\(\int \frac {x^{11}}{1+3 x^4+x^8} \, dx\) [368]

Optimal result

Integrand size = 16, antiderivative size = 62 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {x^4}{4}-\frac {1}{40} \left (15-7 \sqrt {5}\right ) \log \left (3-\sqrt {5}+2 x^4\right )-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right ) \]

[Out]

1/4*x^4-1/40*ln(2*x^4-5^(1/2)+3)*(15-7*5^(1/2))-1/40*ln(2*x^4+5^(1/2)+3)*(15+7*5^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 62, 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} \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {x^4}{4}-\frac {1}{40} \left (15-7 \sqrt {5}\right ) \log \left (2 x^4-\sqrt {5}+3\right )-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}+3\right ) \]

[In]

Int[x^11/(1 + 3*x^4 + x^8),x]

[Out]

x^4/4 - ((15 - 7*Sqrt[5])*Log[3 - Sqrt[5] + 2*x^4])/40 - ((15 + 7*Sqrt[5])*Log[3 + Sqrt[5] + 2*x^4])/40

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}{4} \text {Subst}\left (\int \frac {x^2}{1+3 x+x^2} \, dx,x,x^4\right ) \\ & = \frac {x^4}{4}+\frac {1}{4} \text {Subst}\left (\int \frac {-1-3 x}{1+3 x+x^2} \, dx,x,x^4\right ) \\ & = \frac {x^4}{4}+\frac {1}{40} \left (-15+7 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right )-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right ) \\ & = \frac {x^4}{4}-\frac {1}{40} \left (15-7 \sqrt {5}\right ) \log \left (3-\sqrt {5}+2 x^4\right )-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {1}{40} \left (10 x^4+\left (-15+7 \sqrt {5}\right ) \log \left (-3+\sqrt {5}-2 x^4\right )-\left (15+7 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right )\right ) \]

[In]

Integrate[x^11/(1 + 3*x^4 + x^8),x]

[Out]

(10*x^4 + (-15 + 7*Sqrt[5])*Log[-3 + Sqrt[5] - 2*x^4] - (15 + 7*Sqrt[5])*Log[3 + Sqrt[5] + 2*x^4])/40

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {x^{4}}{4}-\frac {3 \ln \left (x^{8}+3 x^{4}+1\right )}{8}-\frac {7 \,\operatorname {arctanh}\left (\frac {\left (2 x^{4}+3\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{20}\)	\(38\)
risch	\(\frac {x^{4}}{4}-\frac {3 \ln \left (2 x^{4}-\sqrt {5}+3\right )}{8}+\frac {7 \ln \left (2 x^{4}-\sqrt {5}+3\right ) \sqrt {5}}{40}-\frac {3 \ln \left (2 x^{4}+\sqrt {5}+3\right )}{8}-\frac {7 \ln \left (2 x^{4}+\sqrt {5}+3\right ) \sqrt {5}}{40}\)	\(69\)

[In]

int(x^11/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/4*x^4-3/8*ln(x^8+3*x^4+1)-7/20*arctanh(1/5*(2*x^4+3)*5^(1/2))*5^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {1}{4} \, x^{4} + \frac {7}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{8} + 6 \, x^{4} - \sqrt {5} {\left (2 \, x^{4} + 3\right )} + 7}{x^{8} + 3 \, x^{4} + 1}\right ) - \frac {3}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

[In]

integrate(x^11/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

1/4*x^4 + 7/40*sqrt(5)*log((2*x^8 + 6*x^4 - sqrt(5)*(2*x^4 + 3) + 7)/(x^8 + 3*x^4 + 1)) - 3/8*log(x^8 + 3*x^4
+ 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.97 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {x^{4}}{4} + \left (- \frac {3}{8} + \frac {7 \sqrt {5}}{40}\right ) \log {\left (x^{4} - \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} + \left (- \frac {7 \sqrt {5}}{40} - \frac {3}{8}\right ) \log {\left (x^{4} + \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} \]

[In]

integrate(x**11/(x**8+3*x**4+1),x)

[Out]

x**4/4 + (-3/8 + 7*sqrt(5)/40)*log(x**4 - sqrt(5)/2 + 3/2) + (-7*sqrt(5)/40 - 3/8)*log(x**4 + sqrt(5)/2 + 3/2)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {1}{4} \, x^{4} + \frac {7}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) - \frac {3}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

[In]

integrate(x^11/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

1/4*x^4 + 7/40*sqrt(5)*log((2*x^4 - sqrt(5) + 3)/(2*x^4 + sqrt(5) + 3)) - 3/8*log(x^8 + 3*x^4 + 1)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {1}{4} \, x^{4} + \frac {7}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) - \frac {3}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

[In]

integrate(x^11/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

1/4*x^4 + 7/40*sqrt(5)*log((2*x^4 - sqrt(5) + 3)/(2*x^4 + sqrt(5) + 3)) - 3/8*log(x^8 + 3*x^4 + 1)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {x^{11}}{1+3 x^4+x^8} \, dx=\frac {7\,\sqrt {5}\,\ln \left (x^4-\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{40}-\frac {3\,\ln \left (x^4+\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{8}-\frac {3\,\ln \left (x^4-\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{8}-\frac {7\,\sqrt {5}\,\ln \left (x^4+\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{40}+\frac {x^4}{4} \]

[In]

int(x^11/(3*x^4 + x^8 + 1),x)

[Out]

(7*5^(1/2)*log(x^4 - 5^(1/2)/2 + 3/2))/40 - (3*log(5^(1/2)/2 + x^4 + 3/2))/8 - (3*log(x^4 - 5^(1/2)/2 + 3/2))/
8 - (7*5^(1/2)*log(5^(1/2)/2 + x^4 + 3/2))/40 + x^4/4