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]

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

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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 \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*}

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Mathematica [A]  time = 0.04, 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)

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fricas [A]  time = 0.39, size = 51, normalized size = 0.64 \[ \frac {2 \, {\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{105 \, b^{3} x} \]

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)

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giac [A]  time = 0.16, size = 108, normalized size = 1.35 \[ -\frac {16 \, a^{\frac {7}{2}} \mathrm {sgn}\relax (x)}{105 \, b^{3}} + \frac {2 \, {\left (\frac {7 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a \mathrm {sgn}\relax (x)}{b^{2}} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} \mathrm {sgn}\relax (x)}{b^{2}}\right )}}{105 \, b} \]

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*(7*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a*sgn(
x)/b^2 + 3*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*sgn(x)/b
^2)/b

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maple [A]  time = 0.05, size = 46, normalized size = 0.58 \[ \frac {2 \left (b x +a \right ) \left (15 b^{2} x^{2}-12 a b x +8 a^{2}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{105 b^{3} x} \]

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

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maxima [A]  time = 1.44, size = 42, normalized size = 0.52 \[ \frac {2 \, {\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{105 \, b^{3}} \]

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

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mupad [B]  time = 5.48, size = 51, normalized size = 0.64 \[ \frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (8\,a^3-4\,a^2\,b\,x+3\,a\,b^2\,x^2+15\,b^3\,x^3\right )}{105\,b^3\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a*x^2 + b*x^3)^(1/2),x)

[Out]

(2*(a*x^2 + b*x^3)^(1/2)*(8*a^3 + 15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x))/(105*b^3*x)

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

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)

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