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]

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

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Rubi [A]  time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, 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))

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fricas [A]  time = 0.43, size = 69, normalized size = 0.88 \[ \frac {1}{15} \cdot 25^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{75} \cdot 25^{\frac {1}{6}} {\left (2 \cdot 25^{\frac {2}{3}} \sqrt {3} x - 5 \cdot 25^{\frac {1}{3}} \sqrt {3}\right )}\right ) - \frac {1}{150} \cdot 25^{\frac {2}{3}} \log \left (5 \, x^{2} - 25^{\frac {2}{3}} x + 5 \cdot 25^{\frac {1}{3}}\right ) + \frac {1}{75} \cdot 25^{\frac {2}{3}} \log \left (5 \, x + 25^{\frac {2}{3}}\right ) \]

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))

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giac [A]  time = 0.95, size = 58, normalized size = 0.74 \[ \frac {1}{15} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {1}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (2 \, x - 5^{\frac {1}{3}}\right )}\right ) - \frac {1}{30} \cdot 5^{\frac {1}{3}} \log \left (x^{2} - 5^{\frac {1}{3}} x + 5^{\frac {2}{3}}\right ) + \frac {1}{15} \cdot 5^{\frac {1}{3}} \log \left ({\left | x + 5^{\frac {1}{3}} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(abs(x + 5^(1/3)))

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maple [A]  time = 0.00, size = 54, normalized size = 0.69 \[ \frac {5^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,5^{\frac {2}{3}} x}{5}-1\right )}{3}\right )}{15}+\frac {5^{\frac {1}{3}} \ln \left (x +5^{\frac {1}{3}}\right )}{15}-\frac {5^{\frac {1}{3}} \ln \left (x^{2}-5^{\frac {1}{3}} x +5^{\frac {2}{3}}\right )}{30} \]

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))

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maxima [A]  time = 0.98, size = 57, normalized size = 0.73 \[ \frac {1}{15} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {1}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (2 \, x - 5^{\frac {1}{3}}\right )}\right ) - \frac {1}{30} \cdot 5^{\frac {1}{3}} \log \left (x^{2} - 5^{\frac {1}{3}} x + 5^{\frac {2}{3}}\right ) + \frac {1}{15} \cdot 5^{\frac {1}{3}} \log \left (x + 5^{\frac {1}{3}}\right ) \]

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))

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mupad [B]  time = 0.27, size = 70, normalized size = 0.90 \[ \frac {5^{1/3}\,\ln \left (x+5^{1/3}\right )}{15}+\frac {5^{1/3}\,\ln \left (x+\frac {5^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{30}-\frac {5^{1/3}\,\ln \left (x-\frac {5^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{30} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 + 5),x)

[Out]

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

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sympy [A]  time = 0.30, size = 73, normalized size = 0.94 \[ \frac {\sqrt [3]{5} \log {\left (x + \sqrt [3]{5} \right )}}{15} - \frac {\sqrt [3]{5} \log {\left (x^{2} - \sqrt [3]{5} x + 5^{\frac {2}{3}} \right )}}{30} + \frac {\sqrt {3} \sqrt [3]{5} \operatorname {atan}{\left (\frac {2 \sqrt {3} \cdot 5^{\frac {2}{3}} x}{15} - \frac {\sqrt {3}}{3} \right )}}{15} \]

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

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