3.14.65 \(\int \frac {x^4}{1+x^6} \, dx\) [1365]

Optimal. Leaf size=80 \[ -\frac {1}{6} \tan ^{-1}\left (\sqrt {3}-2 x\right )+\frac {1}{3} \tan ^{-1}(x)+\frac {1}{6} \tan ^{-1}\left (\sqrt {3}+2 x\right )+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \]

[Out]

1/3*arctan(x)+1/6*arctan(2*x-3^(1/2))+1/6*arctan(2*x+3^(1/2))+1/12*ln(1+x^2-x*3^(1/2))*3^(1/2)-1/12*ln(1+x^2+x
*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {301, 648, 632, 210, 642, 209} \begin {gather*} -\frac {1}{6} \text {ArcTan}\left (\sqrt {3}-2 x\right )+\frac {\text {ArcTan}(x)}{3}+\frac {1}{6} \text {ArcTan}\left (2 x+\sqrt {3}\right )+\frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(1 + x^6),x]

[Out]

-1/6*ArcTan[Sqrt[3] - 2*x] + ArcTan[x]/3 + ArcTan[Sqrt[3] + 2*x]/6 + Log[1 - Sqrt[3]*x + x^2]/(4*Sqrt[3]) - Lo
g[1 + Sqrt[3]*x + x^2]/(4*Sqrt[3])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 301

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(
2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 +
2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))*Int[1/(r^2 + s^2*x^2), x] +
Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 632

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

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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]

Rubi steps

\begin {align*} \int \frac {x^4}{1+x^6} \, dx &=\frac {1}{3} \int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx+\frac {1}{3} \int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx+\frac {1}{3} \int \frac {1}{1+x^2} \, dx\\ &=\frac {1}{3} \tan ^{-1}(x)+\frac {1}{12} \int \frac {1}{1-\sqrt {3} x+x^2} \, dx+\frac {1}{12} \int \frac {1}{1+\sqrt {3} x+x^2} \, dx+\frac {\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx}{4 \sqrt {3}}-\frac {\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx}{4 \sqrt {3}}\\ &=\frac {1}{3} \tan ^{-1}(x)+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 x\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 x\right )\\ &=-\frac {1}{6} \tan ^{-1}\left (\sqrt {3}-2 x\right )+\frac {1}{3} \tan ^{-1}(x)+\frac {1}{6} \tan ^{-1}\left (\sqrt {3}+2 x\right )+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 73, normalized size = 0.91 \begin {gather*} \frac {1}{12} \left (-2 \tan ^{-1}\left (\sqrt {3}-2 x\right )+4 \tan ^{-1}(x)+2 \tan ^{-1}\left (\sqrt {3}+2 x\right )+\sqrt {3} \log \left (1-\sqrt {3} x+x^2\right )-\sqrt {3} \log \left (1+\sqrt {3} x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(1 + x^6),x]

[Out]

(-2*ArcTan[Sqrt[3] - 2*x] + 4*ArcTan[x] + 2*ArcTan[Sqrt[3] + 2*x] + Sqrt[3]*Log[1 - Sqrt[3]*x + x^2] - Sqrt[3]
*Log[1 + Sqrt[3]*x + x^2])/12

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Maple [A]
time = 0.17, size = 61, normalized size = 0.76

method	result	size
risch	\(\frac {\arctan \left (x \right )}{3}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{3}-\textit {\_R} +x \right )\right )}{6}\)	\(33\)
default	\(\frac {\arctan \left (x \right )}{3}+\frac {\arctan \left (2 x -\sqrt {3}\right )}{6}+\frac {\arctan \left (2 x +\sqrt {3}\right )}{6}+\frac {\ln \left (1+x^{2}-\sqrt {3}\, x \right ) \sqrt {3}}{12}-\frac {\ln \left (1+x^{2}+\sqrt {3}\, x \right ) \sqrt {3}}{12}\)	\(61\)
meijerg	\(\frac {x^{5} \sqrt {3}\, \ln \left (1-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\left (x^{6}\right )^{\frac {1}{6}}\right )}{3 \left (x^{6}\right )^{\frac {5}{6}}}-\frac {x^{5} \sqrt {3}\, \ln \left (1+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {5}{6}}}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

1/3*arctan(x)+1/6*arctan(2*x-3^(1/2))+1/6*arctan(2*x+3^(1/2))+1/12*ln(1+x^2-3^(1/2)*x)*3^(1/2)-1/12*ln(1+x^2+3
^(1/2)*x)*3^(1/2)

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Maxima [A]
time = 0.50, size = 60, normalized size = 0.75 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{6} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{6} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{3} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^6+1),x, algorithm="maxima")

[Out]

-1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/6*arctan(2*x + sqrt(3)) + 1
/6*arctan(2*x - sqrt(3)) + 1/3*arctan(x)

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Fricas [A]
time = 0.36, size = 94, normalized size = 1.18 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \log \left (16 \, x^{2} + 16 \, \sqrt {3} x + 16\right ) + \frac {1}{12} \, \sqrt {3} \log \left (16 \, x^{2} - 16 \, \sqrt {3} x + 16\right ) + \frac {1}{3} \, \arctan \left (x\right ) - \frac {1}{3} \, \arctan \left (-2 \, x + \sqrt {3} + 2 \, \sqrt {x^{2} - \sqrt {3} x + 1}\right ) - \frac {1}{3} \, \arctan \left (-2 \, x - \sqrt {3} + 2 \, \sqrt {x^{2} + \sqrt {3} x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^6+1),x, algorithm="fricas")

[Out]

-1/12*sqrt(3)*log(16*x^2 + 16*sqrt(3)*x + 16) + 1/12*sqrt(3)*log(16*x^2 - 16*sqrt(3)*x + 16) + 1/3*arctan(x) -
1/3*arctan(-2*x + sqrt(3) + 2*sqrt(x^2 - sqrt(3)*x + 1)) - 1/3*arctan(-2*x - sqrt(3) + 2*sqrt(x^2 + sqrt(3)*x
+ 1))

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Sympy [A]
time = 0.09, size = 68, normalized size = 0.85 \begin {gather*} \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} + \frac {\operatorname {atan}{\left (x \right )}}{3} + \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{6} + \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(x**6+1),x)

[Out]

sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + atan(x)/3 + atan(2*x - sqrt(3))/
6 + atan(2*x + sqrt(3))/6

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Giac [A]
time = 1.25, size = 60, normalized size = 0.75 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{6} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{6} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{3} \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^6+1),x, algorithm="giac")

[Out]

-1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/6*arctan(2*x + sqrt(3)) + 1
/6*arctan(2*x - sqrt(3)) + 1/3*arctan(x)

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Mupad [B]
time = 1.05, size = 53, normalized size = 0.66 \begin {gather*} \frac {\mathrm {atan}\left (x\right )}{3}-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^6 + 1),x)

[Out]

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

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