3.178 \(\int x \sqrt{a+b x} \, dx\)

Optimal. Leaf size=34 \[ \frac{2 (a+b x)^{5/2}}{5 b^2}-\frac{2 a (a+b x)^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0086508, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{2 (a+b x)^{5/2}}{5 b^2}-\frac{2 a (a+b x)^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0113039, size = 24, normalized size = 0.71 \[ \frac{2 (a+b x)^{3/2} (3 b x-2 a)}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + b*x],x]

[Out]

(2*(a + b*x)^(3/2)*(-2*a + 3*b*x))/(15*b^2)

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Maple [A]  time = 0.003, size = 21, normalized size = 0.6 \begin{align*} -{\frac{-6\,bx+4\,a}{15\,{b}^{2}} \left ( bx+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(b*x+a)^(3/2)*(-3*b*x+2*a)/b^2

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Maxima [A]  time = 0.954078, size = 35, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{\frac{5}{2}}}{5 \, b^{2}} - \frac{2 \,{\left (b x + a\right )}^{\frac{3}{2}} a}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.6055, size = 70, normalized size = 2.06 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^2*x^2 + a*b*x - 2*a^2)*sqrt(b*x + a)/b^2

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Sympy [B]  time = 1.04818, size = 202, normalized size = 5.94 \begin{align*} - \frac{4 a^{\frac{9}{2}} \sqrt{1 + \frac{b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac{4 a^{\frac{9}{2}}}{15 a^{2} b^{2} + 15 a b^{3} x} - \frac{2 a^{\frac{7}{2}} b x \sqrt{1 + \frac{b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac{4 a^{\frac{7}{2}} b x}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac{8 a^{\frac{5}{2}} b^{2} x^{2} \sqrt{1 + \frac{b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} + \frac{6 a^{\frac{3}{2}} b^{3} x^{3} \sqrt{1 + \frac{b x}{a}}}{15 a^{2} b^{2} + 15 a b^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a**(9/2)*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 4*a**(9/2)/(15*a**2*b**2 + 15*a*b**3*x) - 2*a**(7/2
)*b*x*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 4*a**(7/2)*b*x/(15*a**2*b**2 + 15*a*b**3*x) + 8*a**(5/2)*
b**2*x**2*sqrt(1 + b*x/a)/(15*a**2*b**2 + 15*a*b**3*x) + 6*a**(3/2)*b**3*x**3*sqrt(1 + b*x/a)/(15*a**2*b**2 +
15*a*b**3*x)

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Giac [A]  time = 1.05599, size = 34, normalized size = 1. \begin{align*} \frac{2 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )}}{15 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)/b^2