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

Optimal. Leaf size=52 \[ \frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2002, 2014} \begin {gather*} \frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3],x]

[Out]

(-4*a*(a*x^2 + b*x^3)^(3/2))/(15*b^2*x^3) + (2*(a*x^2 + b*x^3)^(3/2))/(5*b*x^2)

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 \sqrt {a x^2+b x^3} \, dx &=\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {(2 a) \int \frac {\sqrt {a x^2+b x^3}}{x} \, dx}{5 b}\\ &=-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.60 \begin {gather*} \frac {2 \left (x^2 (a+b x)\right )^{3/2} (3 b x-2 a)}{15 b^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 43, normalized size = 0.83 \begin {gather*} \frac {2 \left (-2 a^2+a b x+3 b^2 x^2\right ) \sqrt {a x^2+b x^3}}{15 b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 39, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 81, normalized size = 1.56 \begin {gather*} \frac {4 \, a^{\frac {5}{2}} \mathrm {sgn}\relax (x)}{15 \, b^{2}} + \frac {2 \, {\left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a \mathrm {sgn}\relax (x)}{b} + \frac {{\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 )} \mathrm {sgn}\relax (x)}{b}\right )}}{15 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*a^(5/2)*sgn(x)/b^2 + 2/15*(5*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a*sgn(x)/b + (3*(b*x + a)^(5/2) - 10*(
b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*sgn(x)/b)/b

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maple [A]  time = 0.05, size = 35, normalized size = 0.67 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-3 b x +2 a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.42, size = 30, normalized size = 0.58 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.30, size = 39, normalized size = 0.75 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-2\,a^2+a\,b\,x+3\,b^2\,x^2\right )}{15\,b^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a x^{2} + b x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x**2 + b*x**3), x)

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