3.4.30 \(\int \frac {x^7}{1+x^4+x^8} \, dx\) [330]

Optimal. Leaf size=37 \[ -\frac {\tan ^{-1}\left (\frac {1+2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1+x^4+x^8\right ) \]

[Out]

1/8*ln(x^8+x^4+1)-1/12*arctan(1/3*(2*x^4+1)*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1371, 648, 632, 210, 642} \begin {gather*} \frac {1}{8} \log \left (x^8+x^4+1\right )-\frac {\text {ArcTan}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/(1 + x^4 + x^8),x]

[Out]

-1/4*ArcTan[(1 + 2*x^4)/Sqrt[3]]/Sqrt[3] + Log[1 + x^4 + x^8]/8

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 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]

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*} \int \frac {x^7}{1+x^4+x^8} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {x}{1+x+x^2} \, dx,x,x^4\right )\\ &=-\left (\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^4\right )\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,x^4\right )\\ &=\frac {1}{8} \log \left (1+x^4+x^8\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^4\right )\\ &=-\frac {\tan ^{-1}\left (\frac {1+2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1+x^4+x^8\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 37, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {1+2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1+x^4+x^8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/(1 + x^4 + x^8),x]

[Out]

-1/4*ArcTan[(1 + 2*x^4)/Sqrt[3]]/Sqrt[3] + Log[1 + x^4 + x^8]/8

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Maple [A]
time = 0.02, size = 31, normalized size = 0.84

method	result	size
default	\(\frac {\ln \left (x^{8}+x^{4}+1\right )}{8}-\frac {\arctan \left (\frac {\left (2 x^{4}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\)	\(31\)
risch	\(\frac {\ln \left (4 x^{8}+4 x^{4}+4\right )}{8}-\frac {\arctan \left (\frac {\left (2 x^{4}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/8*ln(x^8+x^4+1)-1/12*arctan(1/3*(2*x^4+1)*3^(1/2))*3^(1/2)

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Maxima [A]
time = 0.54, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+x^4+1),x, algorithm="maxima")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) + 1/8*log(x^8 + x^4 + 1)

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Fricas [A]
time = 0.34, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+x^4+1),x, algorithm="fricas")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) + 1/8*log(x^8 + x^4 + 1)

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Sympy [A]
time = 0.05, size = 37, normalized size = 1.00 \begin {gather*} \frac {\log {\left (x^{8} + x^{4} + 1 \right )}}{8} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(x**8+x**4+1),x)

[Out]

log(x**8 + x**4 + 1)/8 - sqrt(3)*atan(2*sqrt(3)*x**4/3 + sqrt(3)/3)/12

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Giac [A]
time = 5.36, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{8} + x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+x^4+1),x, algorithm="giac")

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1)) + 1/8*log(x^8 + x^4 + 1)

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Mupad [B]
time = 0.04, size = 32, normalized size = 0.86 \begin {gather*} \frac {\ln \left (x^8+x^4+1\right )}{8}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x^4}{3}+\frac {\sqrt {3}}{3}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^4 + x^8 + 1),x)

[Out]

log(x^4 + x^8 + 1)/8 - (3^(1/2)*atan(3^(1/2)/3 + (2*3^(1/2)*x^4)/3))/12

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