3.259 \(\int \frac{1}{5+x^3} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\log \left (x^2-\sqrt [3]{5} x+5^{2/3}\right )}{6\ 5^{2/3}}+\frac{\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}-\frac{\tan ^{-1}\left (\frac{\sqrt [3]{5}-2 x}{\sqrt{3} \sqrt [3]{5}}\right )}{\sqrt{3} 5^{2/3}} \]

[Out]

-(ArcTan[(5^(1/3) - 2*x)/(Sqrt[3]*5^(1/3))]/(Sqrt[3]*5^(2/3))) + Log[5^(1/3) + x]/(3*5^(2/3)) - Log[5^(2/3) -
5^(1/3)*x + x^2]/(6*5^(2/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0458739, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {200, 31, 634, 617, 204, 628} \[ -\frac{\log \left (x^2-\sqrt [3]{5} x+5^{2/3}\right )}{6\ 5^{2/3}}+\frac{\log \left (x+\sqrt [3]{5}\right )}{3\ 5^{2/3}}-\frac{\tan ^{-1}\left (\frac{\sqrt [3]{5}-2 x}{\sqrt{3} \sqrt [3]{5}}\right )}{\sqrt{3} 5^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 + x^3)^(-1),x]

[Out]

-(ArcTan[(5^(1/3) - 2*x)/(Sqrt[3]*5^(1/3))]/(Sqrt[3]*5^(2/3))) + Log[5^(1/3) + x]/(3*5^(2/3)) - Log[5^(2/3) -
5^(1/3)*x + x^2]/(6*5^(2/3))

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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}{5+x^3} \, dx &=\frac{\int \frac{1}{\sqrt [3]{5}+x} \, dx}{3\ 5^{2/3}}+\frac{\int \frac{2 \sqrt [3]{5}-x}{5^{2/3}-\sqrt [3]{5} x+x^2} \, dx}{3\ 5^{2/3}}\\ &=\frac{\log \left (\sqrt [3]{5}+x\right )}{3\ 5^{2/3}}-\frac{\int \frac{-\sqrt [3]{5}+2 x}{5^{2/3}-\sqrt [3]{5} x+x^2} \, dx}{6\ 5^{2/3}}+\frac{\int \frac{1}{5^{2/3}-\sqrt [3]{5} x+x^2} \, dx}{2 \sqrt [3]{5}}\\ &=\frac{\log \left (\sqrt [3]{5}+x\right )}{3\ 5^{2/3}}-\frac{\log \left (5^{2/3}-\sqrt [3]{5} x+x^2\right )}{6\ 5^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 x}{\sqrt [3]{5}}\right )}{5^{2/3}}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt [3]{5}-2 x}{\sqrt{3} \sqrt [3]{5}}\right )}{\sqrt{3} 5^{2/3}}+\frac{\log \left (\sqrt [3]{5}+x\right )}{3\ 5^{2/3}}-\frac{\log \left (5^{2/3}-\sqrt [3]{5} x+x^2\right )}{6\ 5^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0226852, size = 71, normalized size = 0.91 \[ \frac{-\log \left (\sqrt [3]{5} x^2-5^{2/3} x+5\right )+2 \log \left (5^{2/3} x+5\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2\ 5^{2/3} x-5}{5 \sqrt{3}}\right )}{6\ 5^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + x^3)^(-1),x]

[Out]

(2*Sqrt[3]*ArcTan[(-5 + 2*5^(2/3)*x)/(5*Sqrt[3])] + 2*Log[5 + 5^(2/3)*x] - Log[5 - 5^(2/3)*x + 5^(1/3)*x^2])/(
6*5^(2/3))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 54, normalized size = 0.7 \begin{align*}{\frac{\ln \left ( \sqrt [3]{5}+x \right ) \sqrt [3]{5}}{15}}-{\frac{\ln \left ({5}^{{\frac{2}{3}}}-\sqrt [3]{5}x+{x}^{2} \right ) \sqrt [3]{5}}{30}}+{\frac{\sqrt [3]{5}\sqrt{3}}{15}\arctan \left ({\frac{\sqrt{3}}{3} \left ({\frac{2\,{5}^{2/3}x}{5}}-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3+5),x)

[Out]

1/15*ln(5^(1/3)+x)*5^(1/3)-1/30*ln(5^(2/3)-5^(1/3)*x+x^2)*5^(1/3)+1/15*5^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/5
*5^(2/3)*x-1))

________________________________________________________________________________________

Maxima [A]  time = 1.42263, size = 77, normalized size = 0.99 \begin{align*} \frac{1}{15} \cdot 5^{\frac{1}{3}} \sqrt{3} \arctan \left (\frac{1}{15} \cdot 5^{\frac{2}{3}} \sqrt{3}{\left (2 \, x - 5^{\frac{1}{3}}\right )}\right ) - \frac{1}{30} \cdot 5^{\frac{1}{3}} \log \left (x^{2} - 5^{\frac{1}{3}} x + 5^{\frac{2}{3}}\right ) + \frac{1}{15} \cdot 5^{\frac{1}{3}} \log \left (x + 5^{\frac{1}{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+5),x, algorithm="maxima")

[Out]

1/15*5^(1/3)*sqrt(3)*arctan(1/15*5^(2/3)*sqrt(3)*(2*x - 5^(1/3))) - 1/30*5^(1/3)*log(x^2 - 5^(1/3)*x + 5^(2/3)
) + 1/15*5^(1/3)*log(x + 5^(1/3))

________________________________________________________________________________________

Fricas [A]  time = 1.66226, size = 242, normalized size = 3.1 \begin{align*} \frac{1}{15} \cdot 25^{\frac{1}{6}} \sqrt{3} \arctan \left (\frac{1}{75} \cdot 25^{\frac{1}{6}}{\left (2 \cdot 25^{\frac{2}{3}} \sqrt{3} x - 5 \cdot 25^{\frac{1}{3}} \sqrt{3}\right )}\right ) - \frac{1}{150} \cdot 25^{\frac{2}{3}} \log \left (5 \, x^{2} - 25^{\frac{2}{3}} x + 5 \cdot 25^{\frac{1}{3}}\right ) + \frac{1}{75} \cdot 25^{\frac{2}{3}} \log \left (5 \, x + 25^{\frac{2}{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+5),x, algorithm="fricas")

[Out]

1/15*25^(1/6)*sqrt(3)*arctan(1/75*25^(1/6)*(2*25^(2/3)*sqrt(3)*x - 5*25^(1/3)*sqrt(3))) - 1/150*25^(2/3)*log(5
*x^2 - 25^(2/3)*x + 5*25^(1/3)) + 1/75*25^(2/3)*log(5*x + 25^(2/3))

________________________________________________________________________________________

Sympy [A]  time = 0.304137, size = 73, normalized size = 0.94 \begin{align*} \frac{\sqrt [3]{5} \log{\left (x + \sqrt [3]{5} \right )}}{15} - \frac{\sqrt [3]{5} \log{\left (x^{2} - \sqrt [3]{5} x + 5^{\frac{2}{3}} \right )}}{30} + \frac{\sqrt{3} \sqrt [3]{5} \operatorname{atan}{\left (\frac{2 \sqrt{3} \cdot 5^{\frac{2}{3}} x}{15} - \frac{\sqrt{3}}{3} \right )}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3+5),x)

[Out]

5**(1/3)*log(x + 5**(1/3))/15 - 5**(1/3)*log(x**2 - 5**(1/3)*x + 5**(2/3))/30 + sqrt(3)*5**(1/3)*atan(2*sqrt(3
)*5**(2/3)*x/15 - sqrt(3)/3)/15

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+5),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError