3.1 \(\int \frac{1}{(2^{2/3}+x) \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=145 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{3} \left (\sqrt [3]{2} x+1\right )}{\sqrt{x^3+1}}\right )}{3 \sqrt{3}}+\frac{2 \sqrt [3]{2} \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

[Out]

(2*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/(3*Sqrt[3]) + (2*2^(1/3)*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1
- x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^
(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.177738, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2134, 218, 2137, 203} \[ \frac{2 \sqrt [3]{2} \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{3} \left (\sqrt [3]{2} x+1\right )}{\sqrt{x^3+1}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((2^(2/3) + x)*Sqrt[1 + x^3]),x]

[Out]

(2*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/(3*Sqrt[3]) + (2*2^(1/3)*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1
- x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^
(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])

Rule 2134

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[2/(3*c), Int[1/Sqrt[a + b*x^3], x], x
] + Dist[1/(3*c), Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 -
4*a*d^3, 0]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 2137

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*e)/d, Subst[Int[
1/(1 + 3*a*x^2), x], x, (1 + (2*d*x)/c)/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{1}{\left (2^{2/3}+x\right ) \sqrt{1+x^3}} \, dx &=\frac{\int \frac{2^{2/3}-2 x}{\left (2^{2/3}+x\right ) \sqrt{1+x^3}} \, dx}{3\ 2^{2/3}}+\frac{1}{3} \sqrt [3]{2} \int \frac{1}{\sqrt{1+x^3}} \, dx\\ &=\frac{2 \sqrt [3]{2} \sqrt{2+\sqrt{3}} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\sqrt{3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac{1-\sqrt{3}+x}{1+\sqrt{3}+x}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1+x^3}}+\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{1+3 x^2} \, dx,x,\frac{1+\sqrt [3]{2} x}{\sqrt{1+x^3}}\right )\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt{3} \left (1+\sqrt [3]{2} x\right )}{\sqrt{1+x^3}}\right )}{3 \sqrt{3}}+\frac{2 \sqrt [3]{2} \sqrt{2+\sqrt{3}} (1+x) \sqrt{\frac{1-x+x^2}{\left (1+\sqrt{3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac{1-\sqrt{3}+x}{1+\sqrt{3}+x}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \sqrt{1+x^3}}\\ \end{align*}

Mathematica [C]  time = 0.191407, size = 148, normalized size = 1.02 \[ \frac{4 i \sqrt{2} \sqrt{\frac{i (x+1)}{\sqrt{3}+3 i}} \sqrt{x^2-x+1} \Pi \left (\frac{2 \sqrt{3}}{i+2 i 2^{2/3}+\sqrt{3}};\sin ^{-1}\left (\frac{\sqrt{-2 i x+\sqrt{3}+i}}{\sqrt{2} \sqrt [4]{3}}\right )|\frac{2 \sqrt{3}}{3 i+\sqrt{3}}\right )}{\left (1+2\ 2^{2/3}-i \sqrt{3}\right ) \sqrt{x^3+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((2^(2/3) + x)*Sqrt[1 + x^3]),x]

[Out]

((4*I)*Sqrt[2]*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*Sqrt[1 - x + x^2]*EllipticPi[(2*Sqrt[3])/(I + (2*I)*2^(2/3) +
Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/((1 + 2*2^(2/3
) - I*Sqrt[3])*Sqrt[1 + x^3])

________________________________________________________________________________________

Maple [A]  time = 0.117, size = 139, normalized size = 1. \begin{align*} 2\,{\frac{3/2-i/2\sqrt{3}}{\sqrt{{x}^{3}+1} \left ({2}^{2/3}-1 \right ) }\sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x-1/2-i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x-1/2+i/2\sqrt{3}}{-3/2+i/2\sqrt{3}}}}{\it EllipticPi} \left ( \sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}},{\frac{-3/2+i/2\sqrt{3}}{{2}^{2/3}-1}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2^(2/3)+x)/(x^3+1)^(1/2),x)

[Out]

2*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x
-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)/(2^(2/3)-1)*EllipticPi(((1+x)/(3/2-1/2*I*3^(1/2)
))^(1/2),(-3/2+1/2*I*3^(1/2))/(2^(2/3)-1),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2^(2/3)+x)/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^3 + 1)*(x + 2^(2/3))), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{3} + 1}{\left (x^{2} - 2^{\frac{2}{3}} x + 2 \cdot 2^{\frac{1}{3}}\right )}}{x^{6} + 5 \, x^{3} + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2^(2/3)+x)/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^3 + 1)*(x^2 - 2^(2/3)*x + 2*2^(1/3))/(x^6 + 5*x^3 + 4), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x + 2^{\frac{2}{3}}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2**(2/3)+x)/(x**3+1)**(1/2),x)

[Out]

Integral(1/(sqrt((x + 1)*(x**2 - x + 1))*(x + 2**(2/3))), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2^(2/3)+x)/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^3 + 1)*(x + 2^(2/3))), x)