3.1.32 \(\int \frac {1}{(b x+c x^2)^{2/3}} \, dx\) [32]

Optimal. Leaf size=322 \[ \frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right )|-7+4 \sqrt {3}\right )}{c (b+2 c x) \left (b x+c x^2\right )^{2/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}} \]

[Out]

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

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Rubi [A]
time = 0.31, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {636, 633, 242, 225} \begin {gather*} \frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+1}{\left (-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1\right )^2}} F\left (\text {ArcSin}\left (\frac {-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+\sqrt {3}+1}{-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{c (b+2 c x) \left (b x+c x^2\right )^{2/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1\right )^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-2/3),x]

[Out]

(2^(1/3)*3^(3/4)*Sqrt[2 - Sqrt[3]]*b^2*(-((c*(b*x + c*x^2))/b^2))^(2/3)*(1 - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^
(1/3))*Sqrt[(1 + 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3) + 2*2^(1/3)*(-((c*x*(b + c*x))/b^2))^(2/3))/(1 - Sqrt[
3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2
))^(1/3))/(1 - Sqrt[3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))], -7 + 4*Sqrt[3]])/(c*(b + 2*c*x)*(b*x + c*x^
2)^(2/3)*Sqrt[-((1 - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))/(1 - Sqrt[3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(
1/3))^2)])

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[(-s)*((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 242

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[3*(Sqrt[b*x^2]/(2*b*x)), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 636

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(b*x + c*x^2)^p/((-c)*((b*x + c*x^2)/b^2))^p, Int[((-c
)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; FreeQ[{b, c}, x] && RationalQ[p] && 3 <= Denominator[p] <= 4

Rubi steps

\begin {align*} \int \frac {1}{\left (b x+c x^2\right )^{2/3}} \, dx &=\frac {\left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \int \frac {1}{\left (-\frac {c x}{b}-\frac {c^2 x^2}{b^2}\right )^{2/3}} \, dx}{\left (b x+c x^2\right )^{2/3}}\\ &=-\frac {\left (\sqrt [3]{2} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b^2 x^2}{c^2}\right )^{2/3}} \, dx,x,-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right )}{c^2 \left (b x+c x^2\right )^{2/3}}\\ &=\frac {\left (3 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \sqrt {-1-\frac {4 c x}{b}-\frac {4 c^2 x^2}{b^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,2^{2/3} \sqrt [3]{-\frac {c x \left (1+\frac {c x}{b}\right )}{b}}\right )}{2^{2/3} \left (-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right ) \left (b x+c x^2\right )^{2/3}}\\ &=\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b^2 \left (-\frac {c \left (b x+c x^2\right )}{b^2}\right )^{2/3} \sqrt {-1-\frac {4 c x}{b}-\frac {4 c^2 x^2}{b^2}} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right )|-7+4 \sqrt {3}\right )}{c (b+2 c x) \left (b x+c x^2\right )^{2/3} \sqrt {-1-\frac {4 c x (b+c x)}{b^2}} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 9.28, size = 43, normalized size = 0.13 \begin {gather*} \frac {3 x \left (1+\frac {c x}{b}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {c x}{b}\right )}{(x (b+c x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-2/3),x]

[Out]

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

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c \,x^{2}+b x \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(-2/3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x)^(-2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b x + c x^{2}\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(-2/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(-2/3), x)

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Mupad [B]
time = 0.22, size = 36, normalized size = 0.11 \begin {gather*} \frac {3\,x\,{\left (\frac {c\,x}{b}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {2}{3};\ \frac {4}{3};\ -\frac {c\,x}{b}\right )}{{\left (c\,x^2+b\,x\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(2/3),x)

[Out]

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

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