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

Optimal. Leaf size=105 \[ -\frac{32 a^3 \left (a x^2+b x^3\right )^{3/2}}{315 b^4 x^3}+\frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b} \]

[Out]

(2*(a*x^2 + b*x^3)^(3/2))/(9*b) - (32*a^3*(a*x^2 + b*x^3)^(3/2))/(315*b^4*x^3) + (16*a^2*(a*x^2 + b*x^3)^(3/2)
)/(105*b^3*x^2) - (4*a*(a*x^2 + b*x^3)^(3/2))/(21*b^2*x)

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Rubi [A]  time = 0.119729, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \[ -\frac{32 a^3 \left (a x^2+b x^3\right )^{3/2}}{315 b^4 x^3}+\frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x}+\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*(a*x^2 + b*x^3)^(3/2))/(9*b) - (32*a^3*(a*x^2 + b*x^3)^(3/2))/(315*b^4*x^3) + (16*a^2*(a*x^2 + b*x^3)^(3/2)
)/(105*b^3*x^2) - (4*a*(a*x^2 + b*x^3)^(3/2))/(21*b^2*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^2 \sqrt{a x^2+b x^3} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac{(2 a) \int x \sqrt{a x^2+b x^3} \, dx}{3 b}\\ &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x}+\frac{\left (8 a^2\right ) \int \sqrt{a x^2+b x^3} \, dx}{21 b^2}\\ &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b}+\frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x}-\frac{\left (16 a^3\right ) \int \frac{\sqrt{a x^2+b x^3}}{x} \, dx}{105 b^3}\\ &=\frac{2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac{32 a^3 \left (a x^2+b x^3\right )^{3/2}}{315 b^4 x^3}+\frac{16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0281838, size = 53, normalized size = 0.5 \[ \frac{2 \left (x^2 (a+b x)\right )^{3/2} \left (24 a^2 b x-16 a^3-30 a b^2 x^2+35 b^3 x^3\right )}{315 b^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*(x^2*(a + b*x))^(3/2)*(-16*a^3 + 24*a^2*b*x - 30*a*b^2*x^2 + 35*b^3*x^3))/(315*b^4*x^3)

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Maple [A]  time = 0.004, size = 57, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -35\,{x}^{3}{b}^{3}+30\,a{b}^{2}{x}^{2}-24\,{a}^{2}xb+16\,{a}^{3} \right ) }{315\,{b}^{4}x}\sqrt{b{x}^{3}+a{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(b*x+a)*(-35*b^3*x^3+30*a*b^2*x^2-24*a^2*b*x+16*a^3)*(b*x^3+a*x^2)^(1/2)/b^4/x

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Maxima [A]  time = 1.00444, size = 72, normalized size = 0.69 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt{b x + a}}{315 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt(b*x + a)/b^4

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Fricas [A]  time = 0.819516, size = 134, normalized size = 1.28 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt{b x^{3} + a x^{2}}}{315 \, b^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt(b*x^3 + a*x^2)/(b^4*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x**2*(a + b*x)), x)

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Giac [A]  time = 1.14418, size = 84, normalized size = 0.8 \begin{align*} \frac{32 \, a^{\frac{9}{2}} \mathrm{sgn}\left (x\right )}{315 \, b^{4}} + \frac{2 \,{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} \mathrm{sgn}\left (x\right )}{315 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/315*a^(9/2)*sgn(x)/b^4 + 2/315*(35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*
(b*x + a)^(3/2)*a^3)*sgn(x)/b^4