3.152 \(\int \frac{1+\sqrt{3}+x}{x \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}-\frac{2}{3} \left (1+\sqrt{3}\right ) \tanh ^{-1}\left (\sqrt{x^3+1}\right ) \]

[Out]

(-2*(1 + Sqrt[3])*ArcTanh[Sqrt[1 + x^3]])/3 + (2*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^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3
] + x)^2]*Sqrt[1 + x^3])

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Rubi [A]  time = 0.0455044, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {1832, 266, 63, 207, 218} \[ \frac{2 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}-\frac{2}{3} \left (1+\sqrt{3}\right ) \tanh ^{-1}\left (\sqrt{x^3+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + Sqrt[3])*ArcTanh[Sqrt[1 + x^3]])/3 + (2*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^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3
] + x)^2]*Sqrt[1 + x^3])

Rule 1832

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rubi steps

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

Mathematica [C]  time = 0.0255547, size = 39, normalized size = 0.31 \[ x \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-x^3\right )-\frac{2}{3} \left (1+\sqrt{3}\right ) \tanh ^{-1}\left (\sqrt{x^3+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + Sqrt[3])*ArcTanh[Sqrt[1 + x^3]])/3 + x*Hypergeometric2F1[1/3, 1/2, 4/3, -x^3]

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Maple [A]  time = 0.026, size = 132, normalized size = 1.1 \begin{align*} 2\,{\frac{3/2-i/2\sqrt{3}}{\sqrt{{x}^{3}+1}}\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 EllipticF} \left ( \sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) }-{\frac{2+2\,\sqrt{3}}{3}{\it Artanh} \left ( \sqrt{{x}^{3}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x+3^(1/2))/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)*EllipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3
/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))-2/3*arctanh((x^3+1)^(1/2))*(1+3^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^3 + 1)*(x + sqrt(3) + 1)/(x^4 + x), x)

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Sympy [A]  time = 3.12711, size = 56, normalized size = 0.45 \begin{align*} \frac{x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} - \frac{2 \sqrt{3} \operatorname{asinh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} - \frac{2 \operatorname{asinh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3)) - 2*sqrt(3)*asinh(x**(-3/2))/3 - 2
*asinh(x**(-3/2))/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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