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

Optimal. Leaf size=253 \[ \frac {2 \sqrt {2} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}+\frac {2 \left (x^3+1\right )}{\sqrt {x+1} \left (x+\sqrt {3}+1\right ) \sqrt {x^2-x+1}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {809, 303, 218, 1877} \[ \frac {2 \left (x^3+1\right )}{\sqrt {x+1} \left (x+\sqrt {3}+1\right ) \sqrt {x^2-x+1}}+\frac {2 \sqrt {2} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 + x^3))/(Sqrt[1 + x]*(1 + Sqrt[3] + x)*Sqrt[1 - x + x^2]) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*Sqrt[1 + x]*Sqrt[
(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(Sq
rt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 - x + x^2]) + (2*Sqrt[2]*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 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 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 809

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((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[(f + g*x)*(a*d + c*e*x^
3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[m, p] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx &=\frac {\sqrt {1+x^3} \int \frac {x}{\sqrt {1+x^3}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {\sqrt {1+x^3} \int \frac {1-\sqrt {3}+x}{\sqrt {1+x^3}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}+\frac {\left (\sqrt {2 \left (2-\sqrt {3}\right )} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}}\\ &=\frac {2 \left (1+x^3\right )}{\sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}+\frac {2 \sqrt {2} \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*}

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Mathematica [C]  time = 1.00, size = 375, normalized size = 1.48 \[ \frac {(x+1)^{3/2} \left (\frac {12 \sqrt {-\frac {i}{\sqrt {3}+3 i}} \left (x^2-x+1\right )}{(x+1)^2}+\frac {i \sqrt {2} \left (\sqrt {3}+3 i\right ) \sqrt {\frac {-\frac {6 i}{x+1}+\sqrt {3}+3 i}{\sqrt {3}+3 i}} \sqrt {\frac {\frac {6 i}{x+1}+\sqrt {3}-3 i}{\sqrt {3}-3 i}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {x+1}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {x+1}}+\frac {3 \sqrt {2} \left (1-i \sqrt {3}\right ) \sqrt {\frac {-\frac {6 i}{x+1}+\sqrt {3}+3 i}{\sqrt {3}+3 i}} \sqrt {\frac {\frac {6 i}{x+1}+\sqrt {3}-3 i}{\sqrt {3}-3 i}} E\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {x+1}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {x+1}}\right )}{6 \sqrt {-\frac {i}{\sqrt {3}+3 i}} \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x)^(3/2)*((12*Sqrt[(-I)/(3*I + Sqrt[3])]*(1 - x + x^2))/(1 + x)^2 + (3*Sqrt[2]*(1 - I*Sqrt[3])*Sqrt[(3*I
+ Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3*I + Sqrt[3])]*EllipticE
[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x] + (I*Sqrt[
2]*(3*I + Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I)/(1 + x))
/(-3*I + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[
3])])/Sqrt[1 + x]))/(6*Sqrt[(-I)/(3*I + Sqrt[3])]*Sqrt[1 - x + x^2])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1} x}{x^{3} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 275, normalized size = 1.09 \[ \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (-3+i \sqrt {3}\right ) \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \left (i \sqrt {3}\, \EllipticE \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+3 \EllipticE \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-i \sqrt {3}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-\EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )\right )}{2 x^{3}+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/((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 \[ \int \frac {x}{\sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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