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

Optimal. Leaf size=88 \[ \frac{3 b x^{4/3} \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 \left (a+b \sqrt [3]{x}\right )}+\frac{a x \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{a+b \sqrt [3]{x}} \]

[Out]

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

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Rubi [A]  time = 0.0398336, 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, 646, 43} \[ \frac{3 b x^{4/3} \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 \left (a+b \sqrt [3]{x}\right )}+\frac{a x \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{a+b \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]*x)/(a + b*x^(1/3)) + (3*b*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]*x
^(4/3))/(4*(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 646

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[(d + e*x)^m*(b/2 + c*x)^(2*p), 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]

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

Mathematica [A]  time = 0.0091801, size = 43, normalized size = 0.49 \[ \frac{\sqrt{\left (a+b \sqrt [3]{x}\right )^2} \left (4 a x+3 b x^{4/3}\right )}{4 \left (a+b \sqrt [3]{x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.001, size = 43, normalized size = 0.5 \begin{align*}{\frac{1}{4}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 3\,b{x}^{4/3}+4\,ax \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))^(1/2),x)

[Out]

1/4*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(3*b*x^(4/3)+4*a*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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01913, size = 28, normalized size = 0.32 \begin{align*} \frac{3}{4} \, b x^{\frac{4}{3}} + a 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))^(1/2),x, algorithm="fricas")

[Out]

3/4*b*x^(4/3) + a*x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac{2}{3}}}\, 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))**(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 = 1.11565, size = 35, normalized size = 0.4 \begin{align*} \frac{3}{4} \, b x^{\frac{4}{3}} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right ) + a 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))^(1/2),x, algorithm="giac")

[Out]

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