3.2272 \(\int \frac{1}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\log \left (\sqrt [3]{b} \sqrt{x}-\sqrt [3]{a+b x^{3/2}}\right )}{b^{2/3}}-\frac{2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}} \]

[Out]

(-2*ArcTan[(1 + (2*b^(1/3)*Sqrt[x])/(a + b*x^(3/2))^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) - Log[b^(1/3)*Sqrt[x] -
 (a + b*x^(3/2))^(1/3)]/b^(2/3)

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Rubi [A]  time = 0.0946295, antiderivative size = 140, normalized size of antiderivative = 1.71, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {243, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{2 \log \left (1-\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{3 b^{2/3}}+\frac{\log \left (\frac{b^{2/3} x}{\left (a+b x^{3/2}\right )^{2/3}}+\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1\right )}{3 b^{2/3}}-\frac{2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(3/2))^(-2/3),x]

[Out]

(-2*ArcTan[(1 + (2*b^(1/3)*Sqrt[x])/(a + b*x^(3/2))^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) - (2*Log[1 - (b^(1/3)*S
qrt[x])/(a + b*x^(3/2))^(1/3)])/(3*b^(2/3)) + Log[1 + (b^(2/3)*x)/(a + b*x^(3/2))^(2/3) + (b^(1/3)*Sqrt[x])/(a
 + b*x^(3/2))^(1/3)]/(3*b^(2/3))

Rule 243

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[n]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 292

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [C]  time = 0.0095632, size = 52, normalized size = 0.63 \[ \frac{x \left (\frac{b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};-\frac{b x^{3/2}}{a}\right )}{\left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(3/2))^(-2/3),x]

[Out]

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

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{x}^{{\frac{3}{2}}} \right ) ^{-{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [C]  time = 1.08668, size = 39, normalized size = 0.48 \begin{align*} \frac{2 x \Gamma \left (\frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle |{\frac{b x^{\frac{3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*gamma(2/3)*hyper((2/3, 2/3), (5/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(5/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^(3/2) + a)^(-2/3), x)