\(\int \frac {x}{(d+e x^4) (a+b x^4+c x^8)^2} \, dx\) [113]

Optimal result

Integrand size = 25, antiderivative size = 759 \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {x^2 \left (c \left (b^2-2 a c\right ) d-b \left (b^2-3 a c\right ) e+c \left (b c d-b^2 e+2 a c e\right ) x^4\right )}{4 a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x^4+c x^8\right )}+\frac {\sqrt {c} \left (b^4 d e^2-2 a c \left (6 c^2 d^3-5 a \sqrt {b^2-4 a c} e^3-c d e \left (\sqrt {b^2-4 a c} d-14 a e\right )\right )-b^3 e \left (2 c d^2-e \left (\sqrt {b^2-4 a c} d-3 a e\right )\right )+b^2 \left (c^2 d^3-3 a \sqrt {b^2-4 a c} e^3-c d e \left (2 \sqrt {b^2-4 a c} d+3 a e\right )\right )-b c \left (a e^2 \left (\sqrt {b^2-4 a c} d-16 a e\right )-c d^2 \left (\sqrt {b^2-4 a c} d+20 a e\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{4 \sqrt {2} a \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )^2}-\frac {\sqrt {c} \left (b^4 d e^2+b^2 \left (c^2 d^3+3 a \sqrt {b^2-4 a c} e^3+c d e \left (2 \sqrt {b^2-4 a c} d-3 a e\right )\right )-b^3 e \left (2 c d^2+e \left (\sqrt {b^2-4 a c} d+3 a e\right )\right )-2 a c \left (6 c^2 d^3+5 a \sqrt {b^2-4 a c} e^3+c d e \left (\sqrt {b^2-4 a c} d+14 a e\right )\right )-b c \left (c d^2 \left (\sqrt {b^2-4 a c} d-20 a e\right )-a e^2 \left (\sqrt {b^2-4 a c} d+16 a e\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{4 \sqrt {2} a \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}} \left (c d^2-e (b d-a e)\right )^2}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x^2}{\sqrt {d}}\right )}{2 \sqrt {d} \left (c d^2-b d e+a e^2\right )^2} \] Output:

1/4*x^2*(c*(-2*a*c+b^2)*d-b*(-3*a*c+b^2)*e+c*(2*a*c*e-b^2*e+b*c*d)*x^4)/a/ 
(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(c*x^8+b*x^4+a)+1/8*c^(1/2)*(b^4*d*e^2-2* 
a*c*(6*c^2*d^3-5*a*(-4*a*c+b^2)^(1/2)*e^3-c*d*e*((-4*a*c+b^2)^(1/2)*d-14*a 
*e))-b^3*e*(2*c*d^2-e*((-4*a*c+b^2)^(1/2)*d-3*a*e))+b^2*(c^2*d^3-3*a*(-4*a 
*c+b^2)^(1/2)*e^3-c*d*e*(2*(-4*a*c+b^2)^(1/2)*d+3*a*e))-b*c*(a*e^2*((-4*a* 
c+b^2)^(1/2)*d-16*a*e)-c*d^2*((-4*a*c+b^2)^(1/2)*d+20*a*e)))*arctan(2^(1/2 
)*c^(1/2)*x^2/(b-(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/a/(-4*a*c+b^2)^(3/2)/( 
b-(-4*a*c+b^2)^(1/2))^(1/2)/(a*e^2-b*d*e+c*d^2)^2-1/8*c^(1/2)*(b^4*d*e^2+b 
^2*(c^2*d^3+3*a*(-4*a*c+b^2)^(1/2)*e^3+c*d*e*(2*(-4*a*c+b^2)^(1/2)*d-3*a*e 
))-b^3*e*(2*c*d^2+e*((-4*a*c+b^2)^(1/2)*d+3*a*e))-2*a*c*(6*c^2*d^3+5*a*(-4 
*a*c+b^2)^(1/2)*e^3+c*d*e*((-4*a*c+b^2)^(1/2)*d+14*a*e))-b*c*(c*d^2*((-4*a 
*c+b^2)^(1/2)*d-20*a*e)-a*e^2*((-4*a*c+b^2)^(1/2)*d+16*a*e)))*arctan(2^(1/ 
2)*c^(1/2)*x^2/(b+(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/a/(-4*a*c+b^2)^(3/2)/ 
(b+(-4*a*c+b^2)^(1/2))^(1/2)/(c*d^2-e*(-a*e+b*d))^2+1/2*e^(7/2)*arctan(e^( 
1/2)*x^2/d^(1/2))/d^(1/2)/(a*e^2-b*d*e+c*d^2)^2
 

Mathematica [A] (verified)

Time = 2.86 (sec) , antiderivative size = 716, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\frac {\frac {2 \left (c d^2+e (-b d+a e)\right ) x^2 \left (b^3 e-b c \left (3 a e+c d x^4\right )+2 a c^2 \left (d-e x^4\right )+b^2 c \left (-d+e x^4\right )\right )}{a \left (-b^2+4 a c\right ) \left (a+b x^4+c x^8\right )}+\frac {\sqrt {2} \sqrt {c} \left (b^4 d e^2+2 a c \left (-6 c^2 d^3+5 a \sqrt {b^2-4 a c} e^3+c d e \left (\sqrt {b^2-4 a c} d-14 a e\right )\right )+b^3 e \left (-2 c d^2+e \left (\sqrt {b^2-4 a c} d-3 a e\right )\right )+b^2 \left (c^2 d^3-3 a \sqrt {b^2-4 a c} e^3-c d e \left (2 \sqrt {b^2-4 a c} d+3 a e\right )\right )+b c \left (a e^2 \left (-\sqrt {b^2-4 a c} d+16 a e\right )+c d^2 \left (\sqrt {b^2-4 a c} d+20 a e\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{a \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \sqrt {c} \left (-b^4 d e^2-b^2 \left (c^2 d^3+3 a \sqrt {b^2-4 a c} e^3+c d e \left (2 \sqrt {b^2-4 a c} d-3 a e\right )\right )+b^3 e \left (2 c d^2+e \left (\sqrt {b^2-4 a c} d+3 a e\right )\right )+2 a c \left (6 c^2 d^3+5 a \sqrt {b^2-4 a c} e^3+c d e \left (\sqrt {b^2-4 a c} d+14 a e\right )\right )+b c \left (c d^2 \left (\sqrt {b^2-4 a c} d-20 a e\right )-a e^2 \left (\sqrt {b^2-4 a c} d+16 a e\right )\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{a \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}+\frac {4 e^{7/2} \arctan \left (\frac {\sqrt {e} x^2}{\sqrt {d}}\right )}{\sqrt {d}}}{8 \left (c d^2+e (-b d+a e)\right )^2} \] Input:

Integrate[x/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

((2*(c*d^2 + e*(-(b*d) + a*e))*x^2*(b^3*e - b*c*(3*a*e + c*d*x^4) + 2*a*c^ 
2*(d - e*x^4) + b^2*c*(-d + e*x^4)))/(a*(-b^2 + 4*a*c)*(a + b*x^4 + c*x^8) 
) + (Sqrt[2]*Sqrt[c]*(b^4*d*e^2 + 2*a*c*(-6*c^2*d^3 + 5*a*Sqrt[b^2 - 4*a*c 
]*e^3 + c*d*e*(Sqrt[b^2 - 4*a*c]*d - 14*a*e)) + b^3*e*(-2*c*d^2 + e*(Sqrt[ 
b^2 - 4*a*c]*d - 3*a*e)) + b^2*(c^2*d^3 - 3*a*Sqrt[b^2 - 4*a*c]*e^3 - c*d* 
e*(2*Sqrt[b^2 - 4*a*c]*d + 3*a*e)) + b*c*(a*e^2*(-(Sqrt[b^2 - 4*a*c]*d) + 
16*a*e) + c*d^2*(Sqrt[b^2 - 4*a*c]*d + 20*a*e)))*ArcTan[(Sqrt[2]*Sqrt[c]*x 
^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(a*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 
 - 4*a*c]]) + (Sqrt[2]*Sqrt[c]*(-(b^4*d*e^2) - b^2*(c^2*d^3 + 3*a*Sqrt[b^2 
 - 4*a*c]*e^3 + c*d*e*(2*Sqrt[b^2 - 4*a*c]*d - 3*a*e)) + b^3*e*(2*c*d^2 + 
e*(Sqrt[b^2 - 4*a*c]*d + 3*a*e)) + 2*a*c*(6*c^2*d^3 + 5*a*Sqrt[b^2 - 4*a*c 
]*e^3 + c*d*e*(Sqrt[b^2 - 4*a*c]*d + 14*a*e)) + b*c*(c*d^2*(Sqrt[b^2 - 4*a 
*c]*d - 20*a*e) - a*e^2*(Sqrt[b^2 - 4*a*c]*d + 16*a*e)))*ArcTan[(Sqrt[2]*S 
qrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(a*(b^2 - 4*a*c)^(3/2)*Sqrt[b + 
Sqrt[b^2 - 4*a*c]]) + (4*e^(7/2)*ArcTan[(Sqrt[e]*x^2)/Sqrt[d]])/Sqrt[d])/( 
8*(c*d^2 + e*(-(b*d) + a*e))^2)
 

Rubi [A] (verified)

Time = 2.23 (sec) , antiderivative size = 676, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1814, 1567, 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.

\(\displaystyle \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx\)

\(\Big \downarrow \) 1814

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (e x^4+d\right ) \left (c x^8+b x^4+a\right )^2}dx^2\)

\(\Big \downarrow \) 1567

\(\displaystyle \frac {1}{2} \int \left (\frac {e^4}{\left (c d^2-b e d+a e^2\right )^2 \left (e x^4+d\right )}-\frac {\left (c e x^4-c d+b e\right ) e^2}{\left (c d^2-b e d+a e^2\right )^2 \left (c x^8+b x^4+a\right )}+\frac {-c e x^4+c d-b e}{\left (c d^2-b e d+a e^2\right ) \left (c x^8+b x^4+a\right )^2}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {\sqrt {c} e^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}} \left (a e^2-b d e+c d^2\right )^2}-\frac {\sqrt {c} e^2 \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right )}{\sqrt {2} \sqrt {\sqrt {b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )^2}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {8 a b c e-12 a c^2 d+b^3 (-e)+b^2 c d}{\sqrt {b^2-4 a c}}+2 a c e+b^2 (-e)+b c d\right )}{2 \sqrt {2} a \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (-\frac {8 a b c e-12 a c^2 d+b^3 (-e)+b^2 c d}{\sqrt {b^2-4 a c}}+2 a c e+b^2 (-e)+b c d\right )}{2 \sqrt {2} a \left (b^2-4 a c\right ) \sqrt {\sqrt {b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x^2}{\sqrt {d}}\right )}{\sqrt {d} \left (a e^2-b d e+c d^2\right )^2}+\frac {x^2 \left (c x^4 \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d-b^3 e+b^2 c d\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right ) \left (a e^2-b d e+c d^2\right )}\right )\)

Input:

Int[x/((d + e*x^4)*(a + b*x^4 + c*x^8)^2),x]
 

Output:

((x^2*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e + c*(b*c*d - b^2*e + 2*a*c* 
e)*x^4))/(2*a*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a + b*x^4 + c*x^8)) - 
 (Sqrt[c]*e^2*(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c 
]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b - Sqrt[b^2 - 4*a*c]]* 
(c*d^2 - b*d*e + a*e^2)^2) + (Sqrt[c]*(b*c*d - b^2*e + 2*a*c*e + (b^2*c*d 
- 12*a*c^2*d - b^3*e + 8*a*b*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[ 
c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b - 
Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)) - (Sqrt[c]*e^2*(e + (2*c*d - b 
*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4* 
a*c]]])/(Sqrt[2]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)^2) + 
(Sqrt[c]*(b*c*d - b^2*e + 2*a*c*e - (b^2*c*d - 12*a*c^2*d - b^3*e + 8*a*b* 
c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4 
*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 - b 
*d*e + a*e^2)) + (e^(7/2)*ArcTan[(Sqrt[e]*x^2)/Sqrt[d]])/(Sqrt[d]*(c*d^2 - 
 b*d*e + a*e^2)^2))/2
 

Defintions of rubi rules used

rule 1567

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

rule 1814

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e 
_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Sub 
st[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + b*x^(n/k) + c*x^(2*(n/k)))^ 
p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, p, q}, x] && EqQ[n2, 
 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]
 

rule 2009

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

Maple [A] (verified)

Time = 3.52 (sec) , antiderivative size = 859, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {\frac {\frac {c \left (2 a^{2} c \,e^{3}-a \,b^{2} e^{3}-a b c d \,e^{2}+2 a \,c^{2} d^{2} e +b^{3} d \,e^{2}-2 b^{2} c \,d^{2} e +b \,c^{2} d^{3}\right ) x^{6}}{2 a \left (4 a c -b^{2}\right )}+\frac {\left (3 a^{2} b c \,e^{3}-2 a^{2} c^{2} e^{2} d -a \,b^{3} e^{3}-2 a \,b^{2} c d \,e^{2}+5 a b \,c^{2} d^{2} e -2 a \,c^{3} d^{3}+b^{4} d \,e^{2}-2 b^{3} c \,d^{2} e +b^{2} c^{2} d^{3}\right ) x^{2}}{2 a \left (4 a c -b^{2}\right )}}{c \,x^{8}+b \,x^{4}+a}+\frac {2 c \left (-\frac {\left (10 a^{2} c \,e^{3} \sqrt {-4 a c +b^{2}}-3 a \,b^{2} e^{3} \sqrt {-4 a c +b^{2}}-a b c d \,e^{2} \sqrt {-4 a c +b^{2}}+2 a \,c^{2} d^{2} e \sqrt {-4 a c +b^{2}}+b^{3} d \,e^{2} \sqrt {-4 a c +b^{2}}-2 b^{2} c \,d^{2} e \sqrt {-4 a c +b^{2}}+b \,c^{2} d^{3} \sqrt {-4 a c +b^{2}}+16 a^{2} b c \,e^{3}-28 a^{2} c^{2} e^{2} d -3 a \,b^{3} e^{3}-3 a \,b^{2} c d \,e^{2}+20 a b \,c^{2} d^{2} e -12 a \,c^{3} d^{3}+b^{4} d \,e^{2}-2 b^{3} c \,d^{2} e +b^{2} c^{2} d^{3}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (10 a^{2} c \,e^{3} \sqrt {-4 a c +b^{2}}-3 a \,b^{2} e^{3} \sqrt {-4 a c +b^{2}}-a b c d \,e^{2} \sqrt {-4 a c +b^{2}}+2 a \,c^{2} d^{2} e \sqrt {-4 a c +b^{2}}+b^{3} d \,e^{2} \sqrt {-4 a c +b^{2}}-2 b^{2} c \,d^{2} e \sqrt {-4 a c +b^{2}}+b \,c^{2} d^{3} \sqrt {-4 a c +b^{2}}-16 a^{2} b c \,e^{3}+28 a^{2} c^{2} e^{2} d +3 a \,b^{3} e^{3}+3 a \,b^{2} c d \,e^{2}-20 a b \,c^{2} d^{2} e +12 a \,c^{3} d^{3}-b^{4} d \,e^{2}+2 b^{3} c \,d^{2} e -b^{2} c^{2} d^{3}\right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{a \left (4 a c -b^{2}\right )}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {e^{4} \arctan \left (\frac {e \,x^{2}}{\sqrt {d e}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {d e}}\)	\(859\)
risch	\(\text {Expression too large to display}\)	\(19601\)

Input:

int(x/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2/(a*e^2-b*d*e+c*d^2)^2*((1/2*c*(2*a^2*c*e^3-a*b^2*e^3-a*b*c*d*e^2+2*a* 
c^2*d^2*e+b^3*d*e^2-2*b^2*c*d^2*e+b*c^2*d^3)/a/(4*a*c-b^2)*x^6+1/2*(3*a^2* 
b*c*e^3-2*a^2*c^2*d*e^2-a*b^3*e^3-2*a*b^2*c*d*e^2+5*a*b*c^2*d^2*e-2*a*c^3* 
d^3+b^4*d*e^2-2*b^3*c*d^2*e+b^2*c^2*d^3)/a/(4*a*c-b^2)*x^2)/(c*x^8+b*x^4+a 
)+2/a/(4*a*c-b^2)*c*(-1/8*(10*a^2*c*e^3*(-4*a*c+b^2)^(1/2)-3*a*b^2*e^3*(-4 
*a*c+b^2)^(1/2)-a*b*c*d*e^2*(-4*a*c+b^2)^(1/2)+2*a*c^2*d^2*e*(-4*a*c+b^2)^ 
(1/2)+b^3*d*e^2*(-4*a*c+b^2)^(1/2)-2*b^2*c*d^2*e*(-4*a*c+b^2)^(1/2)+b*c^2* 
d^3*(-4*a*c+b^2)^(1/2)+16*a^2*b*c*e^3-28*a^2*c^2*e^2*d-3*a*b^3*e^3-3*a*b^2 
*c*d*e^2+20*a*b*c^2*d^2*e-12*a*c^3*d^3+b^4*d*e^2-2*b^3*c*d^2*e+b^2*c^2*d^3 
)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x 
^2*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))+1/8*(10*a^2*c*e^3*(-4*a*c+b^ 
2)^(1/2)-3*a*b^2*e^3*(-4*a*c+b^2)^(1/2)-a*b*c*d*e^2*(-4*a*c+b^2)^(1/2)+2*a 
*c^2*d^2*e*(-4*a*c+b^2)^(1/2)+b^3*d*e^2*(-4*a*c+b^2)^(1/2)-2*b^2*c*d^2*e*( 
-4*a*c+b^2)^(1/2)+b*c^2*d^3*(-4*a*c+b^2)^(1/2)-16*a^2*b*c*e^3+28*a^2*c^2*e 
^2*d+3*a*b^3*e^3+3*a*b^2*c*d*e^2-20*a*b*c^2*d^2*e+12*a*c^3*d^3-b^4*d*e^2+2 
*b^3*c*d^2*e-b^2*c^2*d^3)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2 
))*c)^(1/2)*arctan(c*x^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))))+1/2*e 
^4/(a*e^2-b*d*e+c*d^2)^2/(d*e)^(1/2)*arctan(e*x^2/(d*e)^(1/2))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x/(e*x**4+d)/(c*x**8+b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x/(e*x^4+d)/(c*x^8+b*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. 37269 vs. \(2 (678) = 1356\).

Time = 9.00 (sec) , antiderivative size = 37269, normalized size of antiderivative = 49.10 \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x/(e*x^4+d)/(c*x^8+b*x^4+a)^2,x, algorithm="giac")
 

Output:

1/2*e^4*arctan(e*x^2/sqrt(d*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2* 
a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*sqrt(d*e)) + 1/32*((2*a^2*b^7*c^8 - 4 
0*a^3*b^5*c^9 + 224*a^4*b^3*c^10 - 384*a^5*b*c^11 - sqrt(2)*sqrt(b^2 - 4*a 
*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^7*c^6 + 20*sqrt(2)*sqrt(b^2 - 4* 
a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^5*c^7 + 2*sqrt(2)*sqrt(b^2 - 4* 
a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^6*c^7 - 112*sqrt(2)*sqrt(b^2 - 
4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^3*c^8 - 32*sqrt(2)*sqrt(b^2 - 
 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^4*c^8 - sqrt(2)*sqrt(b^2 - 4 
*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^5*c^8 + 192*sqrt(2)*sqrt(b^2 - 
 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5*b*c^9 + 96*sqrt(2)*sqrt(b^2 - 
4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^2*c^9 + 16*sqrt(2)*sqrt(b^2 - 
 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^3*c^9 - 48*sqrt(2)*sqrt(b^2 
- 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b*c^10 - 2*(b^2 - 4*a*c)*a^2* 
b^5*c^8 + 32*(b^2 - 4*a*c)*a^3*b^3*c^9 - 96*(b^2 - 4*a*c)*a^4*b*c^10)*d^11 
 - 2*(6*a^2*b^8*c^7 - 116*a^3*b^6*c^8 + 640*a^4*b^4*c^9 - 1088*a^5*b^2*c^1 
0 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^8*c^ 
5 + 58*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^6*c 
^6 + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^7*c 
^6 - 320*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^4 
*c^7 - 92*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {x}{\left (d+e x^4\right ) \left (a+b x^4+c x^8\right )^2} \, dx=\int \frac {x}{\left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )^{2}}d x \] Input:

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

Output:

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