3.2122 \(\int (a+b \sqrt{x})^2 x^2 \, dx\)

Optimal. Leaf size=32 \[ \frac{a^2 x^3}{3}+\frac{4}{7} a b x^{7/2}+\frac{b^2 x^4}{4} \]

[Out]

(a^2*x^3)/3 + (4*a*b*x^(7/2))/7 + (b^2*x^4)/4

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Rubi [A]  time = 0.0214081, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{a^2 x^3}{3}+\frac{4}{7} a b x^{7/2}+\frac{b^2 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^2*x^2,x]

[Out]

(a^2*x^3)/3 + (4*a*b*x^(7/2))/7 + (b^2*x^4)/4

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^2 \, dx &=2 \operatorname{Subst}\left (\int x^5 (a+b x)^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^5+2 a b x^6+b^2 x^7\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}+\frac{4}{7} a b x^{7/2}+\frac{b^2 x^4}{4}\\ \end{align*}

Mathematica [A]  time = 0.0157046, size = 28, normalized size = 0.88 \[ \frac{1}{84} x^3 \left (28 a^2+48 a b \sqrt{x}+21 b^2 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^2*x^2,x]

[Out]

(x^3*(28*a^2 + 48*a*b*Sqrt[x] + 21*b^2*x))/84

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Maple [A]  time = 0., size = 25, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}{a}^{2}}{3}}+{\frac{4\,ab}{7}{x}^{{\frac{7}{2}}}}+{\frac{{b}^{2}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*a^2+4/7*a*b*x^(7/2)+1/4*b^2*x^4

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Maxima [B]  time = 0.965583, size = 132, normalized size = 4.12 \begin{align*} \frac{{\left (b \sqrt{x} + a\right )}^{8}}{4 \, b^{6}} - \frac{10 \,{\left (b \sqrt{x} + a\right )}^{7} a}{7 \, b^{6}} + \frac{10 \,{\left (b \sqrt{x} + a\right )}^{6} a^{2}}{3 \, b^{6}} - \frac{4 \,{\left (b \sqrt{x} + a\right )}^{5} a^{3}}{b^{6}} + \frac{5 \,{\left (b \sqrt{x} + a\right )}^{4} a^{4}}{2 \, b^{6}} - \frac{2 \,{\left (b \sqrt{x} + a\right )}^{3} a^{5}}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*sqrt(x) + a)^8/b^6 - 10/7*(b*sqrt(x) + a)^7*a/b^6 + 10/3*(b*sqrt(x) + a)^6*a^2/b^6 - 4*(b*sqrt(x) + a)^
5*a^3/b^6 + 5/2*(b*sqrt(x) + a)^4*a^4/b^6 - 2/3*(b*sqrt(x) + a)^3*a^5/b^6

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Fricas [A]  time = 1.46816, size = 61, normalized size = 1.91 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + \frac{4}{7} \, a b x^{\frac{7}{2}} + \frac{1}{3} \, a^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 4/7*a*b*x^(7/2) + 1/3*a^2*x^3

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Sympy [A]  time = 0.411889, size = 27, normalized size = 0.84 \begin{align*} \frac{a^{2} x^{3}}{3} + \frac{4 a b x^{\frac{7}{2}}}{7} + \frac{b^{2} x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**3/3 + 4*a*b*x**(7/2)/7 + b**2*x**4/4

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Giac [A]  time = 1.12686, size = 32, normalized size = 1. \begin{align*} \frac{1}{4} \, b^{2} x^{4} + \frac{4}{7} \, a b x^{\frac{7}{2}} + \frac{1}{3} \, a^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 4/7*a*b*x^(7/2) + 1/3*a^2*x^3