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

Optimal. Leaf size=189 \[ \frac{a^3 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{9 a^2 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{9 a b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{3 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

[Out]

(9*a*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1/3))/(a + b/x^(1/3)) + (9*a^2*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^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(
1/3)) + (3*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*Log[x^(1/3)])/(a + b/x^(1/3))

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Rubi [A]  time = 0.0906669, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1341, 1355, 263, 43} \[ \frac{a^3 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{9 a^2 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{9 a b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{3 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1/3))/(a + b/x^(1/3)) + (9*a^2*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^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(
1/3)) + (3*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*Log[x^(1/3)])/(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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0291729, size = 77, normalized size = 0.41 \[ \frac{\sqrt{\frac{\left (a \sqrt [3]{x}+b\right )^2}{x^{2/3}}} \left (9 a^2 b x+2 a^3 x^{4/3}+18 a b^2 x^{2/3}+2 b^3 \sqrt [3]{x} \log (x)\right )}{2 \left (a \sqrt [3]{x}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*x^(1/3))^2/x^(2/3)]*(18*a*b^2*x^(2/3) + 9*a^2*b*x + 2*a^3*x^(4/3) + 2*b^3*x^(1/3)*Log[x]))/(2*(b
+ a*x^(1/3)))

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Maple [A]  time = 0.005, size = 69, normalized size = 0.4 \begin{align*}{\frac{x}{2} \left ({ \left ({a}^{2}{x}^{{\frac{2}{3}}}+2\,ab\sqrt [3]{x}+{b}^{2} \right ){x}^{-{\frac{2}{3}}}} \right ) ^{{\frac{3}{2}}} \left ( 9\,{x}^{2/3}{a}^{2}b+18\,a{b}^{2}\sqrt [3]{x}+2\,{b}^{3}\ln \left ( x \right ) +2\,{a}^{3}x \right ) \left ( b+a\sqrt [3]{x} \right ) ^{-3}} \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))^(3/2),x)

[Out]

1/2*((a^2*x^(2/3)+2*a*b*x^(1/3)+b^2)/x^(2/3))^(3/2)*x*(9*x^(2/3)*a^2*b+18*a*b^2*x^(1/3)+2*b^3*ln(x)+2*a^3*x)/(
b+a*x^(1/3))^3

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Maxima [A]  time = 1.03323, size = 41, normalized size = 0.22 \begin{align*} a^{3} x + b^{3} \log \left (x\right ) + \frac{9}{2} \, a^{2} b x^{\frac{2}{3}} + 9 \, a b^{2} x^{\frac{1}{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))^(3/2),x, algorithm="maxima")

[Out]

a^3*x + b^3*log(x) + 9/2*a^2*b*x^(2/3) + 9*a*b^2*x^(1/3)

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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))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

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

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Giac [A]  time = 1.17087, size = 107, normalized size = 0.57 \begin{align*} a^{3} x \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + b^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + \frac{9}{2} \, a^{2} b x^{\frac{2}{3}} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + 9 \, a b^{2} x^{\frac{1}{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))^(3/2),x, algorithm="giac")

[Out]

a^3*x*sgn(a*x + b*x^(2/3))*sgn(x) + b^3*log(abs(x))*sgn(a*x + b*x^(2/3))*sgn(x) + 9/2*a^2*b*x^(2/3)*sgn(a*x +
b*x^(2/3))*sgn(x) + 9*a*b^2*x^(1/3)*sgn(a*x + b*x^(2/3))*sgn(x)