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

Optimal. Leaf size=153 \[ -\frac{2}{9} (e-2 f) \tanh ^{-1}\left (\frac{(1-x)^2}{3 \sqrt{1-x^3}}\right )-\frac{2 \sqrt{2+\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} (e+f) F\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

[Out]

(-2*(e - 2*f)*ArcTanh[(1 - x)^2/(3*Sqrt[1 - x^3])])/9 - (2*Sqrt[2 + Sqrt[3]]*(e + f)*(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 + Sqrt[3] - x)^2]*Sqrt[1 - x^3])

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Rubi [A]  time = 0.160596, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2139, 218, 2138, 206} \[ -\frac{2}{9} (e-2 f) \tanh ^{-1}\left (\frac{(1-x)^2}{3 \sqrt{1-x^3}}\right )-\frac{2 \sqrt{2+\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} (e+f) F\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e - 2*f)*ArcTanh[(1 - x)^2/(3*Sqrt[1 - x^3])])/9 - (2*Sqrt[2 + Sqrt[3]]*(e + f)*(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 + Sqrt[3] - x)^2]*Sqrt[1 - x^3])

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]

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 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 206

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

Rubi steps

\begin{align*} \int \frac{e+f x}{(2+x) \sqrt{1-x^3}} \, dx &=\frac{1}{6} (e-2 f) \int \frac{2-2 x}{(2+x) \sqrt{1-x^3}} \, dx+\frac{1}{3} (e+f) \int \frac{1}{\sqrt{1-x^3}} \, dx\\ &=-\frac{2 \sqrt{2+\sqrt{3}} (e+f) (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{1}{3} (2 (e-2 f)) \operatorname{Subst}\left (\int \frac{1}{9-x^2} \, dx,x,\frac{(1-x)^2}{\sqrt{1-x^3}}\right )\\ &=-\frac{2}{9} (e-2 f) \tanh ^{-1}\left (\frac{(1-x)^2}{3 \sqrt{1-x^3}}\right )-\frac{2 \sqrt{2+\sqrt{3}} (e+f) (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*}

Mathematica [C]  time = 0.271002, size = 271, normalized size = 1.77 \[ \frac{2 \sqrt{\frac{2}{3}} \sqrt{\frac{i (x-1)}{\sqrt{3}-3 i}} \left (3 f \sqrt{2 i x+\sqrt{3}+i} \left (i \sqrt{3} x+x+i \sqrt{3}-1\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{-2 i x+\sqrt{3}-i}}{\sqrt{2} \sqrt [4]{3}}\right )|\frac{2 \sqrt{3}}{-3 i+\sqrt{3}}\right )-2 \sqrt{3} \sqrt{-2 i x+\sqrt{3}-i} \sqrt{x^2+x+1} (e-2 f) \Pi \left (\frac{2 \sqrt{3}}{3 i+\sqrt{3}};\sin ^{-1}\left (\frac{\sqrt{-2 i x+\sqrt{3}-i}}{\sqrt{2} \sqrt [4]{3}}\right )|\frac{2 \sqrt{3}}{-3 i+\sqrt{3}}\right )\right )}{\left (\sqrt{3}+3 i\right ) \sqrt{-2 i x+\sqrt{3}-i} \sqrt{1-x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[2/3]*Sqrt[(I*(-1 + x))/(-3*I + Sqrt[3])]*(3*f*Sqrt[I + Sqrt[3] + (2*I)*x]*(-1 + I*Sqrt[3] + x + I*Sqrt
[3]*x)*EllipticF[ArcSin[Sqrt[-I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(-3*I + Sqrt[3])] - 2*Sqr
t[3]*(e - 2*f)*Sqrt[-I + Sqrt[3] - (2*I)*x]*Sqrt[1 + x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + Sqrt[3]), ArcSin[S
qrt[-I + Sqrt[3] - (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(-3*I + Sqrt[3])]))/((3*I + Sqrt[3])*Sqrt[-I + Sqr
t[3] - (2*I)*x]*Sqrt[1 - x^3])

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Maple [A]  time = 0.007, size = 246, normalized size = 1.6 \begin{align*}{-{\frac{2\,i}{3}}f\sqrt{3}\sqrt{i \left ( x+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}\sqrt{{\frac{x-1}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}}\sqrt{-i \left ( x+{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}{\it EllipticF} \left ({\frac{\sqrt{3}}{3}\sqrt{i \left ( x+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}},\sqrt{{\frac{i\sqrt{3}}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{-{x}^{3}+1}}}}-{\frac{{\frac{2\,i}{3}} \left ( e-2\,f \right ) \sqrt{3}}{{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}\sqrt{i \left ( x+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}\sqrt{{\frac{x-1}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}}\sqrt{-i \left ( x+{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}{\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{i \left ( x+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}},{\frac{i\sqrt{3}}{{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}},\sqrt{{\frac{i\sqrt{3}}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{-{x}^{3}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I*f*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-3/2+1/2*I*3^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3
^(1/2))*3^(1/2))^(1/2)/(-x^3+1)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),(I*3^(1/2)
/(-3/2+1/2*I*3^(1/2)))^(1/2))-2/3*I*(e-2*f)*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-3/2+1/2*I
*3^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(-x^3+1)^(1/2)/(3/2+1/2*I*3^(1/2))*EllipticPi(1/3*3^
(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),I*3^(1/2)/(3/2+1/2*I*3^(1/2)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(
1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f x + e}{\sqrt{-x^{3} + 1}{\left (x + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{3} + 1}{\left (f x + e\right )}}{x^{4} + 2 \, x^{3} - x - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e + f x}{\sqrt{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 2\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f x + e}{\sqrt{-x^{3} + 1}{\left (x + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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