\(\int \frac {x^8}{3+4 x^3+x^6} \, dx\) [152]

Optimal result

Integrand size = 16, antiderivative size = 28 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {x^3}{3}+\frac {1}{6} \log \left (1+x^3\right )-\frac {3}{2} \log \left (3+x^3\right ) \]

[Out]

1/3*x^3+1/6*ln(x^3+1)-3/2*ln(x^3+3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1371, 717, 646, 31} \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {x^3}{3}+\frac {1}{6} \log \left (x^3+1\right )-\frac {3}{2} \log \left (x^3+3\right ) \]

[In]

Int[x^8/(3 + 4*x^3 + x^6),x]

[Out]

x^3/3 + Log[1 + x^3]/6 - (3*Log[3 + x^3])/2

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 717

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

Rule 1371

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x^2}{3+4 x+x^2} \, dx,x,x^3\right ) \\ & = \frac {x^3}{3}+\frac {1}{3} \text {Subst}\left (\int \frac {-3-4 x}{3+4 x+x^2} \, dx,x,x^3\right ) \\ & = \frac {x^3}{3}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^3\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1}{3+x} \, dx,x,x^3\right ) \\ & = \frac {x^3}{3}+\frac {1}{6} \log \left (1+x^3\right )-\frac {3}{2} \log \left (3+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {x^3}{3}+\frac {1}{6} \log \left (1+x^3\right )-\frac {3}{2} \log \left (3+x^3\right ) \]

[In]

Integrate[x^8/(3 + 4*x^3 + x^6),x]

[Out]

x^3/3 + Log[1 + x^3]/6 - (3*Log[3 + x^3])/2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {x^{3}}{3}+\frac {\ln \left (x^{3}+1\right )}{6}-\frac {3 \ln \left (x^{3}+3\right )}{2}\)	\(23\)
risch	\(\frac {x^{3}}{3}+\frac {\ln \left (x^{3}+1\right )}{6}-\frac {3 \ln \left (x^{3}+3\right )}{2}\)	\(23\)
norman	\(\frac {x^{3}}{3}+\frac {\ln \left (x +1\right )}{6}-\frac {3 \ln \left (x^{3}+3\right )}{2}+\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(32\)
parallelrisch	\(\frac {x^{3}}{3}+\frac {\ln \left (x +1\right )}{6}-\frac {3 \ln \left (x^{3}+3\right )}{2}+\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(32\)

[In]

int(x^8/(x^6+4*x^3+3),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3+1/6*ln(x^3+1)-3/2*ln(x^3+3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {1}{3} \, x^{3} - \frac {3}{2} \, \log \left (x^{3} + 3\right ) + \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

[In]

integrate(x^8/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

1/3*x^3 - 3/2*log(x^3 + 3) + 1/6*log(x^3 + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {x^{3}}{3} + \frac {\log {\left (x^{3} + 1 \right )}}{6} - \frac {3 \log {\left (x^{3} + 3 \right )}}{2} \]

[In]

integrate(x**8/(x**6+4*x**3+3),x)

[Out]

x**3/3 + log(x**3 + 1)/6 - 3*log(x**3 + 3)/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {1}{3} \, x^{3} - \frac {3}{2} \, \log \left (x^{3} + 3\right ) + \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

[In]

integrate(x^8/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

1/3*x^3 - 3/2*log(x^3 + 3) + 1/6*log(x^3 + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {1}{3} \, x^{3} - \frac {3}{2} \, \log \left ({\left | x^{3} + 3 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) \]

[In]

integrate(x^8/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

1/3*x^3 - 3/2*log(abs(x^3 + 3)) + 1/6*log(abs(x^3 + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^8}{3+4 x^3+x^6} \, dx=\frac {\ln \left (x^3+1\right )}{6}-\frac {3\,\ln \left (x^3+3\right )}{2}+\frac {x^3}{3} \]

[In]

int(x^8/(4*x^3 + x^6 + 3),x)

[Out]

log(x^3 + 1)/6 - (3*log(x^3 + 3))/2 + x^3/3