\(\int \frac {x}{a+b x^4+c x^8} \, dx\) [50]

Optimal result

Integrand size = 16, antiderivative size = 154 \[ \int \frac {x}{a+b x^4+c x^8} \, dx=\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.86 \[ \int \frac {x}{a+b x^4+c x^8} \, dx=\frac {\sqrt {c} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1695, 1406, 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.

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

\(\Big \downarrow \) 1695

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

\(\Big \downarrow \) 1406

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((Sqrt[2]*Sqrt[c]*ArcTan[(Sqrt[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]*Sqrt[c]*ArcT 
an[(Sqrt[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]]))/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 1406

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 
2 - 4*a*c, 2]}, Simp[c/q   Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q   I 
nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c 
, 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 = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 c^{2} a^{3}-8 c \,a^{2} b^{2}+a \,b^{4}\right ) \textit {\_Z}^{4}+\left (-4 a b c +b^{3}\right ) \textit {\_Z}^{2}+c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (4 a b c -b^{3}\right ) \textit {\_R}^{2}-c \right ) x^{2}+\left (4 a^{2} b c -b^{3} a \right ) \textit {\_R}^{3}-2 a c \textit {\_R} \right )\right )}{4}\)	\(99\)
default	\(2 c \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )\)	\(121\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (119) = 238\).

Time = 0.09 (sec) , antiderivative size = 619, normalized size of antiderivative = 4.02 \[ \int \frac {x}{a+b x^4+c x^8} \, dx=-\frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c - \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c - \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) - \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c + \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c + \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) \] Input:

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

Output:

-1/4*sqrt(1/2)*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^ 
2 - 4*a^2*c))*log(c*x^2 + 1/2*sqrt(1/2)*(b^2 - 4*a*c - (a*b^3 - 4*a^2*b*c) 
/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^ 
3*c))/(a*b^2 - 4*a^2*c))) + 1/4*sqrt(1/2)*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqr 
t(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2*c))*log(c*x^2 - 1/2*sqrt(1/2)*(b^2 - 
4*a*c - (a*b^3 - 4*a^2*b*c)/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(-(b + (a*b^2 - 4 
*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2*c))) - 1/4*sqrt(1/2)*sqrt( 
-(b - (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2*c))*log(c* 
x^2 + 1/2*sqrt(1/2)*(b^2 - 4*a*c + (a*b^3 - 4*a^2*b*c)/sqrt(a^2*b^2 - 4*a^ 
3*c))*sqrt(-(b - (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2 
*c))) + 1/4*sqrt(1/2)*sqrt(-(b - (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c) 
)/(a*b^2 - 4*a^2*c))*log(c*x^2 - 1/2*sqrt(1/2)*(b^2 - 4*a*c + (a*b^3 - 4*a 
^2*b*c)/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(-(b - (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 
 - 4*a^3*c))/(a*b^2 - 4*a^2*c)))
 

Sympy [A] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.57 \[ \int \frac {x}{a+b x^4+c x^8} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (4096 a^{3} c^{2} - 2048 a^{2} b^{2} c + 256 a b^{4}\right ) + t^{2} \left (- 64 a b c + 16 b^{3}\right ) + c, \left ( t \mapsto t \log {\left (x^{2} + \frac {256 t^{3} a^{2} b c - 64 t^{3} a b^{3} + 8 t a c - 4 t b^{2}}{c} \right )} \right )\right )} \] Input:

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

Output:

RootSum(_t**4*(4096*a**3*c**2 - 2048*a**2*b**2*c + 256*a*b**4) + _t**2*(-6 
4*a*b*c + 16*b**3) + c, Lambda(_t, _t*log(x**2 + (256*_t**3*a**2*b*c - 64* 
_t**3*a*b**3 + 8*_t*a*c - 4*_t*b**2)/c)))
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1028 vs. \(2 (119) = 238\).

Time = 1.28 (sec) , antiderivative size = 1028, normalized size of antiderivative = 6.68 \[ \int \frac {x}{a+b x^4+c x^8} \, dx =\text {Too large to display} \] Input:

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

Mupad [B] (verification not implemented)

Time = 19.89 (sec) , antiderivative size = 1105, normalized size of antiderivative = 7.18 \[ \int \frac {x}{a+b x^4+c x^8} \, dx =\text {Too large to display} \] Input:

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

Output:

atan((b^4*x^2*1i + b*x^2*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^ 
(1/2)*1i + a^2*c^2*x^2*8i - a*b^2*c*x^2*6i)/(128*a^2*b^5*(-(b^3 + (b^6 - 6 
4*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/2) - 4*a*b*c)/(32*a*b^4 + 512* 
a^3*c^2 - 256*a^2*b^2*c))^(3/2) - 64*a^3*c^2*(-(b^3 + (b^6 - 64*a^3*c^3 + 
48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/2) - 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 25 
6*a^2*b^2*c))^(1/2) + 16*a^2*b^2*c*(-(b^3 + (b^6 - 64*a^3*c^3 + 48*a^2*b^2 
*c^2 - 12*a*b^4*c)^(1/2) - 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2* 
c))^(1/2) - 1024*a^3*b^3*c*(-(b^3 + (b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 1 
2*a*b^4*c)^(1/2) - 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c))^(3/2 
) + 2048*a^4*b*c^2*(-(b^3 + (b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4* 
c)^(1/2) - 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c))^(3/2)))*(-(b 
^3 + (b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/2) - 4*a*b*c)/(32 
*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c))^(1/2)*2i + atan((b^4*x^2*1i - b*x^2 
*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/2)*1i + a^2*c^2*x^2*8 
i - a*b^2*c*x^2*6i)/(128*a^2*b^5*(((b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12 
*a*b^4*c)^(1/2) - b^3 + 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c)) 
^(3/2) - 64*a^3*c^2*(((b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/ 
2) - b^3 + 4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c))^(1/2) + 16*a 
^2*b^2*c*(((b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)^(1/2) - b^3 + 
4*a*b*c)/(32*a*b^4 + 512*a^3*c^2 - 256*a^2*b^2*c))^(1/2) - 1024*a^3*b^3...
 

Reduce [F]

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

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

Output:

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