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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 144, normalized size of antiderivative = 0.99, 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, 325, 218} \[ -\frac {x^3+1}{2 x^2 \sqrt {x+1} \sqrt {x^2-x+1}}-\frac {\sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {1}{x^3 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx &=\frac {\sqrt {1+x^3} \int \frac {1}{x^3 \sqrt {1+x^3}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}\\ &=-\frac {1+x^3}{2 x^2 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {\sqrt {1+x^3} \int \frac {1}{\sqrt {1+x^3}} \, dx}{4 \sqrt {1+x} \sqrt {1-x+x^2}}\\ &=-\frac {1+x^3}{2 x^2 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {\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 )}{2 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.65, size = 171, normalized size = 1.17 \[ \frac {-\frac {6 \sqrt {x+1} \left (x^2-x+1\right )}{x^2}-\frac {i (x+1) \sqrt {1+\frac {6 i}{\left (\sqrt {3}-3 i\right ) (x+1)}} \sqrt {6-\frac {36 i}{\left (\sqrt {3}+3 i\right ) (x+1)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {x+1}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{\sqrt {3}+3 i}}}}{12 \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*Sqrt[1 + x]*(1 - x + x^2))/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])])/(12*Sqrt[1 - x + x^2])

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x^{6} + x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^2 - x + 1)*sqrt(x + 1)/(x^6 + x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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maple [B]  time = 0.02, size = 259, normalized size = 1.77 \[ \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (-2 x^{3}+i \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \sqrt {3}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, x^{2} \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-2\right )}{4 \left (x^{3}+1\right ) x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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