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

Optimal. Leaf size=39 \[ -\frac{1}{6} \log \left (x^6-x^3+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^3}{\sqrt{3}}\right )}{\sqrt{3}}+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.0629689, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1594, 1474, 800, 634, 618, 204, 628} \[ -\frac{1}{6} \log \left (x^6-x^3+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^3}{\sqrt{3}}\right )}{\sqrt{3}}+\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)/(x - x^4 + x^7),x]

[Out]

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

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

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 634

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 618

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 204

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

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{1+x^3}{x-x^4+x^7} \, dx &=\int \frac{1+x^3}{x \left (1-x^3+x^6\right )} \, dx\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1+x}{x \left (1-x+x^2\right )} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{1}{x}+\frac{2-x}{1-x+x^2}\right ) \, dx,x,x^3\right )\\ &=\log (x)+\frac{1}{3} \operatorname{Subst}\left (\int \frac{2-x}{1-x+x^2} \, dx,x,x^3\right )\\ &=\log (x)-\frac{1}{6} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,x^3\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^3\right )\\ &=\log (x)-\frac{1}{6} \log \left (1-x^3+x^6\right )-\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^3\right )\\ &=\frac{\tan ^{-1}\left (\frac{-1+2 x^3}{\sqrt{3}}\right )}{\sqrt{3}}+\log (x)-\frac{1}{6} \log \left (1-x^3+x^6\right )\\ \end{align*}

Mathematica [C]  time = 0.0104177, size = 55, normalized size = 1.41 \[ \log (x)-\frac{1}{3} \text{RootSum}\left [\text{$\#$1}^6-\text{$\#$1}^3+1\& ,\frac{\text{$\#$1}^3 \log (x-\text{$\#$1})-2 \log (x-\text{$\#$1})}{2 \text{$\#$1}^3-1}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)/(x - x^4 + x^7),x]

[Out]

Log[x] - RootSum[1 - #1^3 + #1^6 & , (-2*Log[x - #1] + Log[x - #1]*#1^3)/(-1 + 2*#1^3) & ]/3

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Maple [A]  time = 0.004, size = 35, normalized size = 0.9 \begin{align*} \ln \left ( x \right ) -{\frac{\ln \left ({x}^{6}-{x}^{3}+1 \right ) }{6}}+{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,{x}^{3}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x^{5} - 2 \, x^{2}}{x^{6} - x^{3} + 1}\,{d x} + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((x^5 - 2*x^2)/(x^6 - x^3 + 1), x) + log(x)

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Fricas [A]  time = 1.49832, size = 107, normalized size = 2.74 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{3} - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.14817, size = 41, normalized size = 1.05 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{6} - x^{3} + 1 \right )}}{6} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{3}}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09442, size = 47, normalized size = 1.21 \begin{align*} \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{3} - 1\right )}\right ) - \frac{1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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