3.3.35 \(\int \frac {\sqrt {a x^2+b x^3}}{x} \, dx\) [235]

Optimal. Leaf size=25 \[ \frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 b x^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2039} \begin {gather*} \frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x,x]

[Out]

(2*(a*x^2 + b*x^3)^(3/2))/(3*b*x^3)

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

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Mathematica [A]
time = 0.01, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 \left (x^2 (a+b x)\right )^{3/2}}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x,x]

[Out]

(2*(x^2*(a + b*x))^(3/2))/(3*b*x^3)

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Maple [A]
time = 0.36, size = 27, normalized size = 1.08

method	result	size
risch	\(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (b x +a \right )}{3 x b}\)	\(25\)
gosper	\(\frac {2 \left (b x +a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 b x}\)	\(27\)
default	\(\frac {2 \left (b x +a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 b x}\)	\(27\)
trager	\(\frac {2 \left (b x +a \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{3 b x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 12, normalized size = 0.48 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x + a)^(3/2)/b

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Fricas [A]
time = 1.35, size = 26, normalized size = 1.04 \begin {gather*} \frac {2 \, \sqrt {b x^{3} + a x^{2}} {\left (b x + a\right )}}{3 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2*(a + b*x))/x, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (21) = 42\).
time = 1.52, size = 50, normalized size = 2.00 \begin {gather*} -\frac {2 \, a^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{3 \, b} + \frac {2 \, {\left (3 \, \sqrt {b x + a} a \mathrm {sgn}\left (x\right ) + {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} \mathrm {sgn}\left (x\right )\right )}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\sqrt {b\,x^3+a\,x^2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2)/x,x)

[Out]

int((a*x^2 + b*x^3)^(1/2)/x, x)

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