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

Optimal. Leaf size=144 \[ \frac{2\ 3^{3/4} \sqrt{2+\sqrt{3}} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (x+1)^{3/2} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{5 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )}+\frac{2}{5} x \sqrt{x^2-x+1} \sqrt{x+1} \]

[Out]

(2*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/5 + (2*3^(3/4)*Sqrt[2 + Sqrt[3]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 -
x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(5*Sqrt
[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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Rubi [A]  time = 0.0324381, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {713, 195, 218} \[ \frac{2}{5} x \sqrt{x^2-x+1} \sqrt{x+1}+\frac{2\ 3^{3/4} \sqrt{2+\sqrt{3}} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (x+1)^{3/2} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{5 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x]*Sqrt[1 - x + x^2],x]

[Out]

(2*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/5 + (2*3^(3/4)*Sqrt[2 + Sqrt[3]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 -
x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(5*Sqrt
[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[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 \sqrt{1+x} \sqrt{1-x+x^2} \, dx &=\frac{\left (\sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \sqrt{1+x^3} \, dx}{\sqrt{1+x^3}}\\ &=\frac{2}{5} x \sqrt{1+x} \sqrt{1-x+x^2}+\frac{\left (3 \sqrt{1+x} \sqrt{1-x+x^2}\right ) \int \frac{1}{\sqrt{1+x^3}} \, dx}{5 \sqrt{1+x^3}}\\ &=\frac{2}{5} x \sqrt{1+x} \sqrt{1-x+x^2}+\frac{2\ 3^{3/4} \sqrt{2+\sqrt{3}} (1+x)^{3/2} \sqrt{1-x+x^2} \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 )}{5 \sqrt{\frac{1+x}{\left (1+\sqrt{3}+x\right )^2}} \left (1+x^3\right )}\\ \end{align*}

Mathematica [C]  time = 0.506257, size = 169, normalized size = 1.17 \[ \frac{2 x \sqrt{x+1} \left (x^2-x+1\right )+\frac{i (x+1) \sqrt{1+\frac{6 i}{\left (\sqrt{3}-3 i\right ) (x+1)}} \sqrt{6-\frac{36 i}{\left (\sqrt{3}+3 i\right ) (x+1)}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{\sqrt{3}+3 i}}}{\sqrt{x+1}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )}{\sqrt{-\frac{i}{\sqrt{3}+3 i}}}}{5 \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[1 + x]*(1 - x + x^2) + (I*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[6 - (36*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])])/(5*Sqrt[1 - x + x^2])

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Maple [B]  time = 0.565, size = 252, normalized size = 1.8 \begin{align*} -{\frac{1}{5\,{x}^{3}+5}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( 3\,i\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 ) \sqrt{3}-9\,\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 ) -2\,{x}^{4}-2\,x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(3*I*(-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))*3^(1/2)-9*(-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))-2*x^4-2*x)/(
x^3+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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+x)^(1/2)*(x^2-x+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(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 (\sqrt{x^{2} - x + 1} \sqrt{x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(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 \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+x)^(1/2)*(x^2-x+1)^(1/2),x, algorithm="giac")

[Out]

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