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

Optimal. Leaf size=148 \[ \frac {4}{9} \tan ^{-1}\left (\frac {(1-x)^2}{3 \sqrt {x^3-1}}\right )-\frac {2 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2139, 219, 2138, 203} \[ \frac {4}{9} \tan ^{-1}\left (\frac {(1-x)^2}{3 \sqrt {x^3-1}}\right )-\frac {2 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}} \]

Antiderivative was successfully verified.

[In]

Int[x/((2 + x)*Sqrt[-1 + x^3]),x]

[Out]

(4*ArcTan[(1 - x)^2/(3*Sqrt[-1 + x^3])])/9 - (2*Sqrt[2 - Sqrt[3]]*(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*3^(1/4)*Sqrt[-((1 - x)/(1 - Sqr
t[3] - x)^2)]*Sqrt[-1 + x^3])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 219

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 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 2138

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(-2*e)/d, Subst[Int
[1/(9 - a*x^2), x], x, (1 + (f*x)/e)^2/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 + 8*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rule 2139

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*d*e + c*f)/(3*c
*d), Int[1/Sqrt[a + b*x^3], x], x] + Dist[(d*e - c*f)/(3*c*d), Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x]
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && (EqQ[b*c^3 - 4*a*d^3, 0] || EqQ[b*c^3 + 8*a*d^3,
0]) && NeQ[2*d*e + c*f, 0]

Rubi steps

\begin {align*} \int \frac {x}{(2+x) \sqrt {-1+x^3}} \, dx &=\frac {1}{3} \int \frac {1}{\sqrt {-1+x^3}} \, dx-\frac {1}{3} \int \frac {2-2 x}{(2+x) \sqrt {-1+x^3}} \, dx\\ &=-\frac {2 \sqrt {2-\sqrt {3}} (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 )}{3 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}+\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{9+x^2} \, dx,x,\frac {(1-x)^2}{\sqrt {-1+x^3}}\right )\\ &=\frac {4}{9} \tan ^{-1}\left (\frac {(1-x)^2}{3 \sqrt {-1+x^3}}\right )-\frac {2 \sqrt {2-\sqrt {3}} (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 )}{3 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 193, normalized size = 1.30 \[ \frac {2 \sqrt {\frac {1-x}{1+\sqrt [3]{-1}}} \left (\frac {\left (x+\sqrt [3]{-1}\right ) \sqrt {\frac {(-1)^{2/3} x+\sqrt [3]{-1}}{1+\sqrt [3]{-1}}} F\left (\sin ^{-1}\left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}}+\frac {2 i \sqrt {x^2+x+1} \Pi \left (\frac {2 \sqrt {3}}{3 i+\sqrt {3}};\sin ^{-1}\left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right )|\sqrt [3]{-1}\right )}{\sqrt [3]{-1}-2}\right )}{\sqrt {x^3-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[(1 - x)/(1 + (-1)^(1/3))]*((((-1)^(1/3) + x)*Sqrt[((-1)^(1/3) + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]*Ellipt
icF[ArcSin[Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]
+ ((2*I)*Sqrt[1 + x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + Sqrt[3]), ArcSin[Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1
/3))]], (-1)^(1/3)])/(-2 + (-1)^(1/3))))/Sqrt[-1 + x^3]

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fricas [F]  time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{3} - 1} x}{x^{4} + 2 \, x^{3} - x - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 240, normalized size = 1.62 \[ \frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}-\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticPi \left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {1}{2}+\frac {i \sqrt {3}}{6}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.53, size = 208, normalized size = 1.41 \[ -\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (3\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-2\,\Pi \left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )}{3\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x^3 - 1)^(1/2)*(x + 2)),x)

[Out]

-((3^(1/2)*1i + 3)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^
(1/2)*1i)/2 + 3/2))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(3*ellipticF(asin((-(x - 1)/((3^(1/2)*1i)/2
+ 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)) - 2*ellipticPi((3^(1/2)*1i)/6 + 1/2, asin((-(x
- 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2))))/(3*(((3^(1/2)*1i)/2 -
1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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