3.133 \(\int x \sqrt{b \sqrt [3]{x}+a x} \, dx\)

Optimal. Leaf size=213 \[ -\frac{6 b^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 a^{13/4} \sqrt{a x+b \sqrt [3]{x}}}-\frac{36 b^2 x^{2/3} \sqrt{a x+b \sqrt [3]{x}}}{385 a^2}+\frac{12 b^3 \sqrt{a x+b \sqrt [3]{x}}}{77 a^3}+\frac{4 b x^{4/3} \sqrt{a x+b \sqrt [3]{x}}}{55 a}+\frac{2}{5} x^2 \sqrt{a x+b \sqrt [3]{x}} \]

[Out]

(12*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^3) - (36*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a^2) + (4*b*x^(4/3)*Sqrt
[b*x^(1/3) + a*x])/(55*a) + (2*x^2*Sqrt[b*x^(1/3) + a*x])/5 - (6*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
^(13/4)*Sqrt[b*x^(1/3) + a*x])

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Rubi [A]  time = 0.299155, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2018, 2021, 2024, 2011, 329, 220} \[ -\frac{36 b^2 x^{2/3} \sqrt{a x+b \sqrt [3]{x}}}{385 a^2}-\frac{6 b^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 a^{13/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{12 b^3 \sqrt{a x+b \sqrt [3]{x}}}{77 a^3}+\frac{4 b x^{4/3} \sqrt{a x+b \sqrt [3]{x}}}{55 a}+\frac{2}{5} x^2 \sqrt{a x+b \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[b*x^(1/3) + a*x],x]

[Out]

(12*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^3) - (36*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a^2) + (4*b*x^(4/3)*Sqrt
[b*x^(1/3) + a*x])/(55*a) + (2*x^2*Sqrt[b*x^(1/3) + a*x])/5 - (6*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
^(13/4)*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 2021

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

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 x \sqrt{b \sqrt [3]{x}+a x} \, dx &=3 \operatorname{Subst}\left (\int x^5 \sqrt{b x+a x^3} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}+\frac{1}{5} (2 b) \operatorname{Subst}\left (\int \frac{x^6}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}-\frac{\left (18 b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{55 a}\\ &=-\frac{36 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^2}+\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}+\frac{\left (18 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^2}\\ &=\frac{12 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^3}-\frac{36 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^2}+\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}-\frac{\left (6 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^3}\\ &=\frac{12 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^3}-\frac{36 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^2}+\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}-\frac{\left (6 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^3 \sqrt{b \sqrt [3]{x}+a x}}\\ &=\frac{12 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^3}-\frac{36 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^2}+\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}-\frac{\left (12 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^3 \sqrt{b \sqrt [3]{x}+a x}}\\ &=\frac{12 b^3 \sqrt{b \sqrt [3]{x}+a x}}{77 a^3}-\frac{36 b^2 x^{2/3} \sqrt{b \sqrt [3]{x}+a x}}{385 a^2}+\frac{4 b x^{4/3} \sqrt{b \sqrt [3]{x}+a x}}{55 a}+\frac{2}{5} x^2 \sqrt{b \sqrt [3]{x}+a x}-\frac{6 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^{13/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.098597, size = 118, normalized size = 0.55 \[ \frac{2 \sqrt{a x+b \sqrt [3]{x}} \left (\sqrt{\frac{a x^{2/3}}{b}+1} \left (14 a^2 b x^{4/3}+77 a^3 x^2-18 a b^2 x^{2/3}+45 b^3\right )-45 b^3 \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{a x^{2/3}}{b}\right )\right )}{385 a^3 \sqrt{\frac{a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[b*x^(1/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(Sqrt[1 + (a*x^(2/3))/b]*(45*b^3 - 18*a*b^2*x^(2/3) + 14*a^2*b*x^(4/3) + 77*a^3*x^2)
- 45*b^3*Hypergeometric2F1[-1/2, 1/4, 5/4, -((a*x^(2/3))/b)]))/(385*a^3*Sqrt[1 + (a*x^(2/3))/b])

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Maple [A]  time = 0.013, size = 198, normalized size = 0.9 \begin{align*}{\frac{2\,{x}^{2}}{5}\sqrt{b\sqrt [3]{x}+ax}}+{\frac{4\,b}{55\,a}{x}^{{\frac{4}{3}}}\sqrt{b\sqrt [3]{x}+ax}}-{\frac{36\,{b}^{2}}{385\,{a}^{2}}{x}^{{\frac{2}{3}}}\sqrt{b\sqrt [3]{x}+ax}}+{\frac{12\,{b}^{3}}{77\,{a}^{3}}\sqrt{b\sqrt [3]{x}+ax}}-{\frac{6\,{b}^{4}}{77\,{a}^{4}}\sqrt{-ab}\sqrt{{a \left ( \sqrt [3]{x}+{\frac{1}{a}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{a}{\sqrt{-ab}} \left ( \sqrt [3]{x}-{\frac{\sqrt{-ab}}{a}} \right ) }}\sqrt{-{a\sqrt [3]{x}{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{a \left ( \sqrt [3]{x}+{\frac{1}{a}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{b\sqrt [3]{x}+ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^2*(b*x^(1/3)+a*x)^(1/2)+4/55*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)/a-36/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/2)/
a^2+12/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^3-6/77/a^4*b^4*(-a*b)^(1/2)*((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^
(1/2)*(-2*(x^(1/3)-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)/(b*x^(1/3)+a*x)^(1/
2)*EllipticF(((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3))*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*x + b*x^(1/3))*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(a*x + b*x**(1/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3))*x, x)