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

Optimal. Leaf size=80 \[ \frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac{8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3} \]

[Out]

(16*a^2*(a*x^2 + b*x^3)^(5/2))/(315*b^3*x^5) - (8*a*(a*x^2 + b*x^3)^(5/2))/(63*b^2*x^4) + (2*(a*x^2 + b*x^3)^(
5/2))/(9*b*x^3)

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Rubi [A]  time = 0.132562, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ \frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac{8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x,x]

[Out]

(16*a^2*(a*x^2 + b*x^3)^(5/2))/(315*b^3*x^5) - (8*a*(a*x^2 + b*x^3)^(5/2))/(63*b^2*x^4) + (2*(a*x^2 + b*x^3)^(
5/2))/(9*b*x^3)

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} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac{(4 a) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{9 b}\\ &=-\frac{8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}+\frac{\left (8 a^2\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{63 b^2}\\ &=\frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac{8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}\\ \end{align*}

Mathematica [A]  time = 0.0224445, size = 47, normalized size = 0.59 \[ \frac{2 x (a+b x)^3 \left (8 a^2-20 a b x+35 b^2 x^2\right )}{315 b^3 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x,x]

[Out]

(2*x*(a + b*x)^3*(8*a^2 - 20*a*b*x + 35*b^2*x^2))/(315*b^3*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.004, size = 46, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 35\,{b}^{2}{x}^{2}-20\,abx+8\,{a}^{2} \right ) }{315\,{b}^{3}{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,x)

[Out]

2/315*(b*x+a)*(35*b^2*x^2-20*a*b*x+8*a^2)*(b*x^3+a*x^2)^(3/2)/b^3/x^3

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Maxima [A]  time = 1.00761, size = 72, normalized size = 0.9 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b*x + a)/b^3

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Fricas [A]  time = 0.829455, size = 134, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x^{3} + a x^{2}}}{315 \, b^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b*x^3 + a*x^2)/(b^3*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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28919, size = 144, normalized size = 1.8 \begin{align*} -\frac{16 \, a^{\frac{9}{2}} \mathrm{sgn}\left (x\right )}{315 \, b^{3}} + \frac{2 \,{\left (\frac{3 \,{\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 )} a \mathrm{sgn}\left (x\right )}{b^{2}} + \frac{{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} \mathrm{sgn}\left (x\right )}{b^{2}}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/315*a^(9/2)*sgn(x)/b^3 + 2/315*(3*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)*a*s
gn(x)/b^2 + (35*(b*x + a)^(9/2) - 135*(b*x + a)^(7/2)*a + 189*(b*x + a)^(5/2)*a^2 - 105*(b*x + a)^(3/2)*a^3)*s
gn(x)/b^2)/b