3.2124 \(\int (a+b \sqrt{x})^2 \, dx\)

Optimal. Leaf size=27 \[ a^2 x+\frac{4}{3} a b x^{3/2}+\frac{b^2 x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0160736, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 43} \[ a^2 x+\frac{4}{3} a b x^{3/2}+\frac{b^2 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^2,x]

[Out]

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

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/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+b \sqrt{x}\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x (a+b x)^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x+2 a b x^2+b^2 x^3\right ) \, dx,x,\sqrt{x}\right )\\ &=a^2 x+\frac{4}{3} a b x^{3/2}+\frac{b^2 x^2}{2}\\ \end{align*}

Mathematica [A]  time = 0.0117133, size = 27, normalized size = 1. \[ a^2 x+\frac{4}{3} a b x^{3/2}+\frac{b^2 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^2,x]

[Out]

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

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Maple [A]  time = 0.001, size = 22, normalized size = 0.8 \begin{align*}{a}^{2}x+{\frac{4\,ab}{3}{x}^{{\frac{3}{2}}}}+{\frac{{b}^{2}{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.944884, size = 28, normalized size = 1.04 \begin{align*} \frac{1}{2} \, b^{2} x^{2} + \frac{4}{3} \, a b x^{\frac{3}{2}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2 + 4/3*a*b*x^(3/2) + a^2*x

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Fricas [A]  time = 1.44104, size = 53, normalized size = 1.96 \begin{align*} \frac{1}{2} \, b^{2} x^{2} + \frac{4}{3} \, a b x^{\frac{3}{2}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2 + 4/3*a*b*x^(3/2) + a^2*x

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Sympy [A]  time = 0.143745, size = 24, normalized size = 0.89 \begin{align*} a^{2} x + \frac{4 a b x^{\frac{3}{2}}}{3} + \frac{b^{2} x^{2}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x + 4*a*b*x**(3/2)/3 + b**2*x**2/2

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Giac [A]  time = 1.09529, size = 28, normalized size = 1.04 \begin{align*} \frac{1}{2} \, b^{2} x^{2} + \frac{4}{3} \, a b x^{\frac{3}{2}} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2 + 4/3*a*b*x^(3/2) + a^2*x