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

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

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Rubi [A]  time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.69 \begin {gather*} \frac {2 x (a+b x)^3 (5 b x-2 a)}{35 b^2 \sqrt {x^2 (a+b x)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 6.44, size = 60, normalized size = 1.15 \begin {gather*} -\frac {2 \left (x^2 (a+b x)\right )^{3/2} \left (2 a^3-a^2 b x-8 a b^2 x^2-5 b^3 x^3\right )}{35 b^2 x^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 50, normalized size = 0.96 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.18, size = 136, normalized size = 2.62 \begin {gather*} \frac {4 \, a^{\frac {7}{2}} \mathrm {sgn}\relax (x)}{35 \, b^{2}} + \frac {2 \, {\left (\frac {35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} \mathrm {sgn}\relax (x)}{b} + \frac {14 \, {\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 )} a \mathrm {sgn}\relax (x)}{b} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} \mathrm {sgn}\relax (x)}{b}\right )}}{105 \, b} \end {gather*}

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*(35*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2*sgn(x)/b + 14*(3*(b*x + a)^(5/2)
- 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a*sgn(x)/b + 3*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*
(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*sgn(x)/b)/b

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maple [A]  time = 0.04, size = 35, normalized size = 0.67 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-5 b x +2 a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}} \end {gather*}

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 = 1.47, size = 41, normalized size = 0.79 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 5.17, size = 36, normalized size = 0.69 \begin {gather*} -\frac {2\,\left (2\,a-5\,b\,x\right )\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2}{35\,b^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(3/2)/x^2,x)

[Out]

-(2*(2*a - 5*b*x)*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2)/(35*b^2*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

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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