\(\int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 376 \[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\frac {x}{c}+\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} c^{5/4} \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \] Output:

x/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^(3/4)/c^(5/4)/(-b-(-4*a*c+b^2)^(1/2))^(3/4)+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^(3/4)/c^(5/4)/(-b+(-4*a*c+b^2)^(1/2))^(3/4)+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^(3/4)/c^(5/4)/(-b-(-4*a*c+b^2)^(1/2))^(3/4)+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^(3/4)/c^(5/4)/(-b+(-4*a*c+b^2)^(1/2))^(3/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.19 \[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\frac {x}{c}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {a \log (x-\text {$\#$1})+b \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{4 c} \] Input:

Integrate[(c + a/x^8 + b/x^4)^(-1),x]
 

Output:

x/c - RootSum[a + b*#1^4 + c*#1^8 & , (a*Log[x - #1] + b*Log[x - #1]*#1^4) 
/(b*#1^3 + 2*c*#1^7) & ]/(4*c)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1679, 1703, 1752, 756, 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.

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

\(\Big \downarrow \) 1679

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

\(\Big \downarrow \) 1703

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + a/x^8 + b/x^4)^(-1),x]
 

Output:

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

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 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 756

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

rule 1679

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^( 
2*n*p)*(c + b/x^n + a/x^(2*n))^p, x] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] 
 && LtQ[n, 0] && IntegerQ[p]
 

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 1752

Int[((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[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   I 
nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 
, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 
 - 4*a*c] ||  !IGtQ[n/2, 0])
 

Maple [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.16

method	result	size
default	\(\frac {x}{c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (-\textit {\_R}^{4} b -a \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 c}\)	\(59\)
risch	\(\frac {x}{c}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (-\textit {\_R}^{4} b -a \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 c}\)	\(59\)

Input:

int(1/(c+a/x^8+b/x^4),x,method=_RETURNVERBOSE)
 

Output:

x/c+1/4/c*sum((-_R^4*b-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. 4001 vs. \(2 (296) = 592\).

Time = 0.36 (sec) , antiderivative size = 4001, normalized size of antiderivative = 10.64 \[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\text {Too large to display} \] Input:

integrate(1/(c+a/x^8+b/x^4),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\int { \frac {1}{c + \frac {b}{x^{4}} + \frac {a}{x^{8}}} \,d x } \] Input:

integrate(1/(c+a/x^8+b/x^4),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [F]

\[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\int { \frac {1}{c + \frac {b}{x^{4}} + \frac {a}{x^{8}}} \,d x } \] Input:

integrate(1/(c+a/x^8+b/x^4),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.47 (sec) , antiderivative size = 10382, normalized size of antiderivative = 27.61 \[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\text {Too large to display} \] Input:

int(1/(c + a/x^8 + b/x^4),x)
 

Output:

atan(((((16*(a^3*b^6 - 4*a^6*c^3 - 7*a^4*b^4*c + 13*a^5*b^2*c^2))/c - (4*x 
*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 1 
20*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c 
*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^4*c^9 + b^8*c^5 - 16*a*b^6*c^6 + 96 
*a^2*b^4*c^7 - 256*a^3*b^2*c^8)))^(3/4)*(4096*a^5*b*c^6 + 256*a^3*b^5*c^4 
- 2048*a^4*b^3*c^5))/c)*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c 
^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 
 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^4*c^9 + b^8* 
c^5 - 16*a*b^6*c^6 + 96*a^2*b^4*c^7 - 256*a^3*b^2*c^8)))^(1/4) - (4*x*(a^4 
*b^4 + 2*a^6*c^2 - 4*a^5*b^2*c))/c)*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) 
+ 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2 
)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^ 
4*c^9 + b^8*c^5 - 16*a*b^6*c^6 + 96*a^2*b^4*c^7 - 256*a^3*b^2*c^8)))^(1/4) 
*1i - (((16*(a^3*b^6 - 4*a^6*c^3 - 7*a^4*b^4*c + 13*a^5*b^2*c^2))/c + (4*x 
*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 1 
20*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c 
*(-(4*a*c - b^2)^5)^(1/2))/(512*(256*a^4*c^9 + b^8*c^5 - 16*a*b^6*c^6 + 96 
*a^2*b^4*c^7 - 256*a^3*b^2*c^8)))^(3/4)*(4096*a^5*b*c^6 + 256*a^3*b^5*c^4 
- 2048*a^4*b^3*c^5))/c)*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c 
^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2...
 

Reduce [F]

\[ \int \frac {1}{c+\frac {a}{x^8}+\frac {b}{x^4}} \, dx=\int \frac {1}{c +\frac {a}{x^{8}}+\frac {b}{x^{4}}}d x \] Input:

int(1/(c+a/x^8+b/x^4),x)
 

Output:

int(1/(c+a/x^8+b/x^4),x)