3.370 \(\int \frac {x^7}{1+3 x^4+x^8} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (2 x^4-\sqrt {5}+3\right )+\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}+3\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, 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 = {1357, 632, 31} \[ \frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (2 x^4-\sqrt {5}+3\right )+\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}+3\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^7/(1 + 3*x^4 + x^8),x]

[Out]

((5 - 3*Sqrt[5])*Log[3 - Sqrt[5] + 2*x^4])/40 + ((5 + 3*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 632

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 1357

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+3 x^4+x^8} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {x}{1+3 x+x^2} \, dx,x,x^4\right )\\ &=\frac {1}{40} \left (5-3 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right )+\frac {1}{40} \left (5+3 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right )\\ &=\frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (3-\sqrt {5}+2 x^4\right )+\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.96 \[ \frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (-2 x^4+\sqrt {5}-3\right )+\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}+3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(1 + 3*x^4 + x^8),x]

[Out]

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

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fricas [A]  time = 0.90, size = 56, normalized size = 1.02 \[ \frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{8} + 6 \, x^{4} + \sqrt {5} {\left (2 \, x^{4} + 3\right )} + 7}{x^{8} + 3 \, x^{4} + 1}\right ) + \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.50, size = 45, normalized size = 0.82 \[ -\frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) + \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 33, normalized size = 0.60 \[ \frac {3 \sqrt {5}\, \arctanh \left (\frac {\left (2 x^{4}+3\right ) \sqrt {5}}{5}\right )}{20}+\frac {\ln \left (x^{8}+3 x^{4}+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(x^8+3*x^4+1),x)

[Out]

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

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maxima [A]  time = 1.17, size = 45, normalized size = 0.82 \[ -\frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) + \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.36, size = 59, normalized size = 1.07 \[ \frac {\ln \left (x^4-\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{8}+\frac {\ln \left (x^4+\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{8}-\frac {3\,\sqrt {5}\,\ln \left (x^4-\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{40}+\frac {3\,\sqrt {5}\,\ln \left (x^4+\frac {\sqrt {5}}{2}+\frac {3}{2}\right )}{40} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(3*x^4 + x^8 + 1),x)

[Out]

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

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sympy [A]  time = 0.14, size = 53, normalized size = 0.96 \[ \left (\frac {1}{8} - \frac {3 \sqrt {5}}{40}\right ) \log {\left (x^{4} - \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} + \left (\frac {1}{8} + \frac {3 \sqrt {5}}{40}\right ) \log {\left (x^{4} + \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(x**8+3*x**4+1),x)

[Out]

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

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