3.2134 \(\int (a+b \sqrt{x})^3 \, dx\)

Optimal. Leaf size=38 \[ \frac{2 \left (a+b \sqrt{x}\right )^5}{5 b^2}-\frac{a \left (a+b \sqrt{x}\right )^4}{2 b^2} \]

[Out]

-(a*(a + b*Sqrt[x])^4)/(2*b^2) + (2*(a + b*Sqrt[x])^5)/(5*b^2)

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Rubi [A]  time = 0.01534, antiderivative size = 38, 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} \[ \frac{2 \left (a+b \sqrt{x}\right )^5}{5 b^2}-\frac{a \left (a+b \sqrt{x}\right )^4}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^3,x]

[Out]

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

Mathematica [A]  time = 0.0157507, size = 28, normalized size = 0.74 \[ -\frac{\left (a-4 b \sqrt{x}\right ) \left (a+b \sqrt{x}\right )^4}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^3,x]

[Out]

-((a - 4*b*Sqrt[x])*(a + b*Sqrt[x])^4)/(10*b^2)

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Maple [A]  time = 0.002, size = 33, normalized size = 0.9 \begin{align*}{\frac{2\,{b}^{3}}{5}{x}^{{\frac{5}{2}}}}+{\frac{3\,a{b}^{2}{x}^{2}}{2}}+2\,{a}^{2}b{x}^{3/2}+{a}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.976543, size = 43, normalized size = 1.13 \begin{align*} \frac{2}{5} \, b^{3} x^{\frac{5}{2}} + \frac{3}{2} \, a b^{2} x^{2} + 2 \, a^{2} b x^{\frac{3}{2}} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.49886, size = 80, normalized size = 2.11 \begin{align*} \frac{3}{2} \, a b^{2} x^{2} + a^{3} x + \frac{2}{5} \,{\left (b^{3} x^{2} + 5 \, a^{2} b x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.65707, size = 39, normalized size = 1.03 \begin{align*} a^{3} x + 2 a^{2} b x^{\frac{3}{2}} + \frac{3 a b^{2} x^{2}}{2} + \frac{2 b^{3} x^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/2))**3,x)

[Out]

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

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Giac [A]  time = 1.50075, size = 43, normalized size = 1.13 \begin{align*} \frac{2}{5} \, b^{3} x^{\frac{5}{2}} + \frac{3}{2} \, a b^{2} x^{2} + 2 \, a^{2} b x^{\frac{3}{2}} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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