3.156 \(\int \frac{1}{x^4 (3+4 x^3+x^6)} \, dx\)

Optimal. Leaf size=34 \[ -\frac{1}{9 x^3}+\frac{1}{6} \log \left (x^3+1\right )-\frac{1}{54} \log \left (x^3+3\right )-\frac{4 \log (x)}{9} \]

[Out]

-1/(9*x^3) - (4*Log[x])/9 + Log[1 + x^3]/6 - Log[3 + x^3]/54

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Rubi [A]  time = 0.0318425, antiderivative size = 34, normalized size of antiderivative = 1., 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, 709, 800} \[ -\frac{1}{9 x^3}+\frac{1}{6} \log \left (x^3+1\right )-\frac{1}{54} \log \left (x^3+3\right )-\frac{4 \log (x)}{9} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(3 + 4*x^3 + x^6)),x]

[Out]

-1/(9*x^3) - (4*Log[x])/9 + Log[1 + x^3]/6 - Log[3 + x^3]/54

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]]

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]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (3+4 x^3+x^6\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (3+4 x+x^2\right )} \, dx,x,x^3\right )\\ &=-\frac{1}{9 x^3}+\frac{1}{9} \operatorname{Subst}\left (\int \frac{-4-x}{x \left (3+4 x+x^2\right )} \, dx,x,x^3\right )\\ &=-\frac{1}{9 x^3}+\frac{1}{9} \operatorname{Subst}\left (\int \left (-\frac{4}{3 x}+\frac{3}{2 (1+x)}-\frac{1}{6 (3+x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{1}{9 x^3}-\frac{4 \log (x)}{9}+\frac{1}{6} \log \left (1+x^3\right )-\frac{1}{54} \log \left (3+x^3\right )\\ \end{align*}

Mathematica [A]  time = 0.0072617, size = 34, normalized size = 1. \[ -\frac{1}{9 x^3}+\frac{1}{6} \log \left (x^3+1\right )-\frac{1}{54} \log \left (x^3+3\right )-\frac{4 \log (x)}{9} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(3 + 4*x^3 + x^6)),x]

[Out]

-1/(9*x^3) - (4*Log[x])/9 + Log[1 + x^3]/6 - Log[3 + x^3]/54

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Maple [A]  time = 0.008, size = 36, normalized size = 1.1 \begin{align*} -{\frac{1}{9\,{x}^{3}}}-{\frac{4\,\ln \left ( x \right ) }{9}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{6}}-{\frac{\ln \left ({x}^{3}+3 \right ) }{54}}+{\frac{\ln \left ( 1+x \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(x^6+4*x^3+3),x)

[Out]

-1/9/x^3-4/9*ln(x)+1/6*ln(x^2-x+1)-1/54*ln(x^3+3)+1/6*ln(1+x)

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Maxima [A]  time = 1.16239, size = 38, normalized size = 1.12 \begin{align*} -\frac{1}{9 \, x^{3}} - \frac{1}{54} \, \log \left (x^{3} + 3\right ) + \frac{1}{6} \, \log \left (x^{3} + 1\right ) - \frac{4}{27} \, \log \left (x^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

-1/9/x^3 - 1/54*log(x^3 + 3) + 1/6*log(x^3 + 1) - 4/27*log(x^3)

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Fricas [A]  time = 1.42205, size = 96, normalized size = 2.82 \begin{align*} -\frac{x^{3} \log \left (x^{3} + 3\right ) - 9 \, x^{3} \log \left (x^{3} + 1\right ) + 24 \, x^{3} \log \left (x\right ) + 6}{54 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

-1/54*(x^3*log(x^3 + 3) - 9*x^3*log(x^3 + 1) + 24*x^3*log(x) + 6)/x^3

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Sympy [A]  time = 0.17, size = 29, normalized size = 0.85 \begin{align*} - \frac{4 \log{\left (x \right )}}{9} + \frac{\log{\left (x^{3} + 1 \right )}}{6} - \frac{\log{\left (x^{3} + 3 \right )}}{54} - \frac{1}{9 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(x**6+4*x**3+3),x)

[Out]

-4*log(x)/9 + log(x**3 + 1)/6 - log(x**3 + 3)/54 - 1/(9*x**3)

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Giac [A]  time = 1.10187, size = 49, normalized size = 1.44 \begin{align*} \frac{4 \, x^{3} - 3}{27 \, x^{3}} - \frac{1}{54} \, \log \left ({\left | x^{3} + 3 \right |}\right ) + \frac{1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) - \frac{4}{9} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

1/27*(4*x^3 - 3)/x^3 - 1/54*log(abs(x^3 + 3)) + 1/6*log(abs(x^3 + 1)) - 4/9*log(abs(x))