3.245 \(\int \frac{(a x^2+b x^3)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac{2 \left (a x^2+b x^3\right )^{5/2}}{5 b x^5} \]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(5*b*x^5)

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Rubi [A]  time = 0.0406793, antiderivative size = 25, normalized size of antiderivative = 1., 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 = {2014} \[ \frac{2 \left (a x^2+b x^3\right )^{5/2}}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^3,x]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(5*b*x^5)

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

Mathematica [A]  time = 0.0128377, size = 23, normalized size = 0.92 \[ \frac{2 \left (x^2 (a+b x)\right )^{5/2}}{5 b x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^3,x]

[Out]

(2*(x^2*(a + b*x))^(5/2))/(5*b*x^5)

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Maple [A]  time = 0.002, size = 27, normalized size = 1.1 \begin{align*}{\frac{2\,bx+2\,a}{5\,b{x}^{3}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.988703, size = 38, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b x + a}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/b

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Fricas [A]  time = 0.811251, size = 77, normalized size = 3.08 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b x^{3} + a x^{2}}}{5 \, b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x^3 + a*x^2)/(b*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**3, x)

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Giac [B]  time = 1.24223, size = 70, normalized size = 2.8 \begin{align*} -\frac{2 \, a^{\frac{5}{2}} \mathrm{sgn}\left (x\right )}{5 \, b} + \frac{2 \,{\left (5 \,{\left (b x + a\right )}^{\frac{3}{2}} a \mathrm{sgn}\left (x\right ) +{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} \mathrm{sgn}\left (x\right )\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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