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

Optimal. Leaf size=137 \[ \frac{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^5}{2 b^3}-\frac{6 a \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^4}{5 b^3}+\frac{3 a^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^3}{4 b^3} \]

[Out]

(3*a^2*(a + b*x^(1/3))^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(4*b^3) - (6*a*(a + b*x^(1/3))^4*Sqrt[a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3)])/(5*b^3) + ((a + b*x^(1/3))^5*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(2*b^3)

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Rubi [A]  time = 0.0558239, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1341, 645} \[ \frac{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^5}{2 b^3}-\frac{6 a \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^4}{5 b^3}+\frac{3 a^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^3}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*(a + b*x^(1/3))^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(4*b^3) - (6*a*(a + b*x^(1/3))^4*Sqrt[a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3)])/(5*b^3) + ((a + b*x^(1/3))^5*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(2*b^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 645

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*
e, 0] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0328785, size = 65, normalized size = 0.47 \[ \frac{x \sqrt{\left (a+b \sqrt [3]{x}\right )^2} \left (45 a^2 b \sqrt [3]{x}+20 a^3+36 a b^2 x^{2/3}+10 b^3 x\right )}{20 \left (a+b \sqrt [3]{x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^(1/3))^2]*x*(20*a^3 + 45*a^2*b*x^(1/3) + 36*a*b^2*x^(2/3) + 10*b^3*x))/(20*(a + b*x^(1/3)))

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Maple [A]  time = 0.003, size = 65, normalized size = 0.5 \begin{align*}{\frac{1}{20}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 36\,a{b}^{2}{x}^{5/3}+45\,{a}^{2}b{x}^{4/3}+10\,{b}^{3}{x}^{2}+20\,{a}^{3}x \right ) \left ( a+b\sqrt [3]{x} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(36*a*b^2*x^(5/3)+45*a^2*b*x^(4/3)+10*b^3*x^2+20*a^3*x)/(a+b*x^(1/3
))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95631, size = 82, normalized size = 0.6 \begin{align*} \frac{1}{2} \, b^{3} x^{2} + \frac{9}{5} \, a b^{2} x^{\frac{5}{3}} + \frac{9}{4} \, a^{2} b x^{\frac{4}{3}} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x^2 + 9/5*a*b^2*x^(5/3) + 9/4*a^2*b*x^(4/3) + a^3*x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + 2 a b \sqrt [3]{x} + 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+2*a*b*x**(1/3)+b**2*x**(2/3))**(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 = 1.1347, size = 86, normalized size = 0.63 \begin{align*} \frac{1}{2} \, b^{3} x^{2} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right ) + \frac{9}{5} \, a b^{2} x^{\frac{5}{3}} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right ) + \frac{9}{4} \, a^{2} b x^{\frac{4}{3}} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right ) + a^{3} x \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x^2*sgn(b*x^(1/3) + a) + 9/5*a*b^2*x^(5/3)*sgn(b*x^(1/3) + a) + 9/4*a^2*b*x^(4/3)*sgn(b*x^(1/3) + a) +
a^3*x*sgn(b*x^(1/3) + a)