*d^11*e^14 + 30081024*a^11*c^8*d^9*e^16 + 16588800*a^12*c^7*d^ 
7*e^18 + 5554176*a^13*c^6*d^5*e^20 + 991232*a^14*c^5*d^3*e^22)/(512*(a^4*c 
^8*d^18 + a^12*d^2*e^16 + 8*a^11*c*d^4*e^14 + 8*a^5*c^7*d^16*e^2 + 28*a^6* 
c^6*d^14*e^4 + 56*a^7*c^5*d^12*e^6 + 70*a^8*c^4*d^10*e^8 + 56*a^9*c^3*d^8* 
e^10 + 28*a^10*c^2*d^6*e^12)) - (x*(-(49*a^4*e^8*(-a^7*c^3)^(1/2) + 9*c^4* 
d^8*(-a^7*c^3)^(1/2) - 12*a^4*c^5*d^7*e + 252*a^7*c^2*d*e^7 - 156*a^5*c^4* 
d^5*e^3 - 404*a^6*c^3*d^3*e^5 + 68*a*c^3*d^6*e^2*(-a^7*c^3)^(1/2) - 492*a^ 
3*c*d^2*e^6*(-a^7*c^3)^(1/2) + 30*a^2*c^2*d^4*e^4*(-a^7*c^3)^(1/2))/(25...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 3860, normalized size of antiderivative = 7.85 \[ \int \frac {1}{\left (d+e x^2\right )^2 \left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int(1/(e*x^2+d)^2/(c*x^4+a)^2,x)
 

Output:

(36*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)* 
x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**2*c*d**4*e**3 + 36*c**(1/4)*a**(3/4)*sq 
rt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sq 
rt(2)))*a**2*c*d**3*e**4*x**2 + 4*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4) 
*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c**2*d**6* 
e + 4*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c 
)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c**2*d**5*e**2*x**2 + 36*c**(1/4)*a**( 
3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**( 
1/4)*sqrt(2)))*a*c**2*d**4*e**3*x**4 + 36*c**(1/4)*a**(3/4)*sqrt(2)*atan(( 
c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*c* 
*2*d**3*e**4*x**6 + 4*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sq 
rt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c**3*d**6*e*x**4 + 4*c** 
(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c** 
(1/4)*a**(1/4)*sqrt(2)))*c**3*d**5*e**2*x**6 + 14*c**(3/4)*a**(1/4)*sqrt(2 
)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2 
)))*a**3*d**3*e**4 + 14*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)* 
sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**3*d**2*e**5*x**2 - 
24*c**(3/4)*a**(1/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x 
)/(c**(1/4)*a**(1/4)*sqrt(2)))*a**2*c*d**5*e**2 - 24*c**(3/4)*a**(1/4)*sqr 
t(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*...