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

Optimal. Leaf size=139 \[ \frac{2}{9} (e+2 f) \tanh ^{-1}\left (\frac{(x+1)^2}{3 \sqrt{x^3+1}}\right )+\frac{2 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (e-f) F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

[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 = 0.149903, antiderivative size = 139, 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} \[ \frac{2}{9} (e+2 f) \tanh ^{-1}\left (\frac{(x+1)^2}{3 \sqrt{x^3+1}}\right )+\frac{2 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} (e-f) F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

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}{3} (e-f) \int \frac{1}{\sqrt{1+x^3}} \, dx+\frac{1}{6} (e+2 f) \int \frac{2+2 x}{(2-x) \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 = 0.297003, size = 273, normalized size = 1.96 \[ \frac{2 \sqrt{\frac{2}{3}} \sqrt{-\frac{i (x+1)}{\sqrt{3}-3 i}} \left (2 \sqrt{3} \sqrt{2 i x+\sqrt{3}-i} \sqrt{x^2-x+1} (e+2 f) \Pi \left (\frac{2 \sqrt{3}}{3 i+\sqrt{3}};\sin ^{-1}\left (\frac{\sqrt{2 i x+\sqrt{3}-i}}{\sqrt{2} \sqrt [4]{3}}\right )|\frac{2 \sqrt{3}}{-3 i+\sqrt{3}}\right )-3 i f \sqrt{-2 i x+\sqrt{3}+i} \left (\left (\sqrt{3}-i\right ) x-\sqrt{3}-i\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{2 i x+\sqrt{3}-i}}{\sqrt{2} \sqrt [4]{3}}\right )|\frac{2 \sqrt{3}}{-3 i+\sqrt{3}}\right )\right )}{\left (\sqrt{3}+3 i\right ) \sqrt{2 i x+\sqrt{3}-i} \sqrt{x^3+1}} \]

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*I)*f*Sqrt[I + Sqrt[3] - (2*I)*x]*(-I - Sqrt[3] + (-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*Sqrt[3]*(e + 2*f)*Sqrt[-I + Sqrt[3] + (2*I)*x]*Sqrt[1 - x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + Sqrt[3]), Ar
cSin[Sqrt[-I + Sqrt[3] + (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(-3*I + Sqrt[3])]))/((3*I + Sqrt[3])*Sqrt[-I
+ Sqrt[3] + (2*I)*x]*Sqrt[1 + x^3])

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Maple [B]  time = 0.005, size = 246, normalized size = 1.8 \begin{align*} -2\,{\frac{f \left ( 3/2-i/2\sqrt{3} \right ) }{\sqrt{{x}^{3}+1}}\sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x-1/2-i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}}\sqrt{{\frac{x-1/2+i/2\sqrt{3}}{-3/2+i/2\sqrt{3}}}}{\it EllipticF} \left ( \sqrt{{\frac{1+x}{3/2-i/2\sqrt{3}}}},\sqrt{{\frac{-3/2+i/2\sqrt{3}}{-3/2-i/2\sqrt{3}}}} \right ) }+{\frac{ \left ( 2\,e+4\,f \right ) \left ({\frac{3}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }{3}\sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}}\sqrt{{\frac{1}{-{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }}\sqrt{{\frac{1}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) }}{\it EllipticPi} \left ( \sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}},{\frac{1}{2}}-{\frac{i}{6}}\sqrt{3},\sqrt{{\frac{-{\frac{3}{2}}+{\frac{i}{2}}\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*f*(3/2-1/2*I*3^(1/2))*((1+x)/(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(((1+x)/(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))+2/3*(e+2*f)*(3/2-1/2*I*3^(1/2))*((1+x)/(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(((1+x)/(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., 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}{x \sqrt{x^{3} + 1} - 2 \sqrt{x^{3} + 1}}\, dx - \int \frac{f x}{x \sqrt{x^{3} + 1} - 2 \sqrt{x^{3} + 1}}\, 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/(x*sqrt(x**3 + 1) - 2*sqrt(x**3 + 1)), x) - Integral(f*x/(x*sqrt(x**3 + 1) - 2*sqrt(x**3 + 1)), 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)