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

Optimal. Leaf size=134 \[ -\frac{2}{3} e \tanh ^{-1}\left (\sqrt{1-x^3}\right )-\frac{2 \sqrt{2+\sqrt{3}} f (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

[Out]

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

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Rubi [A]  time = 0.0588458, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1832, 266, 63, 206, 12, 218} \[ -\frac{2}{3} e \tanh ^{-1}\left (\sqrt{1-x^3}\right )-\frac{2 \sqrt{2+\sqrt{3}} f (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)/(x*Sqrt[1 - x^3]),x]

[Out]

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

Rule 1832

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

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

Mathematica [C]  time = 0.0186727, size = 34, normalized size = 0.25 \[ f x \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};x^3\right )-\frac{2}{3} e \tanh ^{-1}\left (\sqrt{1-x^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*ArcTanh[Sqrt[1 - x^3]])/3 + f*x*Hypergeometric2F1[1/3, 1/2, 4/3, x^3]

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Maple [A]  time = 0.004, size = 122, normalized size = 0.9 \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{2\,e}{3}{\it Artanh} \left ( \sqrt{-{x}^{3}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)/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*e*arctanh((-x^3+1)^(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} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)/(sqrt(-x^3 + 1)*x), 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} - x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.61708, size = 65, normalized size = 0.49 \begin{align*} e \left (\begin{cases} - \frac{2 \operatorname{acosh}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{for}\: \frac{1}{\left |{x^{3}}\right |} > 1 \\\frac{2 i \operatorname{asin}{\left (\frac{1}{x^{\frac{3}{2}}} \right )}}{3} & \text{otherwise} \end{cases}\right ) + \frac{f x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{1}{2} \\ \frac{4}{3} \end{matrix}\middle |{x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((-2*acosh(x**(-3/2))/3, 1/Abs(x**3) > 1), (2*I*asin(x**(-3/2))/3, True)) + f*x*gamma(1/3)*hyper((1
/3, 1/2), (4/3,), x**3*exp_polar(2*I*pi))/(3*gamma(4/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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