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

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

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Rubi [A]  time = 0.12, antiderivative size = 105, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.04, size = 53, normalized size = 0.50 \begin {gather*} \frac {2 \left (x^2 (a+b x)\right )^{3/2} \left (-16 a^3+24 a^2 b x-30 a b^2 x^2+35 b^3 x^3\right )}{315 b^4 x^3} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.05, size = 66, normalized size = 0.63 \begin {gather*} \frac {2 \sqrt {a x^2+b x^3} \left (-16 a^4+8 a^3 b x-6 a^2 b^2 x^2+5 a b^3 x^3+35 b^4 x^4\right )}{315 b^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[a*x^2 + b*x^3],x]

[Out]

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

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fricas [A]  time = 0.38, size = 62, normalized size = 0.59 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 131, normalized size = 1.25 \begin {gather*} \frac {32 \, a^{\frac {9}{2}} \mathrm {sgn}\relax (x)}{315 \, b^{4}} + \frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a \mathrm {sgn}\relax (x)}{b^{3}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} \mathrm {sgn}\relax (x)}{b^{3}}\right )}}{315 \, b} \end {gather*}

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*(9*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*s
qrt(b*x + a)*a^3)*a*sgn(x)/b^3 + (35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(
b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*sgn(x)/b^3)/b

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maple [A]  time = 0.04, size = 57, normalized size = 0.54 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x} \end {gather*}

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 = 1.46, size = 53, normalized size = 0.50 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 5.49, size = 62, normalized size = 0.59 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-16\,a^4+8\,a^3\,b\,x-6\,a^2\,b^2\,x^2+5\,a\,b^3\,x^3+35\,b^4\,x^4\right )}{315\,b^4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x^2 + b*x^3)^(1/2),x)

[Out]

(2*(a*x^2 + b*x^3)^(1/2)*(35*b^4*x^4 - 16*a^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x))/(315*b^4*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt {x^{2} \left (a + b x\right )}\, dx \end {gather*}

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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