3.3.34 \(\int \sqrt {a x^2+b x^3} \, dx\) [234]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {2027, 2039} \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 2027

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 2039

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] &&
NeQ[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 = 41, normalized size = 0.79 \begin {gather*} \frac {2 \sqrt {x^2 (a+b x)} \left (-2 a^2+a b x+3 b^2 x^2\right )}{15 b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3],x]

[Out]

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

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Maple [A]
time = 0.38, size = 35, normalized size = 0.67

method	result	size
gosper	\(-\frac {2 \left (b x +a \right ) \left (-3 b x +2 a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 b^{2} x}\)	\(35\)
default	\(-\frac {2 \left (b x +a \right ) \left (-3 b x +2 a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 b^{2} x}\)	\(35\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-3 b^{2} x^{2}-a b x +2 a^{2}\right )}{15 x \,b^{2}}\)	\(39\)
trager	\(-\frac {2 \left (-3 b^{2} x^{2}-a b x +2 a^{2}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15 b^{2} x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.29, 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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Fricas [A]
time = 1.56, 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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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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Giac [A]
time = 1.67, size = 81, normalized size = 1.56 \begin {gather*} \frac {4 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{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}\left (x\right )}{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}\left (x\right )}{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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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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