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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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 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 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]

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 = 0.41, size = 185, normalized size = 1.27 \[ \frac {\sqrt {x+1} \left (-\frac {2 \left (x^2-x+1\right )}{x^2}-\frac {3 i \sqrt {2} \sqrt {\frac {-2 i x+\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt {\frac {2 i x+\sqrt {3}-i}{\sqrt {3}-3 i}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {i (x+1)}{3 i+\sqrt {3}}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i (x+1)}{\sqrt {3}+3 i}}}\right )}{4 \sqrt {x^2-x+1}} \]

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])

________________________________________________________________________________________

fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x^{3}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x^{3}}\,{d x} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.03, size = 259, normalized size = 1.77 \[ -\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (2 x^{3}+3 i \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \sqrt {3}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-9 \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+2\right )}{4 \left (x^{3}+1\right ) x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x^{3}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {x+1}\,\sqrt {x^2-x+1}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x + 1} \sqrt {x^{2} - x + 1}}{x^{3}}\, dx \]

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)

________________________________________________________________________________________