3.233 \(\int x \sqrt{a x^2+b x^3} \, dx\)

Optimal. Leaf size=80 \[ \frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^3}-\frac{8 a \left (a x^2+b x^3\right )^{3/2}}{35 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{7 b x} \]

[Out]

(16*a^2*(a*x^2 + b*x^3)^(3/2))/(105*b^3*x^3) - (8*a*(a*x^2 + b*x^3)^(3/2))/(35*b^2*x^2) + (2*(a*x^2 + b*x^3)^(
3/2))/(7*b*x)

________________________________________________________________________________________

Rubi [A]  time = 0.073556, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2016, 2002, 2014} \[ \frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^3}-\frac{8 a \left (a x^2+b x^3\right )^{3/2}}{35 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{7 b x} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a*x^2 + b*x^3],x]

[Out]

(16*a^2*(a*x^2 + b*x^3)^(3/2))/(105*b^3*x^3) - (8*a*(a*x^2 + b*x^3)^(3/2))/(35*b^2*x^2) + (2*(a*x^2 + b*x^3)^(
3/2))/(7*b*x)

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int x \sqrt{a x^2+b x^3} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac{(4 a) \int \sqrt{a x^2+b x^3} \, dx}{7 b}\\ &=-\frac{8 a \left (a x^2+b x^3\right )^{3/2}}{35 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}+\frac{\left (8 a^2\right ) \int \frac{\sqrt{a x^2+b x^3}}{x} \, dx}{35 b^2}\\ &=\frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^3}-\frac{8 a \left (a x^2+b x^3\right )^{3/2}}{35 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}\\ \end{align*}

Mathematica [A]  time = 0.0196859, size = 42, normalized size = 0.52 \[ \frac{2 \left (x^2 (a+b x)\right )^{3/2} \left (8 a^2-12 a b x+15 b^2 x^2\right )}{105 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*(x^2*(a + b*x))^(3/2)*(8*a^2 - 12*a*b*x + 15*b^2*x^2))/(105*b^3*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 46, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 15\,{b}^{2}{x}^{2}-12\,abx+8\,{a}^{2} \right ) }{105\,{b}^{3}x}\sqrt{b{x}^{3}+a{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(b*x+a)*(15*b^2*x^2-12*a*b*x+8*a^2)*(b*x^3+a*x^2)^(1/2)/b^3/x

________________________________________________________________________________________

Maxima [A]  time = 1.01651, size = 57, normalized size = 0.71 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)/b^3

________________________________________________________________________________________

Fricas [A]  time = 0.81328, size = 111, normalized size = 1.39 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{105 \, b^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2)/(b^3*x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{x^{2} \left (a + b x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(x**2*(a + b*x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.15079, size = 68, normalized size = 0.85 \begin{align*} -\frac{16 \, a^{\frac{7}{2}} \mathrm{sgn}\left (x\right )}{105 \, b^{3}} + \frac{2 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} \mathrm{sgn}\left (x\right )}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/105*a^(7/2)*sgn(x)/b^3 + 2/105*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*sgn(x)
/b^3