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

Optimal. Leaf size=146 \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} (x+1)^{3/2} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{2 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )}-\frac{\sqrt{x+1} \sqrt{x^2-x+1}}{2 x^2} \]

[Out]

-(Sqrt[1 + x]*Sqrt[1 - x + x^2])/(2*x^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]])/(2*Sqr
t[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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Rubi [A]  time = 0.0443619, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {915, 277, 218} \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} (x+1)^{3/2} \sqrt{x^2-x+1} \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{2 \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \left (x^3+1\right )}-\frac{\sqrt{x+1} \sqrt{x^2-x+1}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + x]*Sqrt[1 - x + x^2])/(2*x^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]])/(2*Sqr
t[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

Rule 915

Int[((g_.)*(x_))^(n_)*((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[(g*x)^n*(a*d + c*e*x^3)^p,
x], x] /; FreeQ[{a, b, c, d, e, g, m, n, p}, x] && EqQ[m - p, 0] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.410136, size = 185, normalized size = 1.27 \[ \frac{\sqrt{x+1} \left (-\frac{2 \left (x^2-x+1\right )}{x^2}-\frac{3 i \sqrt{2} \sqrt{\frac{-2 i x+\sqrt{3}+i}{\sqrt{3}+3 i}} \sqrt{\frac{2 i x+\sqrt{3}-i}{\sqrt{3}-3 i}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{-\frac{i (x+1)}{\sqrt{3}+3 i}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )}{\sqrt{-\frac{i (x+1)}{\sqrt{3}+3 i}}}\right )}{4 \sqrt{x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.089, size = 259, normalized size = 1.8 \begin{align*} -{\frac{1}{ \left ( 4\,{x}^{3}+4 \right ){x}^{2}}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( 3\,i\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{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{3}{x}^{2}-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 ){x}^{2}+2\,{x}^{3}+2 \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^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x^{3}}\,{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^3,x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 - x + 1)*sqrt(x + 1)/x^3, 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}}, 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^3,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - x + 1} \sqrt{x + 1}}{x^{3}}\,{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^3,x, algorithm="giac")

[Out]

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