\(\int \frac {d+e x^4}{x^3 (a+b x^4+c x^8)} \, dx\) [56]

Optimal result

Integrand size = 25, antiderivative size = 199 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )} \, dx=-\frac {d}{2 a x^2}-\frac {\sqrt {c} \left (d+\frac {b d-2 a e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (d-\frac {b d-2 a e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

-1/2*d/a/x^2-1/4*c^(1/2)*(d+(-2*a*e+b*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/(b-(-4*a*c+b^2)^(1/2 
))^(1/2)-1/4*c^(1/2)*(d-(-2*a*e+b*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/(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 = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.45 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )} \, dx=-\frac {d}{2 a x^2}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b d \log (x-\text {$\#$1})-a e \log (x-\text {$\#$1})+c d \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^2+2 c \text {$\#$1}^6}\&\right ]}{4 a} \] Input:

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

Output:

-1/2*d/(a*x^2) - RootSum[a + b*#1^4 + c*#1^8 & , (b*d*Log[x - #1] - a*e*Lo 
g[x - #1] + c*d*Log[x - #1]*#1^4)/(b*#1^2 + 2*c*#1^6) & ]/(4*a)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1814, 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.

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

\(\Big \downarrow \) 1814

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

\(\Big \downarrow \) 1604

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-(d/(a*x^2)) - ((Sqrt[c]*(d + (b*d - 2*a*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sq 
rt[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]*(d - (b*d - 2*a*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
 

Defintions of rubi rules used

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 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 = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {d}{2 a \,x^{2}}+\frac {2 c \left (-\frac {\left (-b d -\sqrt {-4 a c +b^{2}}\, d +2 a e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-\sqrt {-4 a c +b^{2}}\, d -2 a e +b d \right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{a}\)	\(177\)
risch	\(-\frac {d}{2 a \,x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 c^{2} a^{5}-8 a^{4} b^{2} c +a^{3} b^{4}\right ) \textit {\_Z}^{4}+\left (-4 a^{3} b \,e^{2} c -16 a^{3} d e \,c^{2}+a^{2} b^{3} e^{2}+12 a^{2} b^{2} d e c +12 a^{2} b \,c^{2} d^{2}-2 a \,b^{4} d e -7 a \,b^{3} c \,d^{2}+b^{5} d^{2}\right ) \textit {\_Z}^{2}+a^{2} e^{4} c -2 a b d \,e^{3} c +2 d^{2} e^{2} a \,c^{2}+b^{2} d^{2} e^{2} c -2 d^{3} e b \,c^{2}+d^{4} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-72 c^{2} a^{5}+38 a^{4} b^{2} c -5 a^{3} b^{4}\right ) \textit {\_R}^{4}+\left (15 a^{3} b \,e^{2} c +68 a^{3} d e \,c^{2}-4 a^{2} b^{3} e^{2}-48 a^{2} b^{2} d e c -49 a^{2} b \,c^{2} d^{2}+8 a \,b^{4} d e +28 a \,b^{3} c \,d^{2}-4 b^{5} d^{2}\right ) \textit {\_R}^{2}-4 a^{2} e^{4} c +8 a b d \,e^{3} c -8 d^{2} e^{2} a \,c^{2}-4 b^{2} d^{2} e^{2} c +8 d^{3} e b \,c^{2}-4 d^{4} c^{3}\right ) x^{2}+\left (-4 a^{4} b c e -4 a^{4} c^{2} d +e \,b^{3} a^{3}+5 a^{3} b^{2} c d -d \,b^{4} a^{2}\right ) \textit {\_R}^{3}+\left (a^{3} c \,e^{3}-a^{2} b c d \,e^{2}+a^{2} c^{2} d^{2} e \right ) \textit {\_R} \right )\right )}{4}\)	\(453\)

Input:

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

Output:

-1/2*d/a/x^2+2/a*c*(-1/8*(-b*d-(-4*a*c+b^2)^(1/2)*d+2*a*e)/(-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*(-(-4*a*c+b^2)^(1/2)*d-2*a*e+b*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)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2772 vs. \(2 (157) = 314\).

Time = 0.73 (sec) , antiderivative size = 2772, normalized size of antiderivative = 13.93 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(sqrt(1/2)*a*x^2*sqrt(-(a^2*b*e^2 + (b^3 - 3*a*b*c)*d^2 - 2*(a*b^2 - 2 
*a^2*c)*d*e + (a^3*b^2 - 4*a^4*c)*sqrt(-(4*a^3*b*d*e^3 - a^4*e^4 - (b^4 - 
2*a*b^2*c + a^2*c^2)*d^4 + 4*(a*b^3 - a^2*b*c)*d^3*e - 2*(3*a^2*b^2 - a^3* 
c)*d^2*e^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))*log((3*a*b^2*c*d^2* 
e^2 - 3*a^2*b*c*d*e^3 + a^3*c*e^4 + (b^2*c^2 - a*c^3)*d^4 - (b^3*c + a*b*c 
^2)*d^3*e)*x^2 + 1/2*sqrt(1/2)*((b^5 - 5*a*b^3*c + 4*a^2*b*c^2)*d^3 - (3*a 
*b^4 - 13*a^2*b^2*c + 4*a^3*c^2)*d^2*e + 3*(a^2*b^3 - 4*a^3*b*c)*d*e^2 - ( 
a^3*b^2 - 4*a^4*c)*e^3 - ((a^3*b^4 - 6*a^4*b^2*c + 8*a^5*c^2)*d - (a^4*b^3 
 - 4*a^5*b*c)*e)*sqrt(-(4*a^3*b*d*e^3 - a^4*e^4 - (b^4 - 2*a*b^2*c + a^2*c 
^2)*d^4 + 4*(a*b^3 - a^2*b*c)*d^3*e - 2*(3*a^2*b^2 - a^3*c)*d^2*e^2)/(a^6* 
b^2 - 4*a^7*c)))*sqrt(-(a^2*b*e^2 + (b^3 - 3*a*b*c)*d^2 - 2*(a*b^2 - 2*a^2 
*c)*d*e + (a^3*b^2 - 4*a^4*c)*sqrt(-(4*a^3*b*d*e^3 - a^4*e^4 - (b^4 - 2*a* 
b^2*c + a^2*c^2)*d^4 + 4*(a*b^3 - a^2*b*c)*d^3*e - 2*(3*a^2*b^2 - a^3*c)*d 
^2*e^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))) - sqrt(1/2)*a*x^2*sqrt 
(-(a^2*b*e^2 + (b^3 - 3*a*b*c)*d^2 - 2*(a*b^2 - 2*a^2*c)*d*e + (a^3*b^2 - 
4*a^4*c)*sqrt(-(4*a^3*b*d*e^3 - a^4*e^4 - (b^4 - 2*a*b^2*c + a^2*c^2)*d^4 
+ 4*(a*b^3 - a^2*b*c)*d^3*e - 2*(3*a^2*b^2 - a^3*c)*d^2*e^2)/(a^6*b^2 - 4* 
a^7*c)))/(a^3*b^2 - 4*a^4*c))*log((3*a*b^2*c*d^2*e^2 - 3*a^2*b*c*d*e^3 + a 
^3*c*e^4 + (b^2*c^2 - a*c^3)*d^4 - (b^3*c + a*b*c^2)*d^3*e)*x^2 - 1/2*sqrt 
(1/2)*((b^5 - 5*a*b^3*c + 4*a^2*b*c^2)*d^3 - (3*a*b^4 - 13*a^2*b^2*c + ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-integrate((c*d*x^4 + b*d - a*e)*x/(c*x^8 + b*x^4 + a), x)/a - 1/2*d/(a*x^ 
2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3003 vs. \(2 (157) = 314\).

Time = 1.39 (sec) , antiderivative size = 3003, normalized size of antiderivative = 15.09 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*((sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 8*sqrt(2)*sqrt(b*c 
+ sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)* 
c)*b^3*c^2 - 2*b^4*c^2 + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^ 
3 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 + sqrt(2)*sqrt(b*c + 
 sqrt(b^2 - 4*a*c)*c)*b^2*c^3 + 16*a*b^2*c^3 - 4*sqrt(2)*sqrt(b*c + sqrt(b 
^2 - 4*a*c)*c)*a*c^4 - 32*a^2*c^4 + 2*(b^2 - 4*a*c)*b^2*c^2 - 8*(b^2 - 4*a 
*c)*a*c^3)*d*x^4*abs(a) - (2*a*b^3*c^3 - 8*a^2*b*c^4 - sqrt(2)*sqrt(b^2 - 
4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c + 4*sqrt(2)*sqrt(b^2 - 4*a* 
c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c) 
*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqr 
t(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 - 2*(b^2 - 4*a*c)*a*b*c^3)*d*x^4 + (s 
qrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 
 - 4*a*c)*c)*a*b^3*c - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 2 
*b^5*c + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 + 8*sqrt(2)* 
sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 
4*a*c)*c)*b^3*c^2 + 16*a*b^3*c^2 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)* 
c)*a*b*c^3 - 32*a^2*b*c^3 + 2*(b^2 - 4*a*c)*b^3*c - 8*(b^2 - 4*a*c)*a*b*c^ 
2)*d*abs(a) - (sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4 - 8*sqrt(2)*s 
qrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 
 4*a*c)*c)*a*b^3*c - 2*a*b^4*c + 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c...
 

Mupad [B] (verification not implemented)

Time = 26.22 (sec) , antiderivative size = 15013, normalized size of antiderivative = 75.44 \[ \int \frac {d+e x^4}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

- atan((((-(b^5*d^2 + a^2*b^3*e^2 + a^2*e^2*(-(4*a*c - b^2)^3)^(1/2) + b^2 
*d^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2*d^2 - 2*a*b^4*d*e - 7*a*b^3*c 
*d^2 - a*c*d^2*(-(4*a*c - b^2)^3)^(1/2) - 4*a^3*b*c*e^2 - 16*a^3*c^2*d*e + 
 12*a^2*b^2*c*d*e - 2*a*b*d*e*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16* 
a^5*c^2 - 8*a^4*b^2*c)))^(1/2)*(((-(b^5*d^2 + a^2*b^3*e^2 + a^2*e^2*(-(4*a 
*c - b^2)^3)^(1/2) + b^2*d^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2*d^2 - 
 2*a*b^4*d*e - 7*a*b^3*c*d^2 - a*c*d^2*(-(4*a*c - b^2)^3)^(1/2) - 4*a^3*b* 
c*e^2 - 16*a^3*c^2*d*e + 12*a^2*b^2*c*d*e - 2*a*b*d*e*(-(4*a*c - b^2)^3)^( 
1/2))/(32*(a^3*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)))^(1/2)*(((-(b^5*d^2 + a^2* 
b^3*e^2 + a^2*e^2*(-(4*a*c - b^2)^3)^(1/2) + b^2*d^2*(-(4*a*c - b^2)^3)^(1 
/2) + 12*a^2*b*c^2*d^2 - 2*a*b^4*d*e - 7*a*b^3*c*d^2 - a*c*d^2*(-(4*a*c - 
b^2)^3)^(1/2) - 4*a^3*b*c*e^2 - 16*a^3*c^2*d*e + 12*a^2*b^2*c*d*e - 2*a*b* 
d*e*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)))^( 
1/2)*(4096*a^12*b^6*c^4 - 32768*a^13*b^4*c^5 + 65536*a^14*b^2*c^6) + x^2*( 
9216*a^11*b^5*c^5*d - 1024*a^10*b^7*c^4*d - 24576*a^12*b^3*c^6*d + 1024*a^ 
11*b^6*c^4*e - 8192*a^12*b^4*c^5*e + 16384*a^13*b^2*c^6*e + 16384*a^13*b*c 
^7*d))*(-(b^5*d^2 + a^2*b^3*e^2 + a^2*e^2*(-(4*a*c - b^2)^3)^(1/2) + b^2*d 
^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2*d^2 - 2*a*b^4*d*e - 7*a*b^3*c*d 
^2 - a*c*d^2*(-(4*a*c - b^2)^3)^(1/2) - 4*a^3*b*c*e^2 - 16*a^3*c^2*d*e + 1 
2*a^2*b^2*c*d*e - 2*a*b*d*e*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16...
 

Reduce [F]

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

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

Output:

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