\(\int \frac {x^9}{a+b x^4+c x^8} \, dx\) [46]

Optimal result

Integrand size = 18, antiderivative size = 192 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\frac {x^2}{2 c}-\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

1/2*x^2/c-1/4*(b-(-2*a*c+b^2)/(-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)/c^(3/2)/(b-(-4*a*c+b^2)^(1/2))^(1 
/2)-1/4*(b+(-2*a*c+b^2)/(-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)/c^(3/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.09 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\frac {2 \sqrt {c} x^2-\frac {\sqrt {2} \left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}}}{4 c^{3/2}} \] Input:

Integrate[x^9/(a + b*x^4 + c*x^8),x]
 

Output:

(2*Sqrt[c]*x^2 - (Sqrt[2]*(-b^2 + 2*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqr 
t[2]*Sqrt[c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b 
- Sqrt[b^2 - 4*a*c]]) - (Sqrt[2]*(b^2 - 2*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTa 
n[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*S 
qrt[b + Sqrt[b^2 - 4*a*c]]))/(4*c^(3/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1695, 1442, 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.

\(\displaystyle \int \frac {x^9}{a+b x^4+c x^8} \, dx\)

\(\Big \downarrow \) 1695

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

\(\Big \downarrow \) 1442

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

Int[x^9/(a + b*x^4 + c*x^8),x]
 

Output:

(x^2/c - (((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x 
^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a 
*c]]) + ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x^2 
)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b + Sqrt[b^2 - 4*a*c 
]]))/c)/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 1442

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

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 1695

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

Maple [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {x^{2}}{2 c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 a^{2} c^{3}-8 a \,b^{2} c^{2}+b^{4} c \right ) \textit {\_Z}^{4}+\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}\right ) \textit {\_Z}^{2}+c^{2} a^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-a^{2} c^{2}+a \,b^{2} c \right ) x^{2}+\left (-4 a b \,c^{2}+b^{3} c \right ) \textit {\_R}^{3}+\left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right ) \textit {\_R} \right )}{4 c}\)	\(139\)
default	\(\frac {x^{2}}{2 c}-\frac {\left (-b \sqrt {-4 a c +b^{2}}-2 a c +b^{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-b \sqrt {-4 a c +b^{2}}+2 a c -b^{2}\right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\)	\(176\)

Input:

int(x^9/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

1/2*x^2/c+1/4/c*sum(_R*ln((-a^2*c^2+a*b^2*c)*x^2+(-4*a*b*c^2+b^3*c)*_R^3+( 
2*a^2*c^2-4*a*b^2*c+b^4)*_R),_R=RootOf((16*a^2*c^3-8*a*b^2*c^2+b^4*c)*_Z^4 
+(12*a^2*b*c^2-7*a*b^3*c+b^5)*_Z^2+c^2*a^3))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1071 vs. \(2 (150) = 300\).

Time = 0.09 (sec) , antiderivative size = 1071, normalized size of antiderivative = 5.58 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx =\text {Too large to display} \] Input:

integrate(x^9/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

-1/4*(sqrt(1/2)*c*sqrt(-(b^3 - 3*a*b*c + (b^2*c^3 - 4*a*c^4)*sqrt((b^4 - 2 
*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))*log(-(a*b^2 
 - a^2*c)*x^2 + 1/2*sqrt(1/2)*(b^4 - 5*a*b^2*c + 4*a^2*c^2 - (b^3*c^3 - 4* 
a*b*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))*sqrt(-(b^3 
 - 3*a*b*c + (b^2*c^3 - 4*a*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 
 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))) - sqrt(1/2)*c*sqrt(-(b^3 - 3*a*b*c + ( 
b^2*c^3 - 4*a*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))/ 
(b^2*c^3 - 4*a*c^4))*log(-(a*b^2 - a^2*c)*x^2 - 1/2*sqrt(1/2)*(b^4 - 5*a*b 
^2*c + 4*a^2*c^2 - (b^3*c^3 - 4*a*b*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/ 
(b^2*c^6 - 4*a*c^7)))*sqrt(-(b^3 - 3*a*b*c + (b^2*c^3 - 4*a*c^4)*sqrt((b^4 
 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))) + sqrt 
(1/2)*c*sqrt(-(b^3 - 3*a*b*c - (b^2*c^3 - 4*a*c^4)*sqrt((b^4 - 2*a*b^2*c + 
 a^2*c^2)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 4*a*c^4))*log(-(a*b^2 - a^2*c)* 
x^2 + 1/2*sqrt(1/2)*(b^4 - 5*a*b^2*c + 4*a^2*c^2 + (b^3*c^3 - 4*a*b*c^4)*s 
qrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))*sqrt(-(b^3 - 3*a*b*c 
 - (b^2*c^3 - 4*a*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7 
)))/(b^2*c^3 - 4*a*c^4))) - sqrt(1/2)*c*sqrt(-(b^3 - 3*a*b*c - (b^2*c^3 - 
4*a*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^6 - 4*a*c^7)))/(b^2*c^3 - 
 4*a*c^4))*log(-(a*b^2 - a^2*c)*x^2 - 1/2*sqrt(1/2)*(b^4 - 5*a*b^2*c + 4*a 
^2*c^2 + (b^3*c^3 - 4*a*b*c^4)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^2*c^...
 

Sympy [A] (verification not implemented)

Time = 2.79 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.70 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (4096 a^{2} c^{5} - 2048 a b^{2} c^{4} + 256 b^{4} c^{3}\right ) + t^{2} \cdot \left (192 a^{2} b c^{2} - 112 a b^{3} c + 16 b^{5}\right ) + a^{3}, \left ( t \mapsto t \log {\left (x^{2} + \frac {256 t^{3} a b c^{4} - 64 t^{3} b^{3} c^{3} - 8 t a^{2} c^{2} + 16 t a b^{2} c - 4 t b^{4}}{a^{2} c - a b^{2}} \right )} \right )\right )} + \frac {x^{2}}{2 c} \] Input:

integrate(x**9/(c*x**8+b*x**4+a),x)
 

Output:

RootSum(_t**4*(4096*a**2*c**5 - 2048*a*b**2*c**4 + 256*b**4*c**3) + _t**2* 
(192*a**2*b*c**2 - 112*a*b**3*c + 16*b**5) + a**3, Lambda(_t, _t*log(x**2 
+ (256*_t**3*a*b*c**4 - 64*_t**3*b**3*c**3 - 8*_t*a**2*c**2 + 16*_t*a*b**2 
*c - 4*_t*b**4)/(a**2*c - a*b**2)))) + x**2/(2*c)
 

Maxima [F]

\[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\int { \frac {x^{9}}{c x^{8} + b x^{4} + a} \,d x } \] Input:

integrate(x^9/(c*x^8+b*x^4+a),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2043 vs. \(2 (150) = 300\).

Time = 1.24 (sec) , antiderivative size = 2043, normalized size of antiderivative = 10.64 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:

integrate(x^9/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

1/2*x^2/c + 1/8*(2*a*b^3*c^3 - 8*a^2*b*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqr 
t(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)*sqrt(b*c + sq 
rt(b^2 - 4*a*c)*c)*a*b*c^3 - 2*(b^2 - 4*a*c)*a*b*c^3 - (sqrt(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)*x^4*abs(c) + ( 
2*b^4*c^3 - 8*a*b^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 
4*a*c)*c)*b^4*c + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c) 
*c)*a*b^2*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c 
)*b^3*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2* 
c^3 - 2*(b^2 - 4*a*c)*b^2*c^3)*x^4 - (sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c) 
*c)*a*b^4 - 8*sqrt(2)*sqrt(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)*c)*a^3*c^2 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)* 
c)*a^2*b*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 + 16*a...
 

Mupad [B] (verification not implemented)

Time = 21.88 (sec) , antiderivative size = 5659, normalized size of antiderivative = 29.47 \[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:

int(x^9/(a + b*x^4 + c*x^8),x)
 

Output:

atan(((((16*(a*b^8 + 4*a^5*c^4 - 8*a^2*b^6*c + 20*a^3*b^4*c^2 - 16*a^4*b^2 
*c^3))/c^2 + (((16*(32*a*b^7*c^3 - 256*a^4*b*c^6 - 256*a^2*b^5*c^4 + 576*a 
^3*b^3*c^5))/c^2 - ((256*a*b^6*c^6 - 2048*a^2*b^4*c^7 + 4096*a^3*b^2*c^8)* 
(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4* 
a*c - b^2)^3)^(1/2)))/(2*c^2*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))*(-(b^5 
 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c 
- b^2)^3)^(1/2))/(32*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))^(1/2) - (4*x^2 
*(64*a^4*b*c^5 + 16*a^2*b^5*c^3 - 80*a^3*b^3*c^4))/c^2)*(-(b^5 + b^2*(-(4* 
a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1 
/2))/(32*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))^(1/2))*(-(b^5 + b^2*(-(4*a 
*c - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/ 
2))/(32*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))^(1/2) - (4*x^2*(a^2*b^6 - 2 
*a^5*c^3 - 5*a^3*b^4*c + 6*a^4*b^2*c^2))/c^2)*(-(b^5 + b^2*(-(4*a*c - b^2) 
^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2))/(32*( 
16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))^(1/2)*1i - (((16*(a*b^8 + 4*a^5*c^4 
- 8*a^2*b^6*c + 20*a^3*b^4*c^2 - 16*a^4*b^2*c^3))/c^2 + (((16*(32*a*b^7*c^ 
3 - 256*a^4*b*c^6 - 256*a^2*b^5*c^4 + 576*a^3*b^3*c^5))/c^2 - ((256*a*b^6* 
c^6 - 2048*a^2*b^4*c^7 + 4096*a^3*b^2*c^8)*(b^5 + b^2*(-(4*a*c - b^2)^3)^( 
1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2)))/(2*c^2*(1 
6*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))*(-(b^5 + b^2*(-(4*a*c - b^2)^3)^(1...
 

Reduce [F]

\[ \int \frac {x^9}{a+b x^4+c x^8} \, dx=\int \frac {x^{9}}{c \,x^{8}+b \,x^{4}+a}d x \] Input:

int(x^9/(c*x^8+b*x^4+a),x)
 

Output:

int(x^9/(c*x^8+b*x^4+a),x)