\(\int \frac {d+e x^4}{x^3 (a+b x^4+c x^8)^2} \, dx\) [98]

Optimal result

Integrand size = 25, antiderivative size = 376 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=-\frac {3 b^2 d-10 a c d-a b e}{4 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x^4}{4 a \left (b^2-4 a c\right ) x^2 \left (a+b x^4+c x^8\right )}-\frac {\sqrt {c} \left (3 b^2 d-10 a c d-a b e+\frac {3 b^3 d-16 a b c d-a b^2 e+12 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{4 \sqrt {2} a^2 \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (3 b^2 d-10 a c d-a b e-\frac {3 b^3 d-16 a b c d-a b^2 e+12 a^2 c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{4 \sqrt {2} a^2 \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.26 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.65 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=-\frac {\frac {4 d}{x^2}+\frac {2 x^2 \left (b^3 d+2 a c \left (a e-c d x^4\right )+b^2 \left (-a e+c d x^4\right )-a b c \left (3 d+e x^4\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}+\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {3 b^3 d \log (x-\text {$\#$1})-13 a b c d \log (x-\text {$\#$1})-a b^2 e \log (x-\text {$\#$1})+6 a^2 c e \log (x-\text {$\#$1})+3 b^2 c d \log (x-\text {$\#$1}) \text {$\#$1}^4-10 a c^2 d \log (x-\text {$\#$1}) \text {$\#$1}^4-a b c e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^2+2 c \text {$\#$1}^6}\&\right ]}{b^2-4 a c}}{8 a^2} \] Input:

Integrate[(d + e*x^4)/(x^3*(a + b*x^4 + c*x^8)^2),x]
 

Output:

-1/8*((4*d)/x^2 + (2*x^2*(b^3*d + 2*a*c*(a*e - c*d*x^4) + b^2*(-(a*e) + c* 
d*x^4) - a*b*c*(3*d + e*x^4)))/((b^2 - 4*a*c)*(a + b*x^4 + c*x^8)) + RootS 
um[a + b*#1^4 + c*#1^8 & , (3*b^3*d*Log[x - #1] - 13*a*b*c*d*Log[x - #1] - 
 a*b^2*e*Log[x - #1] + 6*a^2*c*e*Log[x - #1] + 3*b^2*c*d*Log[x - #1]*#1^4 
- 10*a*c^2*d*Log[x - #1]*#1^4 - a*b*c*e*Log[x - #1]*#1^4)/(b*#1^2 + 2*c*#1 
^6) & ]/(b^2 - 4*a*c))/a^2
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1814, 1600, 25, 1604, 1480, 218}

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.

\(\displaystyle \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx\)

\(\Big \downarrow \) 1814

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

\(\Big \downarrow \) 1600

\(\displaystyle \frac {1}{2} \left (\frac {c x^4 (b d-2 a e)-a b e-2 a c d+b^2 d}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}-\frac {\int -\frac {3 c (b d-2 a e) x^4+3 b^2 d-10 a c d-a b e}{x^4 \left (c x^8+b x^4+a\right )}dx^2}{2 a \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {3 c (b d-2 a e) x^4+3 b^2 d-10 a c d-a b e}{x^4 \left (c x^8+b x^4+a\right )}dx^2}{2 a \left (b^2-4 a c\right )}+\frac {c x^4 (b d-2 a e)-a b e-2 a c d+b^2 d}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}\right )\)

\(\Big \downarrow \) 1604

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\int \frac {c \left (3 d b^2-a e b-10 a c d\right ) x^4+3 b^3 d-13 a b c d-a b^2 e+6 a^2 c e}{c x^8+b x^4+a}dx^2}{a}-\frac {-a b e-10 a c d+3 b^2 d}{a x^2}}{2 a \left (b^2-4 a c\right )}+\frac {c x^4 (b d-2 a e)-a b e-2 a c d+b^2 d}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}\right )\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\frac {1}{2} c \left (\frac {12 a^2 c e-a b^2 e-16 a b c d+3 b^3 d}{\sqrt {b^2-4 a c}}-a b e-10 a c d+3 b^2 d\right ) \int \frac {1}{c x^4+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx^2+\frac {1}{2} c \left (-\frac {12 a^2 c e-a b^2 e-16 a b c d+3 b^3 d}{\sqrt {b^2-4 a c}}-a b e-10 a c d+3 b^2 d\right ) \int \frac {1}{c x^4+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx^2}{a}-\frac {-a b e-10 a c d+3 b^2 d}{a x^2}}{2 a \left (b^2-4 a c\right )}+\frac {c x^4 (b d-2 a e)-a b e-2 a c d+b^2 d}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{2} \left (\frac {-\frac {\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {12 a^2 c e-a b^2 e-16 a b c d+3 b^3 d}{\sqrt {b^2-4 a c}}-a b e-10 a c d+3 b^2 d\right )}{\sqrt {2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (-\frac {12 a^2 c e-a b^2 e-16 a b c d+3 b^3 d}{\sqrt {b^2-4 a c}}-a b e-10 a c d+3 b^2 d\right )}{\sqrt {2} \sqrt {\sqrt {b^2-4 a c}+b}}}{a}-\frac {-a b e-10 a c d+3 b^2 d}{a x^2}}{2 a \left (b^2-4 a c\right )}+\frac {c x^4 (b d-2 a e)-a b e-2 a c d+b^2 d}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^4+c x^8\right )}\right )\)

Input:

Int[(d + e*x^4)/(x^3*(a + b*x^4 + c*x^8)^2),x]
 

Output:

((b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^4)/(2*a*(b^2 - 4*a*c)*x^2*(a 
 + b*x^4 + c*x^8)) + (-((3*b^2*d - 10*a*c*d - a*b*e)/(a*x^2)) - ((Sqrt[c]* 
(3*b^2*d - 10*a*c*d - a*b*e + (3*b^3*d - 16*a*b*c*d - a*b^2*e + 12*a^2*c*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]]) + (Sqrt[c]*(3*b^2*d - 10*a*c*d 
 - a*b*e - (3*b^3*d - 16*a*b*c*d - a*b^2*e + 12*a^2*c*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]]))/a)/(2*a*(b^2 - 4*a*c)))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1600

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^2 + c*x^4)^(p + 1) 
*((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2)/(2*a*f*(p + 1)*(b^2 - 4*a 
*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(f*x)^m*(a + b*x^2 + c 
*x^4)^(p + 1)*Simp[d*(b^2*(m + 2*(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) 
- a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x], x] /; 
FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Int 
egerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1604

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) 
/(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1))   Int[(f*x)^(m + 2)*(a + b*x^2 
 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x 
], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ 
m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

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]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\frac {-\frac {c \left (a b e +2 a c d -d \,b^{2}\right ) x^{6}}{2 \left (4 a c -b^{2}\right )}+\frac {\left (2 a^{2} c e -a \,b^{2} e -3 a b c d +b^{3} d \right ) x^{2}}{8 a c -2 b^{2}}}{c \,x^{8}+b \,x^{4}+a}+\frac {2 c \left (-\frac {\left (-a b e \sqrt {-4 a c +b^{2}}-10 \sqrt {-4 a c +b^{2}}\, a c d +3 \sqrt {-4 a c +b^{2}}\, b^{2} d +12 a^{2} c e -a \,b^{2} e -16 a b c d +3 b^{3} d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-a b e \sqrt {-4 a c +b^{2}}-10 \sqrt {-4 a c +b^{2}}\, a c d +3 \sqrt {-4 a c +b^{2}}\, b^{2} d -12 a^{2} c e +a \,b^{2} e +16 a b c d -3 b^{3} d \right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{2 a^{2}}-\frac {d}{2 a^{2} x^{2}}\)	\(387\)
risch	\(\text {Expression too large to display}\)	\(1522\)

Input:

int((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6960 vs. \(2 (332) = 664\).

Time = 9.29 (sec) , antiderivative size = 6960, normalized size of antiderivative = 18.51 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x**4+d)/x**3/(c*x**8+b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\int { \frac {e x^{4} + d}{{\left (c x^{8} + b x^{4} + a\right )}^{2} x^{3}} \,d x } \] Input:

integrate((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x, algorithm="maxima")
 

Output:

1/4*((a*b*c*e - (3*b^2*c - 10*a*c^2)*d)*x^8 - ((3*b^3 - 11*a*b*c)*d - (a*b 
^2 - 2*a^2*c)*e)*x^4 - 2*(a*b^2 - 4*a^2*c)*d)/((a^2*b^2*c - 4*a^3*c^2)*x^1 
0 + (a^2*b^3 - 4*a^3*b*c)*x^6 + (a^3*b^2 - 4*a^4*c)*x^2) + 1/2*integrate(( 
(a*b*c*e - (3*b^2*c - 10*a*c^2)*d)*x^4 - (3*b^3 - 13*a*b*c)*d + (a*b^2 - 6 
*a^2*c)*e)*x/(c*x^8 + b*x^4 + a), x)/(a^2*b^2 - 4*a^3*c)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5103 vs. \(2 (332) = 664\).

Time = 4.72 (sec) , antiderivative size = 5103, normalized size of antiderivative = 13.57 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x, algorithm="giac")
 

Output:

-1/4*(3*b^2*c*d*x^8 - 10*a*c^2*d*x^8 - a*b*c*e*x^8 + 3*b^3*d*x^4 - 11*a*b* 
c*d*x^4 - a*b^2*e*x^4 + 2*a^2*c*e*x^4 + 2*a*b^2*d - 8*a^2*c*d)/((c*x^10 + 
b*x^6 + a*x^2)*(a^2*b^2 - 4*a^3*c)) - 1/16*((3*sqrt(2)*sqrt(b*c + sqrt(b^2 
 - 4*a*c)*c)*b^4*c - 22*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 
- 6*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 - 6*b^4*c^2 + 40*sqrt( 
2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^3 + 20*sqrt(2)*sqrt(b*c + sqrt(b^ 
2 - 4*a*c)*c)*a*b*c^3 + 3*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^3 
+ 44*a*b^2*c^3 - 10*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^4 - 80*a^2 
*c^4 + 6*(b^2 - 4*a*c)*b^2*c^2 - 20*(b^2 - 4*a*c)*a*c^3)*d*x^4*abs(a^2*b^2 
 - 4*a^3*c) - (sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c - 4*sqrt(2) 
*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 
 - 4*a*c)*c)*a*b^2*c^2 - 2*a*b^3*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c 
)*c)*a*b*c^3 + 8*a^2*b*c^3 + 2*(b^2 - 4*a*c)*a*b*c^2)*e*x^4*abs(a^2*b^2 - 
4*a^3*c) + (6*a^2*b^5*c^3 - 44*a^3*b^3*c^4 + 80*a^4*b*c^5 - 3*sqrt(2)*sqrt 
(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^5*c + 22*sqrt(2)*sqrt( 
b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^3*c^2 + 6*sqrt(2)*sqrt( 
b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^4*c^2 - 40*sqrt(2)*sqrt 
(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b*c^3 - 20*sqrt(2)*sqrt( 
b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b^2*c^3 - 3*sqrt(2)*sqrt( 
b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^3*c^3 + 10*sqrt(2)*s...
 

Mupad [B] (verification not implemented)

Time = 31.02 (sec) , antiderivative size = 55298, normalized size of antiderivative = 147.07 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\text {Too large to display} \] Input:

int((d + e*x^4)/(x^3*(a + b*x^4 + c*x^8)^2),x)
 

Output:

atan(((x^2*(16307453952*a^28*c^17*e^5 + 62914560000*a^25*b*c^19*d^5 - 1258 
29120000*a^26*c^19*d^4*e + 145152*a^15*b^21*c^9*d^5 - 5723136*a^16*b^19*c^ 
10*d^5 + 100763136*a^17*b^17*c^11*d^5 - 1041678336*a^18*b^15*c^12*d^5 + 69 
88922880*a^19*b^13*c^13*d^5 - 31716016128*a^20*b^11*c^14*d^5 + 98230861824 
*a^21*b^9*c^15*d^5 - 203931254784*a^22*b^7*c^16*d^5 + 269299482624*a^23*b^ 
5*c^17*d^5 - 201326592000*a^24*b^3*c^18*d^5 - 7680*a^19*b^18*c^8*e^5 + 347 
136*a^20*b^16*c^9*e^5 - 7077888*a^21*b^14*c^10*e^5 + 84738048*a^22*b^12*c^ 
11*e^5 - 651952128*a^23*b^10*c^12*e^5 + 3326607360*a^24*b^8*c^13*e^5 - 112 
23957504*a^25*b^6*c^14*e^5 + 24108859392*a^26*b^4*c^15*e^5 - 29896998912*a 
^27*b^2*c^16*e^5 + 311040*a^16*b^21*c^8*d^3*e^2 - 12337920*a^17*b^19*c^9*d 
^3*e^2 - 241920*a^17*b^20*c^8*d^2*e^3 + 216852480*a^18*b^17*c^10*d^3*e^2 + 
 10083840*a^18*b^18*c^9*d^2*e^3 - 2214543360*a^19*b^15*c^11*d^3*e^2 - 1862 
86080*a^19*b^16*c^10*d^2*e^3 + 14460518400*a^20*b^13*c^12*d^3*e^2 + 200097 
7920*a^20*b^14*c^11*d^2*e^3 - 62456463360*a^21*b^11*c^13*d^3*e^2 - 1376845 
8240*a^21*b^12*c^12*d^2*e^3 + 177529159680*a^22*b^9*c^14*d^3*e^2 + 6292635 
6480*a^22*b^10*c^13*d^2*e^3 - 316271493120*a^23*b^7*c^15*d^3*e^2 - 1910086 
04160*a^23*b^8*c^14*d^2*e^3 + 307274711040*a^24*b^5*c^16*d^3*e^2 + 3713217 
33120*a^24*b^6*c^15*d^2*e^3 - 91855257600*a^25*b^3*c^17*d^3*e^2 - 41951428 
6080*a^25*b^4*c^16*d^2*e^3 + 209882972160*a^26*b^2*c^17*d^2*e^3 - 95126814 
720*a^27*b*c^17*d*e^4 - 103680*a^15*b^22*c^8*d^4*e + 3732480*a^16*b^20*...
 

Reduce [F]

\[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )^2} \, dx=\int \frac {e \,x^{4}+d}{x^{3} \left (c \,x^{8}+b \,x^{4}+a \right )^{2}}d x \] Input:

int((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x)
 

Output:

int((e*x^4+d)/x^3/(c*x^8+b*x^4+a)^2,x)