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

Optimal. Leaf size=216 \[ \frac{39 b^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 a^{17/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{234 b^2 x^{2/3} \sqrt{a x+b \sqrt [3]{x}}}{385 a^3}-\frac{78 b^3 \sqrt{a x+b \sqrt [3]{x}}}{77 a^4}-\frac{26 b x^{4/3} \sqrt{a x+b \sqrt [3]{x}}}{55 a^2}+\frac{2 x^2 \sqrt{a x+b \sqrt [3]{x}}}{5 a} \]

[Out]

(-78*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^4) + (234*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a^3) - (26*b*x^(4/3)*S
qrt[b*x^(1/3) + a*x])/(55*a^2) + (2*x^2*Sqrt[b*x^(1/3) + a*x])/(5*a) + (39*b^(15/4)*(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])/(77*a^(17/4)*Sqrt[b*x^(1/3) + a*x])

________________________________________________________________________________________

Rubi [A]  time = 0.318692, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2018, 2024, 2011, 329, 220} \[ \frac{234 b^2 x^{2/3} \sqrt{a x+b \sqrt [3]{x}}}{385 a^3}+\frac{39 b^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 a^{17/4} \sqrt{a x+b \sqrt [3]{x}}}-\frac{78 b^3 \sqrt{a x+b \sqrt [3]{x}}}{77 a^4}-\frac{26 b x^{4/3} \sqrt{a x+b \sqrt [3]{x}}}{55 a^2}+\frac{2 x^2 \sqrt{a x+b \sqrt [3]{x}}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[b*x^(1/3) + a*x],x]

[Out]

(-78*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^4) + (234*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a^3) - (26*b*x^(4/3)*S
qrt[b*x^(1/3) + a*x])/(55*a^2) + (2*x^2*Sqrt[b*x^(1/3) + a*x])/(5*a) + (39*b^(15/4)*(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])/(77*a^(17/4)*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 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^2}{\sqrt{b \sqrt [3]{x}+a x}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^8}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}-\frac{(13 b) \operatorname{Subst}\left (\int \frac{x^6}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{5 a}\\ &=-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}+\frac{\left (117 b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{55 a^2}\\ &=\frac{234 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^3}-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}-\frac{\left (117 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^3}\\ &=-\frac{78 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^4}+\frac{234 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^3}-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}+\frac{\left (39 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^4}\\ &=-\frac{78 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^4}+\frac{234 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^3}-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}+\frac{\left (39 b^4 \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 )}{77 a^4 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{78 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^4}+\frac{234 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^3}-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}+\frac{\left (78 b^4 \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 )}{77 a^4 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{78 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^4}+\frac{234 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^3}-\frac{26 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a^2}+\frac{2 x^2 \sqrt{b \sqrt [3]{x}+a x}}{5 a}+\frac{39 b^{15/4} \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 )}{77 a^{17/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.066839, size = 124, normalized size = 0.57 \[ \frac{2 \sqrt{a x+b \sqrt [3]{x}} \left (26 a^2 b^2 x^{4/3}-14 a^3 b x^2+77 a^4 x^{8/3}+195 b^4 \sqrt{\frac{a x^{2/3}}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{a x^{2/3}}{b}\right )-78 a b^3 x^{2/3}-195 b^4\right )}{385 a^4 \left (a x^{2/3}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(-195*b^4 - 78*a*b^3*x^(2/3) + 26*a^2*b^2*x^(4/3) - 14*a^3*b*x^2 + 77*a^4*x^(8/3) + 1
95*b^4*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^(2/3))/b)]))/(385*a^4*(b + a*x^(2/3)))

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 164, normalized size = 0.8 \begin{align*}{\frac{1}{385\,{a}^{5}} \left ( 195\,{b}^{4}\sqrt{-ab}\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 ) +52\,{x}^{5/3}{a}^{3}{b}^{2}-28\,{x}^{7/3}{a}^{4}b-156\,x{a}^{2}{b}^{3}+154\,{x}^{3}{a}^{5}-390\,\sqrt [3]{x}a{b}^{4} \right ){\frac{1}{\sqrt{\sqrt [3]{x} \left ( b+a{x}^{{\frac{2}{3}}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/385*(195*b^4*(-a*b)^(1/2)*((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))+52*x^(5/3)*a^3*b^2-28*x^(7/3)*a^4*b-156*x*a^2*b^3+154*x^3*a^5-390*x^(1/3)*a*b^4)/(x^(1/3)*(b+a*x^(2/
3)))^(1/2)/a^5

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} x^{3} - a b x^{\frac{7}{3}} + b^{2} x^{\frac{5}{3}}\right )} \sqrt{a x + b x^{\frac{1}{3}}}}{a^{3} x^{2} + b^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*x^3 - a*b*x^(7/3) + b^2*x^(5/3))*sqrt(a*x + b*x^(1/3))/(a^3*x^2 + b^3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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