\(\int (a+b \sqrt {x})^2 x \, dx\) [2123]

Optimal result

Integrand size = 13, antiderivative size = 32 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {a^2 x^2}{2}+\frac {4}{5} a b x^{5/2}+\frac {b^2 x^3}{3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {a^2 x^2}{2}+\frac {4}{5} a b x^{5/2}+\frac {b^2 x^3}{3} \]

[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 45

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])

Rule 272

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]]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^3 (a+b x)^2 \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {1}{30} x^2 \left (15 a^2+24 a b \sqrt {x}+10 b^2 x\right ) \]

[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

Maple [A] (verified)

Time = 5.80 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {a^{2} x^{2}}{2}+\frac {4 a b \,x^{\frac {5}{2}}}{5}+\frac {b^{2} x^{3}}{3}\)	\(25\)
default	\(\frac {a^{2} x^{2}}{2}+\frac {4 a b \,x^{\frac {5}{2}}}{5}+\frac {b^{2} x^{3}}{3}\)	\(25\)
trager	\(\frac {\left (2 b^{2} x^{2}+3 a^{2} x +2 b^{2} x +3 a^{2}+2 b^{2}\right ) \left (-1+x \right )}{6}+\frac {4 a b \,x^{\frac {5}{2}}}{5}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {1}{3} \, b^{2} x^{3} + \frac {4}{5} \, a b x^{\frac {5}{2}} + \frac {1}{2} \, a^{2} x^{2} \]

[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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {a^{2} x^{2}}{2} + \frac {4 a b x^{\frac {5}{2}}}{5} + \frac {b^{2} x^{3}}{3} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (24) = 48\).

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {{\left (b \sqrt {x} + a\right )}^{6}}{3 \, b^{4}} - \frac {6 \, {\left (b \sqrt {x} + a\right )}^{5} a}{5 \, b^{4}} + \frac {3 \, {\left (b \sqrt {x} + a\right )}^{4} a^{2}}{2 \, b^{4}} - \frac {2 \, {\left (b \sqrt {x} + a\right )}^{3} a^{3}}{3 \, b^{4}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {1}{3} \, b^{2} x^{3} + \frac {4}{5} \, a b x^{\frac {5}{2}} + \frac {1}{2} \, a^{2} x^{2} \]

[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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \left (a+b \sqrt {x}\right )^2 x \, dx=\frac {a^2\,x^2}{2}+\frac {b^2\,x^3}{3}+\frac {4\,a\,b\,x^{5/2}}{5} \]

[In]

int(x*(a + b*x^(1/2))^2,x)

[Out]

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