3.2123 \(\int (a+b \sqrt{x})^2 x \, dx\)

Optimal. Leaf size=32 \[ \frac{a^2 x^2}{2}+\frac{4}{5} a b x^{5/2}+\frac{b^2 x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.0178599, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{a^2 x^2}{2}+\frac{4}{5} a b x^{5/2}+\frac{b^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^2*x,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 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 x \, dx &=2 \operatorname{Subst}\left (\int x^3 (a+b x)^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^3+2 a b x^4+b^2 x^5\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^2}{2}+\frac{4}{5} a b x^{5/2}+\frac{b^2 x^3}{3}\\ \end{align*}

Mathematica [A]  time = 0.0133287, size = 28, normalized size = 0.88 \[ \frac{1}{30} x^2 \left (15 a^2+24 a b \sqrt{x}+10 b^2 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^2*x,x]

[Out]

(x^2*(15*a^2 + 24*a*b*Sqrt[x] + 10*b^2*x))/30

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Maple [A]  time = 0.001, size = 25, normalized size = 0.8 \begin{align*}{\frac{{a}^{2}{x}^{2}}{2}}+{\frac{4\,ab}{5}{x}^{{\frac{5}{2}}}}+{\frac{{b}^{2}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.97089, size = 86, normalized size = 2.69 \begin{align*} \frac{{\left (b \sqrt{x} + a\right )}^{6}}{3 \, b^{4}} - \frac{6 \,{\left (b \sqrt{x} + a\right )}^{5} a}{5 \, b^{4}} + \frac{3 \,{\left (b \sqrt{x} + a\right )}^{4} a^{2}}{2 \, b^{4}} - \frac{2 \,{\left (b \sqrt{x} + a\right )}^{3} a^{3}}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46985, size = 61, normalized size = 1.91 \begin{align*} \frac{1}{3} \, b^{2} x^{3} + \frac{4}{5} \, a b x^{\frac{5}{2}} + \frac{1}{2} \, a^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.225381, size = 27, normalized size = 0.84 \begin{align*} \frac{a^{2} x^{2}}{2} + \frac{4 a b x^{\frac{5}{2}}}{5} + \frac{b^{2} x^{3}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10795, size = 32, normalized size = 1. \begin{align*} \frac{1}{3} \, b^{2} x^{3} + \frac{4}{5} \, a b x^{\frac{5}{2}} + \frac{1}{2} \, a^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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