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

Optimal. Leaf size=88 \[ \frac{3 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{a x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

[Out]

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

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Rubi [A]  time = 0.0525367, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1341, 1355, 14} \[ \frac{3 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{a x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \, dx &=3 \operatorname{Subst}\left (\int \sqrt{a^2+\frac{b^2}{x^2}+\frac{2 a b}{x}} x^2 \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (a b+\frac{b^2}{x}\right ) x^2 \, dx,x,\sqrt [3]{x}\right )}{a b+\frac{b^2}{\sqrt [3]{x}}}\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (b^2 x+a b x^2\right ) \, dx,x,\sqrt [3]{x}\right )}{a b+\frac{b^2}{\sqrt [3]{x}}}\\ &=\frac{3 b^2 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}+\frac{a \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} x}{a+\frac{b}{\sqrt [3]{x}}}\\ \end{align*}

Mathematica [A]  time = 0.0147499, size = 49, normalized size = 0.56 \[ \frac{\sqrt{\frac{\left (a \sqrt [3]{x}+b\right )^2}{x^{2/3}}} \left (2 a x^{4/3}+3 b x\right )}{2 \left (a \sqrt [3]{x}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 50, normalized size = 0.6 \begin{align*}{\frac{1}{2}\sqrt{{ \left ({a}^{2}{x}^{{\frac{2}{3}}}+2\,ab\sqrt [3]{x}+{b}^{2} \right ){x}^{-{\frac{2}{3}}}}}\sqrt [3]{x} \left ( 3\,{x}^{2/3}b+2\,ax \right ) \left ( b+a\sqrt [3]{x} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968008, size = 14, normalized size = 0.16 \begin{align*} a x + \frac{3}{2} \, b x^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + 3/2*b*x^(2/3)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15607, size = 46, normalized size = 0.52 \begin{align*} a x \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + \frac{3}{2} \, b x^{\frac{2}{3}} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x*sgn(a*x + b*x^(2/3))*sgn(x) + 3/2*b*x^(2/3)*sgn(a*x + b*x^(2/3))*sgn(x)