\(\int \frac {x^{10}}{a+b x^4+c x^8} \, dx\) [54]

Optimal result

Integrand size = 18, antiderivative size = 381 \[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\frac {x^3}{3 c}-\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{7/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{7/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}}}+\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{7/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{7/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:

1/3*x^3/c-1/4*(b+(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x 
/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/c^(7/4)/(-b-(-4*a*c+b^2)^(1/2))^(1 
/4)-1/4*(b-(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x/(-b+( 
-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/c^(7/4)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)+1/ 
4*(b+(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b-(-4*a* 
c+b^2)^(1/2))^(1/4))*2^(1/4)/c^(7/4)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)+1/4*(b- 
(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b+(-4*a*c+b^2 
)^(1/2))^(1/4))*2^(1/4)/c^(7/4)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)
 

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.18 \[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\frac {4 x^3-3 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {a \log (x-\text {$\#$1})+b \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{12 c} \] Input:

Integrate[x^10/(a + b*x^4 + c*x^8),x]
 

Output:

(4*x^3 - 3*RootSum[a + b*#1^4 + c*#1^8 & , (a*Log[x - #1] + b*Log[x - #1]* 
#1^4)/(b*#1 + 2*c*#1^5) & ])/(12*c)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 344, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1703, 27, 1834, 27, 827, 218, 221}

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.

\(\displaystyle \int \frac {x^{10}}{a+b x^4+c x^8} \, dx\)

\(\Big \downarrow \) 1703

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^3}{3 c}-\frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )+\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )}{c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x^3}{3 c}-\frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )+\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{c}\)

Input:

Int[x^10/(a + b*x^4 + c*x^8),x]
 

Output:

x^3/(3*c) - ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4 
)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4 
*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4) 
]/(2*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4))) + (b - (b^2 - 2*a*c) 
/Sqrt[b^2 - 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^( 
1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4 
)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqrt 
[b^2 - 4*a*c])^(1/4))))/c
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 221

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

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 1703

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x 
_Symbol] :> Simp[d^(2*n - 1)*(d*x)^(m - 2*n + 1)*((a + b*x^n + c*x^(2*n))^( 
p + 1)/(c*(m + 2*n*p + 1))), x] - Simp[d^(2*n)/(c*(m + 2*n*p + 1))   Int[(d 
*x)^(m - 2*n)*Simp[a*(m - 2*n + 1) + b*(m + n*(p - 1) + 1)*x^n, x]*(a + b*x 
^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2*n] && N 
eQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1, 0 
] && IntegerQ[p]
 

rule 1834

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + 
 (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + 
 (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 
 - (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ 
[{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n 
, 0]
 

Maple [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.17

method	result	size
default	\(\frac {x^{3}}{3 c}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{6} b +\textit {\_R}^{2} a \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 c}\)	\(63\)
risch	\(\frac {x^{3}}{3 c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-\textit {\_R}^{6} b -\textit {\_R}^{2} a \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 c}\)	\(65\)

Input:

int(x^10/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

1/3*x^3/c-1/4/c*sum((_R^6*b+_R^2*a)/(2*_R^7*c+_R^3*b)*ln(x-_R),_R=RootOf(_ 
Z^8*c+_Z^4*b+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7003 vs. \(2 (299) = 598\).

Time = 0.84 (sec) , antiderivative size = 7003, normalized size of antiderivative = 18.38 \[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:

integrate(x^10/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\text {Timed out} \] Input:

integrate(x**10/(c*x**8+b*x**4+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\int { \frac {x^{10}}{c x^{8} + b x^{4} + a} \,d x } \] Input:

integrate(x^10/(c*x^8+b*x^4+a),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/3*x^3/c - integrate((b*x^6 + a*x^2)/(c*x^8 + b*x^4 + a), x)/c
 

Giac [F]

\[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\int { \frac {x^{10}}{c x^{8} + b x^{4} + a} \,d x } \] Input:

integrate(x^10/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

integrate(x^10/(c*x^8 + b*x^4 + a), x)
 

Mupad [B] (verification not implemented)

Time = 21.36 (sec) , antiderivative size = 12709, normalized size of antiderivative = 33.36 \[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:

int(x^10/(a + b*x^4 + c*x^8),x)
 

Output:

atan(((((8192*a^6*b*c^6 - 256*a^3*b^7*c^3 + 2560*a^4*b^5*c^4 - 8192*a^5*b^ 
3*c^5)/c^3 - (4*x*(-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 
 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b 
^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 5*a*b 
^4*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^4*c^11 + b^8*c^7 - 16*a*b^6*c^8 
 + 96*a^2*b^4*c^9 - 256*a^3*b^2*c^10)))^(1/4)*(8192*a^6*c^8 - 256*a^3*b^6* 
c^5 + 2560*a^4*b^4*c^6 - 8192*a^5*b^2*c^7))/c^3)*(-(b^11 + b^6*(-(4*a*c - 
b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4 
*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*( 
-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^ 
4*c^11 + b^8*c^7 - 16*a*b^6*c^8 + 96*a^2*b^4*c^9 - 256*a^3*b^2*c^10)))^(3/ 
4) + (4*x*(a^5*b^5 - 5*a^6*b^3*c + 5*a^7*b*c^2))/c^3)*(-(b^11 + b^6*(-(4*a 
*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 28 
0*a^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2* 
c^2*(-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(2 
56*a^4*c^11 + b^8*c^7 - 16*a*b^6*c^8 + 96*a^2*b^4*c^9 - 256*a^3*b^2*c^10)) 
)^(1/4)*1i - (((8192*a^6*b*c^6 - 256*a^3*b^7*c^3 + 2560*a^4*b^5*c^4 - 8192 
*a^5*b^3*c^5)/c^3 + (4*x*(-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5* 
b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3*c^3*(-(4* 
a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^2)^5)^(1/...
 

Reduce [F]

\[ \int \frac {x^{10}}{a+b x^4+c x^8} \, dx=\int \frac {x^{10}}{c \,x^{8}+b \,x^{4}+a}d x \] Input:

int(x^10/(c*x^8+b*x^4+a),x)
 

Output:

int(x^10/(c*x^8+b*x^4+a),x)