3.161 \(\int \frac{x}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac{5 b^{3/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{2 a^{9/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{5 \sqrt{a x+b \sqrt [3]{x}}}{a^2}-\frac{3 x}{a \sqrt{a x+b \sqrt [3]{x}}} \]

[Out]

(-3*x)/(a*Sqrt[b*x^(1/3) + a*x]) + (5*Sqrt[b*x^(1/3) + a*x])/a^2 - (5*b^(3/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt
[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(
2*a^(9/4)*Sqrt[b*x^(1/3) + a*x])

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Rubi [A]  time = 0.208698, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2018, 2022, 2024, 2011, 329, 220} \[ -\frac{5 b^{3/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{2 a^{9/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{5 \sqrt{a x+b \sqrt [3]{x}}}{a^2}-\frac{3 x}{a \sqrt{a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-3*x)/(a*Sqrt[b*x^(1/3) + a*x]) + (5*Sqrt[b*x^(1/3) + a*x])/a^2 - (5*b^(3/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt
[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(
2*a^(9/4)*Sqrt[b*x^(1/3) + a*x])

Rule 2018

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

Rule 2022

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2024

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

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{x}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^5}{\left (b x+a x^3\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{3 x}{a \sqrt{b \sqrt [3]{x}+a x}}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a}\\ &=-\frac{3 x}{a \sqrt{b \sqrt [3]{x}+a x}}+\frac{5 \sqrt{b \sqrt [3]{x}+a x}}{a^2}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a^2}\\ &=-\frac{3 x}{a \sqrt{b \sqrt [3]{x}+a x}}+\frac{5 \sqrt{b \sqrt [3]{x}+a x}}{a^2}-\frac{\left (5 b \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{2 a^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{3 x}{a \sqrt{b \sqrt [3]{x}+a x}}+\frac{5 \sqrt{b \sqrt [3]{x}+a x}}{a^2}-\frac{\left (5 b \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{a^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{3 x}{a \sqrt{b \sqrt [3]{x}+a x}}+\frac{5 \sqrt{b \sqrt [3]{x}+a x}}{a^2}-\frac{5 b^{3/4} \left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right ) \sqrt{\frac{b+a x^{2/3}}{\left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{2 a^{9/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.0632735, size = 82, normalized size = 0.55 \[ \frac{\sqrt{a x+b \sqrt [3]{x}} \left (-5 b \sqrt{\frac{a x^{2/3}}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{a x^{2/3}}{b}\right )+2 a x^{2/3}+5 b\right )}{a^2 \left (a x^{2/3}+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(Sqrt[b*x^(1/3) + a*x]*(5*b + 2*a*x^(2/3) - 5*b*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*
x^(2/3))/b)]))/(a^2*(b + a*x^(2/3)))

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Maple [A]  time = 0.01, size = 185, normalized size = 1.2 \begin{align*}{\frac{1}{2\,{a}^{3}} \left ( -5\,\sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{a\sqrt [3]{x}}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}\sqrt{\sqrt [3]{x} \left ( b+a{x}^{2/3} \right ) }b+6\,\sqrt{b\sqrt [3]{x}+ax}\sqrt [3]{x}ab+4\,\sqrt [3]{x}\sqrt{\sqrt [3]{x} \left ( b+a{x}^{2/3} \right ) }ab+4\,\sqrt{\sqrt [3]{x} \left ( b+a{x}^{2/3} \right ) }x{a}^{2} \right ){\frac{1}{\sqrt [3]{x}}} \left ( b+a{x}^{{\frac{2}{3}}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-5*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(
-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/
2)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*b+6*(b*x^(1/3)+a*x)^(1/2)*x^(1/3)*a*b+4*x^(1/3)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)
*a*b+4*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*x*a^2)/x^(1/3)/(b+a*x^(2/3))/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{4} x^{3} + 3 \, a^{2} b^{2} x^{\frac{5}{3}} - 2 \, a b^{3} x -{\left (2 \, a^{3} b x^{2} - b^{4}\right )} x^{\frac{1}{3}}\right )} \sqrt{a x + b x^{\frac{1}{3}}}}{a^{6} x^{4} + 2 \, a^{3} b^{3} x^{2} + b^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^3 + 3*a^2*b^2*x^(5/3) - 2*a*b^3*x - (2*a^3*b*x^2 - b^4)*x^(1/3))*sqrt(a*x + b*x^(1/3))/(a^6*x^
4 + 2*a^3*b^3*x^2 + b^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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