### 3.2589 $$\int \frac{1}{(b e-c e x)^{2/3} (b^2+b c x+c^2 x^2)^{2/3}} \, dx$$

Optimal. Leaf size=71 $\frac{x \left (1-\frac{c^3 x^3}{b^3}\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};\frac{c^3 x^3}{b^3}\right )}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}}$

[Out]

(x*(1 - (c^3*x^3)/b^3)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, (c^3*x^3)/b^3])/((b*e - c*e*x)^(2/3)*(b^2 + b*c*
x + c^2*x^2)^(2/3))

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Rubi [A]  time = 0.0391936, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {713, 246, 245} $\frac{x \left (1-\frac{c^3 x^3}{b^3}\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};\frac{c^3 x^3}{b^3}\right )}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*(1 - (c^3*x^3)/b^3)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, (c^3*x^3)/b^3])/((b*e - c*e*x)^(2/3)*(b^2 + b*c*
x + c^2*x^2)^(2/3))

Rule 713

Int[((d_.) + (e_.)*(x_))^(m_)*((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[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && IGtQ[m - p + 1, 0] &&  !Intege
rQ[p]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx &=\frac{\left (b^3 e-c^3 e x^3\right )^{2/3} \int \frac{1}{\left (b^3 e-c^3 e x^3\right )^{2/3}} \, dx}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}}\\ &=\frac{\left (1-\frac{c^3 x^3}{b^3}\right )^{2/3} \int \frac{1}{\left (1-\frac{c^3 x^3}{b^3}\right )^{2/3}} \, dx}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}}\\ &=\frac{x \left (1-\frac{c^3 x^3}{b^3}\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};\frac{c^3 x^3}{b^3}\right )}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}}\\ \end{align*}

Mathematica [B]  time = 0.217441, size = 258, normalized size = 3.63 $-\frac{3 \left (-\sqrt{3} \sqrt{-b^2}+b+2 c x\right ) \left (\frac{-\sqrt{3} \sqrt{-b^2} c x+3 b^2+\sqrt{3} \sqrt{-b^2} b+3 b c x}{\sqrt{3} \sqrt{-b^2} c x+3 b^2-\sqrt{3} \sqrt{-b^2} b+3 b c x}\right )^{2/3} \sqrt [3]{e (b-c x)} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};\frac{4 \sqrt{3} \sqrt{-b^2} (b-c x)}{\left (3 b+\sqrt{3} \sqrt{-b^2}\right ) \left (-b-2 c x+\sqrt{3} \sqrt{-b^2}\right )}\right )}{\left (3 b-\sqrt{3} \sqrt{-b^2}\right ) c e \left (b^2+b c x+c^2 x^2\right )^{2/3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*(e*(b - c*x))^(1/3)*(b - Sqrt[3]*Sqrt[-b^2] + 2*c*x)*((3*b^2 + Sqrt[3]*b*Sqrt[-b^2] + 3*b*c*x - Sqrt[3]*Sq
rt[-b^2]*c*x)/(3*b^2 - Sqrt[3]*b*Sqrt[-b^2] + 3*b*c*x + Sqrt[3]*Sqrt[-b^2]*c*x))^(2/3)*Hypergeometric2F1[1/3,
2/3, 4/3, (4*Sqrt[3]*Sqrt[-b^2]*(b - c*x))/((3*b + Sqrt[3]*Sqrt[-b^2])*(-b + Sqrt[3]*Sqrt[-b^2] - 2*c*x))])/((
3*b - Sqrt[3]*Sqrt[-b^2])*c*e*(b^2 + b*c*x + c^2*x^2)^(2/3))

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Maple [F]  time = 2.985, size = 0, normalized size = 0. \begin{align*} \int{ \left ( -cex+be \right ) ^{-{\frac{2}{3}}} \left ({c}^{2}{x}^{2}+bcx+{b}^{2} \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c^2*x^2 + b*c*x + b^2)^(2/3)*(-c*e*x + b*e)^(2/3)), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(c^2*x^2 + b*c*x + b^2)^(1/3)*(-c*e*x + b*e)^(1/3)/(c^3*e*x^3 - b^3*e), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (- e \left (- b + c x\right )\right )^{\frac{2}{3}} \left (b^{2} + b c x + c^{2} x^{2}\right )^{\frac{2}{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((-e*(-b + c*x))**(2/3)*(b**2 + b*c*x + c**2*x**2)**(2/3)), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c^2*x^2 + b*c*x + b^2)^(2/3)*(-c*e*x + b*e)^(2/3)), x)