3.5.85 \(\int \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \, dx\) [485]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 = {1355, 1369, 14} \begin {gather*} \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}}} \end {gather*}

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 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]]

Rule 1355

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 1369

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]

Rubi steps

\begin {align*} \int \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \, dx &=3 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.01, size = 49, normalized size = 0.56 \begin {gather*} \frac {\left (3 b+2 a \sqrt [3]{x}\right ) \sqrt {\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}} x}{2 \left (b+a \sqrt [3]{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 50, normalized size = 0.57

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {a^{2} x^{\frac {2}{3}}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}}\, x \left (2 a \,x^{\frac {1}{3}}+3 b \right )}{2 b +2 a \,x^{\frac {1}{3}}}\)	\(47\)
default	\(\frac {\sqrt {\frac {a^{2} x^{\frac {2}{3}}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}}\, x^{\frac {1}{3}} \left (3 b \,x^{\frac {2}{3}}+2 a x \right )}{2 b +2 a \,x^{\frac {1}{3}}}\)	\(50\)

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,method=_RETURNVERBOSE)

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

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Maxima [A]
time = 0.29, size = 10, normalized size = 0.11 \begin {gather*} a x + \frac {3}{2} \, b x^{\frac {2}{3}} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {b^{2}}{x^{\frac {2}{3}}}}\, dx \end {gather*}

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 = 3.18, size = 34, normalized size = 0.39 \begin {gather*} 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 {gather*}

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)

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Mupad [B]
time = 1.43, size = 39, normalized size = 0.44 \begin {gather*} \frac {x\,\left (a+\frac {3\,b}{2\,x^{1/3}}\right )\,\sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2\,a\,b}{x^{1/3}}}}{a+\frac {b}{x^{1/3}}} \end {gather*}

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]

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

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