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

Optimal. Leaf size=52 \[ \frac{2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5} \]

[Out]

(-4*a*(a*x^2 + b*x^3)^(5/2))/(35*b^2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)

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Rubi [A]  time = 0.0826054, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ \frac{2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^2,x]

[Out]

(-4*a*(a*x^2 + b*x^3)^(5/2))/(35*b^2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)

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])

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^2} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac{(2 a) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{7 b}\\ &=-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}\\ \end{align*}

Mathematica [A]  time = 0.01721, size = 36, normalized size = 0.69 \[ \frac{2 x (a+b x)^3 (5 b x-2 a)}{35 b^2 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*x)^3*(-2*a + 5*b*x))/(35*b^2*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.003, size = 35, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -5\,bx+2\,a \right ) }{35\,{b}^{2}{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^2,x)

[Out]

-2/35*(b*x+a)*(-5*b*x+2*a)*(b*x^3+a*x^2)^(3/2)/b^2/x^3

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Maxima [A]  time = 1.02657, size = 55, normalized size = 1.06 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{35 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x + a)/b^2

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Fricas [A]  time = 0.817854, size = 105, normalized size = 2.02 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{35 \, b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*x^3 + 8*a*b^2*x^2 + a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2)/(b^2*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^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27493, size = 112, normalized size = 2.15 \begin{align*} \frac{4 \, a^{\frac{7}{2}} \mathrm{sgn}\left (x\right )}{35 \, b^{2}} + \frac{2 \,{\left (\frac{7 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x + a\right )}^{\frac{3}{2}} a\right )} a \mathrm{sgn}\left (x\right )}{b} + \frac{{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} \mathrm{sgn}\left (x\right )}{b}\right )}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/35*a^(7/2)*sgn(x)/b^2 + 2/105*(7*(3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)*a*sgn(x)/b + (15*(b*x + a)^(7/2)
- 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*sgn(x)/b)/b