3.506 \(\int \frac{1}{\sqrt{1+x} \sqrt{1-x+x^2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{2 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}} \]

[Out]

(2*Sqrt[2 + Sqrt[3]]*Sqrt[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 + x^2])

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Rubi [A]  time = 0.0219877, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {713, 218} \[ \frac{2 \sqrt{2+\sqrt{3}} \sqrt{x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2 + Sqrt[3]]*Sqrt[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 + x^2])

Rule 713

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[((d + e*x)^FracPart[p
]*(a + b*x + c*x^2)^FracPart[p])/(a*d + c*e*x^3)^FracPart[p], Int[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && IGtQ[m - p + 1, 0] &&  !Intege
rQ[p]

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{1+x} \sqrt{1-x+x^2}} \, dx &=\frac{\sqrt{1+x^3} \int \frac{1}{\sqrt{1+x^3}} \, dx}{\sqrt{1+x} \sqrt{1-x+x^2}}\\ &=\frac{2 \sqrt{2+\sqrt{3}} \sqrt{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+x^2}}\\ \end{align*}

Mathematica [C]  time = 0.144856, size = 148, normalized size = 1.35 \[ \frac{i (x+1) \sqrt{1+\frac{6 i}{\left (\sqrt{3}-3 i\right ) (x+1)}} \sqrt{\frac{2}{3}-\frac{4 i}{\left (\sqrt{3}+3 i\right ) (x+1)}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{\sqrt{3}+3 i}}}{\sqrt{x+1}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )}{\sqrt{-\frac{i}{\sqrt{3}+3 i}} \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[2/3 - (4*I)/((3*I + Sqrt[3])*(1 + x))]*EllipticF[I*
ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/(Sqrt[(-I)/(3*I + Sqrt[3]
)]*Sqrt[1 - x + x^2])

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Maple [A]  time = 0.651, size = 137, normalized size = 1.3 \begin{align*}{\frac{-i\sqrt{3}+3}{{x}^{3}+1}\sqrt{1+x}\sqrt{{x}^{2}-x+1}\sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{i\sqrt{3}+3}}}\sqrt{{\frac{2\,x-1+i\sqrt{3}}{i\sqrt{3}-3}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}},\sqrt{-{\frac{i\sqrt{3}-3}{i\sqrt{3}+3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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