3.2282 \(\int \frac {x}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=42 \[ \frac {x^2 \sqrt [3]{a+b x^{3/2}} \, _2F_1\left (1,\frac {5}{3};\frac {7}{3};-\frac {b x^{3/2}}{a}\right )}{2 a} \]

[Out]

1/2*x^2*(a+b*x^(3/2))^(1/3)*hypergeom([1, 5/3],[7/3],-b*x^(3/2)/a)/a

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Rubi [A]  time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.36, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {341, 365, 364} \[ \frac {x^2 \left (\frac {b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^(3/2))^(2/3),x]

[Out]

(x^2*(1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^(3/2))/a)])/(2*(a + b*x^(3/2))^(2/3))

Rule 341

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

Rule 364

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b x^{3/2}\right )^{2/3}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right )\\ &=\frac {\left (2 \left (1+\frac {b x^{3/2}}{a}\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx,x,\sqrt {x}\right )}{\left (a+b x^{3/2}\right )^{2/3}}\\ &=\frac {x^2 \left (1+\frac {b x^{3/2}}{a}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 57, normalized size = 1.36 \[ \frac {x^2 \left (\frac {b x^{3/2}}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {7}{3};-\frac {b x^{3/2}}{a}\right )}{2 \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^(3/2))^(2/3),x]

[Out]

(x^2*(1 + (b*x^(3/2))/a)^(2/3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^(3/2))/a)])/(2*(a + b*x^(3/2))^(2/3))

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fricas [F]  time = 3.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{\frac {5}{2}} - a x\right )} {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{b^{2} x^{3} - a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^(5/2) - a*x)*(b*x^(3/2) + a)^(1/3)/(b^2*x^3 - a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^(3/2) + a)^(2/3), x)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (b \,x^{\frac {3}{2}}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^(3/2) + a)^(2/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{{\left (a+b\,x^{3/2}\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^(3/2))^(2/3),x)

[Out]

int(x/(a + b*x^(3/2))^(2/3), x)

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sympy [C]  time = 0.88, size = 41, normalized size = 0.98 \[ \frac {2 x^{2} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{\frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**2*gamma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3))

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