3.394 \(\int \frac {1}{x^5 (1-3 x^4+x^8)} \, dx\)

Optimal. Leaf size=66 \[ -\frac {1}{4 x^4}-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \log \left (-2 x^4-\sqrt {5}+3\right )-\frac {1}{40} \left (15-7 \sqrt {5}\right ) \log \left (-2 x^4+\sqrt {5}+3\right )+3 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1357, 709, 800, 632, 31} \[ -\frac {1}{4 x^4}-\frac {1}{40} \left (15+7 \sqrt {5}\right ) \log \left (-2 x^4-\sqrt {5}+3\right )-\frac {1}{40} \left (15-7 \sqrt {5}\right ) \log \left (-2 x^4+\sqrt {5}+3\right )+3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(1 - 3*x^4 + x^8)),x]

[Out]

-1/(4*x^4) + 3*Log[x] - ((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 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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 {1}{x^5 \left (1-3 x^4+x^8\right )} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-3 x+x^2\right )} \, dx,x,x^4\right )\\ &=-\frac {1}{4 x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {3-x}{x \left (1-3 x+x^2\right )} \, dx,x,x^4\right )\\ &=-\frac {1}{4 x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {3}{x}+\frac {8-3 x}{1-3 x+x^2}\right ) \, dx,x,x^4\right )\\ &=-\frac {1}{4 x^4}+3 \log (x)+\frac {1}{4} \operatorname {Subst}\left (\int \frac {8-3 x}{1-3 x+x^2} \, dx,x,x^4\right )\\ &=-\frac {1}{4 x^4}+3 \log (x)+\frac {1}{40} \left (-15+7 \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 (15+7 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {3}{2}+\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right )\\ &=-\frac {1}{4 x^4}+3 \log (x)-\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*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.92 \[ \frac {1}{40} \left (-\frac {10}{x^4}+\left (7 \sqrt {5}-15\right ) \log \left (-2 x^4+\sqrt {5}+3\right )-\left (15+7 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}-3\right )+120 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(1 - 3*x^4 + x^8)),x]

[Out]

(-10/x^4 + 120*Log[x] + (-15 + 7*Sqrt[5])*Log[3 + Sqrt[5] - 2*x^4] - (15 + 7*Sqrt[5])*Log[-3 + Sqrt[5] + 2*x^4
])/40

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fricas [A]  time = 0.82, size = 76, normalized size = 1.15 \[ \frac {7 \, \sqrt {5} x^{4} \log \left (\frac {2 \, x^{8} - 6 \, x^{4} - \sqrt {5} {\left (2 \, x^{4} - 3\right )} + 7}{x^{8} - 3 \, x^{4} + 1}\right ) - 15 \, x^{4} \log \left (x^{8} - 3 \, x^{4} + 1\right ) + 120 \, x^{4} \log \relax (x) - 10}{40 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^8-3*x^4+1),x, algorithm="fricas")

[Out]

1/40*(7*sqrt(5)*x^4*log((2*x^8 - 6*x^4 - sqrt(5)*(2*x^4 - 3) + 7)/(x^8 - 3*x^4 + 1)) - 15*x^4*log(x^8 - 3*x^4
+ 1) + 120*x^4*log(x) - 10)/x^4

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giac [A]  time = 0.51, size = 66, normalized size = 1.00 \[ \frac {7}{40} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{4} - \sqrt {5} - 3 \right |}}{{\left | 2 \, x^{4} + \sqrt {5} - 3 \right |}}\right ) - \frac {3 \, x^{4} + 1}{4 \, x^{4}} + \frac {3}{4} \, \log \left (x^{4}\right ) - \frac {3}{8} \, \log \left ({\left | x^{8} - 3 \, x^{4} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^8-3*x^4+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 71, normalized size = 1.08 \[ -\frac {7 \sqrt {5}\, \arctanh \left (\frac {\left (2 x^{2}-1\right ) \sqrt {5}}{5}\right )}{20}+\frac {7 \sqrt {5}\, \arctanh \left (\frac {\left (2 x^{2}+1\right ) \sqrt {5}}{5}\right )}{20}+3 \ln \relax (x )-\frac {3 \ln \left (x^{4}-x^{2}-1\right )}{8}-\frac {3 \ln \left (x^{4}+x^{2}-1\right )}{8}-\frac {1}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(x^8-3*x^4+1),x)

[Out]

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

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maxima [A]  time = 1.40, size = 56, normalized size = 0.85 \[ \frac {7}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} - 3}{2 \, x^{4} + \sqrt {5} - 3}\right ) - \frac {1}{4 \, x^{4}} - \frac {3}{8} \, \log \left (x^{8} - 3 \, x^{4} + 1\right ) + \frac {3}{4} \, \log \left (x^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(x^8-3*x^4+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.35, size = 49, normalized size = 0.74 \[ 3\,\ln \relax (x)-\frac {1}{4\,x^4}+\ln \left (x^4-\frac {\sqrt {5}}{2}-\frac {3}{2}\right )\,\left (\frac {7\,\sqrt {5}}{40}-\frac {3}{8}\right )-\ln \left (x^4+\frac {\sqrt {5}}{2}-\frac {3}{2}\right )\,\left (\frac {7\,\sqrt {5}}{40}+\frac {3}{8}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(x^8 - 3*x^4 + 1)),x)

[Out]

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

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sympy [A]  time = 0.19, size = 66, normalized size = 1.00 \[ 3 \log {\relax (x )} + \left (- \frac {3}{8} + \frac {7 \sqrt {5}}{40}\right ) \log {\left (x^{4} - \frac {3}{2} - \frac {\sqrt {5}}{2} \right )} + \left (- \frac {7 \sqrt {5}}{40} - \frac {3}{8}\right ) \log {\left (x^{4} - \frac {3}{2} + \frac {\sqrt {5}}{2} \right )} - \frac {1}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(x**8-3*x**4+1),x)

[Out]

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

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