3.33 \(\int \frac{1}{(b x+c x^2)^{5/3}} \, dx\)

Optimal. Leaf size=384 \[ \frac{\sqrt [3]{2} 3^{3/4} \sqrt{2-\sqrt{3}} b^2 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3} \left (1-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}\right ) \sqrt{\frac{2 \sqrt [3]{2} \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3}+2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}+1}{\left (-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}+\sqrt{3}+1}{-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{c (b+2 c x) \left (b x+c x^2\right )^{5/3} \sqrt{-\frac{1-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}}{\left (-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1\right )^2}}}+\frac{3 (b+2 c x) \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}}{2 c \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{5/3}} \]

[Out]

(3*(b + 2*c*x)*(-((c*(b*x + c*x^2))/b^2))^(5/3))/(2*c*(-((c*x*(b + c*x))/b^2))^(2/3)*(b*x + c*x^2)^(5/3)) + (2
^(1/3)*3^(3/4)*Sqrt[2 - Sqrt[3]]*b^2*(-((c*(b*x + c*x^2))/b^2))^(5/3)*(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)
^(5/3)*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)])

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Rubi [A]  time = 0.457139, antiderivative size = 384, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {622, 619, 199, 236, 219} \[ \frac{3 (b+2 c x) \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}}{2 c \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{5/3}}+\frac{\sqrt [3]{2} 3^{3/4} \sqrt{2-\sqrt{3}} b^2 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3} \left (1-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}\right ) \sqrt{\frac{2 \sqrt [3]{2} \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3}+2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}+1}{\left (-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}+\sqrt{3}+1}{-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{c (b+2 c x) \left (b x+c x^2\right )^{5/3} \sqrt{-\frac{1-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}}{\left (-2^{2/3} \sqrt [3]{-\frac{c x (b+c x)}{b^2}}-\sqrt{3}+1\right )^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-5/3),x]

[Out]

(3*(b + 2*c*x)*(-((c*(b*x + c*x^2))/b^2))^(5/3))/(2*c*(-((c*x*(b + c*x))/b^2))^(2/3)*(b*x + c*x^2)^(5/3)) + (2
^(1/3)*3^(3/4)*Sqrt[2 - Sqrt[3]]*b^2*(-((c*(b*x + c*x^2))/b^2))^(5/3)*(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)
^(5/3)*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 622

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

Rule 619

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 236

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 219

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 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int \frac{1}{\left (b x+c x^2\right )^{5/3}} \, dx &=\frac{\left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3} \int \frac{1}{\left (-\frac{c x}{b}-\frac{c^2 x^2}{b^2}\right )^{5/3}} \, dx}{\left (b x+c x^2\right )^{5/3}}\\ &=-\frac{\left (4 \sqrt [3]{2} b^2 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{b^2 x^2}{c^2}\right )^{5/3}} \, dx,x,-\frac{c}{b}-\frac{2 c^2 x}{b^2}\right )}{c^2 \left (b x+c x^2\right )^{5/3}}\\ &=\frac{3 (b+2 c x) \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}}{2 c \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{5/3}}-\frac{\left (\sqrt [3]{2} b^2 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}\right ) \operatorname{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 )^{5/3}}\\ &=\frac{3 (b+2 c x) \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}}{2 c \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{5/3}}+\frac{\left (3 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3} \sqrt{-1-\frac{4 c x}{b}-\frac{4 c^2 x^2}{b^2}}\right ) \operatorname{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 )^{5/3}}\\ &=\frac{3 (b+2 c x) \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{5/3}}{2 c \left (-\frac{c x (b+c x)}{b^2}\right )^{2/3} \left (b x+c x^2\right )^{5/3}}+\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 )^{5/3} \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 )^{5/3} \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*}

Mathematica [C]  time = 0.0120089, size = 47, normalized size = 0.12 \[ -\frac{3 \left (\frac{c x}{b}+1\right )^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{5}{3};\frac{1}{3};-\frac{c x}{b}\right )}{2 b (x (b+c x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-5/3),x]

[Out]

(-3*(1 + (c*x)/b)^(2/3)*Hypergeometric2F1[-2/3, 5/3, 1/3, -((c*x)/b)])/(2*b*(x*(b + c*x))^(2/3))

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Maple [F]  time = 0.613, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^(5/3),x)

[Out]

int(1/(c*x^2+b*x)^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(5/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(-5/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(5/3),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x)^(1/3)/(c^2*x^4 + 2*b*c*x^3 + b^2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b x + c x^{2}\right )^{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**(5/3),x)

[Out]

Integral((b*x + c*x**2)**(-5/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^(5/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(-5/3), x)