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

Optimal. Leaf size=154 \[ \frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1359, 1093, 205} \[ \frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^4 + c*x^8),x]

[Out]

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

Rule 205

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

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[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 1359

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
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]

Rubi steps

\begin {align*} \int \frac {x}{a+b x^4+c x^8} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{a+b x^2+c x^4} \, dx,x,x^2\right )\\ &=\frac {c \operatorname {Subst}\left (\int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx,x,x^2\right )}{2 \sqrt {b^2-4 a c}}-\frac {c \operatorname {Subst}\left (\int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx,x,x^2\right )}{2 \sqrt {b^2-4 a c}}\\ &=\frac {\sqrt {c} \tan ^{-1}\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} \tan ^{-1}\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}}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 133, normalized size = 0.86 \[ \frac {\sqrt {c} \left (\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^4 + c*x^8),x]

[Out]

(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])

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fricas [B]  time = 0.68, size = 619, normalized size = 4.02 \[ -\frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c - \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c - \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b + \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) - \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c + \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) + \frac {1}{4} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}} \log \left (c x^{2} - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - 4 \, a c + \frac {a b^{3} - 4 \, a^{2} b c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}\right )} \sqrt {-\frac {b - \frac {a b^{2} - 4 \, a^{2} c}{\sqrt {a^{2} b^{2} - 4 \, a^{3} c}}}{a b^{2} - 4 \, a^{2} c}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))) - 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)))

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giac [B]  time = 19.50, size = 1030, normalized size = 6.69 \[ \frac {{\left (\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 + \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 + \sqrt {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 \, {\left (b^{2} - 4 \, a c\right )} b^{2} c - 8 \, {\left (b^{2} - 4 \, a c\right )} a c^{2} - 2 \, {\left (b^{2} - 4 \, a c\right )} b c^{2}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x^{2}}{\sqrt {\frac {b + \sqrt {b^{2} - 4 \, a c}}{c}}}\right )}{8 \, {\left (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}\right )} {\left | c \right |}} + \frac {{\left (\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 - \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 - \sqrt {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 \, {\left (b^{2} - 4 \, a c\right )} b^{2} c + 8 \, {\left (b^{2} - 4 \, a c\right )} a c^{2} + 2 \, {\left (b^{2} - 4 \, a c\right )} b c^{2}\right )} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} x^{2}}{\sqrt {\frac {b - \sqrt {b^{2} - 4 \, a c}}{c}}}\right )}{8 \, {\left (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}\right )} {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 + 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*sq
rt(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 + sqrt(b^2 - 4*a*c)*c)*b^2*c - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr
t(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 - 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 - sqrt(b^2 - 4*a*c)*c)*b^2*c + sqrt(2)*sqr
t(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))

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maple [A]  time = 0.01, size = 120, normalized size = 0.78 \[ -\frac {\sqrt {2}\, c \arctanh \left (\frac {\sqrt {2}\, c \,x^{2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, c \,x^{2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{c x^{8} + b x^{4} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(c*x^8 + b*x^4 + a), x)

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mupad [B]  time = 2.27, size = 1105, normalized size = 7.18 \[ \mathrm {atan}\left (\frac {b^4\,x^2\,1{}\mathrm {i}+b\,x^2\,\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}\,1{}\mathrm {i}+a^2\,c^2\,x^2\,8{}\mathrm {i}-a\,b^2\,c\,x^2\,6{}\mathrm {i}}{128\,a^2\,b^5\,{\left (-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}-64\,a^3\,c^2\,\sqrt {-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}+16\,a^2\,b^2\,c\,\sqrt {-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}-1024\,a^3\,b^3\,c\,{\left (-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}+2048\,a^4\,b\,c^2\,{\left (-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}}\right )\,\sqrt {-\frac {b^3+\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {b^4\,x^2\,1{}\mathrm {i}-b\,x^2\,\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}\,1{}\mathrm {i}+a^2\,c^2\,x^2\,8{}\mathrm {i}-a\,b^2\,c\,x^2\,6{}\mathrm {i}}{128\,a^2\,b^5\,{\left (\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}-64\,a^3\,c^2\,\sqrt {\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}+16\,a^2\,b^2\,c\,\sqrt {\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}-1024\,a^3\,b^3\,c\,{\left (\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}+2048\,a^4\,b\,c^2\,{\left (\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}\right )}^{3/2}}\right )\,\sqrt {\frac {\sqrt {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}-b^3+4\,a\,b\,c}{512\,a^3\,c^2-256\,a^2\,b^2\,c+32\,a\,b^4}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^4 + c*x^8),x)

[Out]

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 - 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 + 51
2*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 - 256*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 - 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) + 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*8i - 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*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))^(3/2) + 2048*a^4*b*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))^(3/2)))*(((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)*2i

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sympy [A]  time = 3.36, size = 88, normalized size = 0.57 \[ \operatorname {RootSum} {\left (t^{4} \left (4096 a^{3} c^{2} - 2048 a^{2} b^{2} c + 256 a b^{4}\right ) + t^{2} \left (- 64 a b c + 16 b^{3}\right ) + c, \left (t \mapsto t \log {\left (x^{2} + \frac {256 t^{3} a^{2} b c - 64 t^{3} a b^{3} + 8 t a c - 4 t b^{2}}{c} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_t**4*(4096*a**3*c**2 - 2048*a**2*b**2*c + 256*a*b**4) + _t**2*(-64*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)))

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