3.2271 \(\int \frac{x^3}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{5 a^2 \log \left (\sqrt [3]{b} \sqrt{x}-\sqrt [3]{a+b x^{3/2}}\right )}{9 b^{8/3}}-\frac{10 a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{8/3}}-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b} \]

[Out]

(-5*a*x*(a + b*x^(3/2))^(1/3))/(9*b^2) + (x^(5/2)*(a + b*x^(3/2))^(1/3))/(3*b) - (10*a^2*ArcTan[(1 + (2*b^(1/3
)*Sqrt[x])/(a + b*x^(3/2))^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(8/3)) - (5*a^2*Log[b^(1/3)*Sqrt[x] - (a + b*x^(3/2))
^(1/3)])/(9*b^(8/3))

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Rubi [A]  time = 0.162139, antiderivative size = 198, normalized size of antiderivative = 1.42, number of steps used = 10, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {341, 321, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{10 a^2 \log \left (1-\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{27 b^{8/3}}+\frac{5 a^2 \log \left (\frac{b^{2/3} x}{\left (a+b x^{3/2}\right )^{2/3}}+\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1\right )}{27 b^{8/3}}-\frac{10 a^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{8/3}}-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^(3/2))^(2/3),x]

[Out]

(-5*a*x*(a + b*x^(3/2))^(1/3))/(9*b^2) + (x^(5/2)*(a + b*x^(3/2))^(1/3))/(3*b) - (10*a^2*ArcTan[(1 + (2*b^(1/3
)*Sqrt[x])/(a + b*x^(3/2))^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(8/3)) - (10*a^2*Log[1 - (b^(1/3)*Sqrt[x])/(a + b*x^(
3/2))^(1/3)])/(27*b^(8/3)) + (5*a^2*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)])/(27*b^(8/3))

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 321

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

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{x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^7}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt{x}\right )}{3 b}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}+\frac{\left (10 a^2\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt{x}\right )}{9 b^2}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}+\frac{\left (10 a^2\right ) \operatorname{Subst}\left (\int \frac{x}{1-b x^3} \, dx,x,\frac{\sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{9 b^2}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}+\frac{\left (10 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt [3]{b} x} \, dx,x,\frac{\sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{27 b^{7/3}}-\frac{\left (10 a^2\right ) \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 )}{27 b^{7/3}}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac{10 a^2 \log \left (1-\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{27 b^{8/3}}+\frac{\left (5 a^2\right ) \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 )}{27 b^{8/3}}-\frac{\left (5 a^2\right ) \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 )}{9 b^{7/3}}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac{10 a^2 \log \left (1-\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{27 b^{8/3}}+\frac{5 a^2 \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 )}{27 b^{8/3}}+\frac{\left (10 a^2\right ) \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 )}{9 b^{8/3}}\\ &=-\frac{5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac{x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac{10 a^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}}{\sqrt{3}}\right )}{9 \sqrt{3} b^{8/3}}-\frac{10 a^2 \log \left (1-\frac{\sqrt [3]{b} \sqrt{x}}{\sqrt [3]{a+b x^{3/2}}}\right )}{27 b^{8/3}}+\frac{5 a^2 \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 )}{27 b^{8/3}}\\ \end{align*}

Mathematica [C]  time = 0.0241633, size = 75, normalized size = 0.54 \[ \frac{x \left (5 a^2 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{b x^{3/2}}{b x^{3/2}+a}\right )-5 a^2-2 a b x^{3/2}+3 b^2 x^3\right )}{9 b^2 \left (a+b x^{3/2}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^(3/2))^(2/3),x]

[Out]

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

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \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(x^3/(a+b*x^(3/2))^(2/3),x)

[Out]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [C]  time = 2.90247, size = 41, normalized size = 0.29 \begin{align*} \frac{2 x^{4} \Gamma \left (\frac{8}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{8}{3} \\ \frac{11}{3} \end{matrix}\middle |{\frac{b x^{\frac{3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{11}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**4*gamma(8/3)*hyper((2/3, 8/3), (11/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(11/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\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(x^3/(a+b*x^(3/2))^(2/3),x, algorithm="giac")

[Out]

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