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

Optimal. Leaf size=108 \[ -\frac{32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \]

[Out]

(-32*a^3*(a*x^2 + b*x^3)^(5/2))/(1155*b^4*x^5) + (16*a^2*(a*x^2 + b*x^3)^(5/2))/(231*b^3*x^4) - (4*a*(a*x^2 +
b*x^3)^(5/2))/(33*b^2*x^3) + (2*(a*x^2 + b*x^3)^(5/2))/(11*b*x^2)

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Rubi [A]  time = 0.141219, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2002, 2016, 2014} \[ -\frac{32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-32*a^3*(a*x^2 + b*x^3)^(5/2))/(1155*b^4*x^5) + (16*a^2*(a*x^2 + b*x^3)^(5/2))/(231*b^3*x^4) - (4*a*(a*x^2 +
b*x^3)^(5/2))/(33*b^2*x^3) + (2*(a*x^2 + b*x^3)^(5/2))/(11*b*x^2)

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 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])

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 \left (a x^2+b x^3\right )^{3/2} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac{(6 a) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x} \, dx}{11 b}\\ &=-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}+\frac{\left (8 a^2\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{33 b^2}\\ &=\frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac{\left (16 a^3\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{231 b^3}\\ &=-\frac{32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac{16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}\\ \end{align*}

Mathematica [A]  time = 0.0270002, size = 58, normalized size = 0.54 \[ \frac{2 x (a+b x)^3 \left (40 a^2 b x-16 a^3-70 a b^2 x^2+105 b^3 x^3\right )}{1155 b^4 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*x*(a + b*x)^3*(-16*a^3 + 40*a^2*b*x - 70*a*b^2*x^2 + 105*b^3*x^3))/(1155*b^4*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.003, size = 57, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -105\,{x}^{3}{b}^{3}+70\,a{b}^{2}{x}^{2}-40\,{a}^{2}xb+16\,{a}^{3} \right ) }{1155\,{b}^{4}{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)

[Out]

-2/1155*(b*x+a)*(-105*b^3*x^3+70*a*b^2*x^2-40*a^2*b*x+16*a^3)*(b*x^3+a*x^2)^(3/2)/b^4/x^3

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Maxima [A]  time = 0.987819, size = 86, normalized size = 0.8 \begin{align*} \frac{2 \,{\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x + a}}{1155 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x + a)/b^4

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Fricas [A]  time = 0.740348, size = 161, normalized size = 1.49 \begin{align*} \frac{2 \,{\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x^{3} + a x^{2}}}{1155 \, b^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2)/
(b^4*x)

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

[Out]

Integral((a*x**2 + b*x**3)**(3/2), x)

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Giac [A]  time = 1.26395, size = 177, normalized size = 1.64 \begin{align*} \frac{32 \, a^{\frac{11}{2}} \mathrm{sgn}\left (x\right )}{1155 \, b^{4}} + \frac{2 \,{\left (\frac{11 \,{\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 )} a \mathrm{sgn}\left (x\right )}{b^{3}} + \frac{{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}\right )} \mathrm{sgn}\left (x\right )}{b^{3}}\right )}}{3465 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1155*a^(11/2)*sgn(x)/b^4 + 2/3465*(11*(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)*a*sgn(x)/b^3 + (315*(b*x + a)^(11/2) - 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(7/
2)*a^2 - 2772*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4)*sgn(x)/b^3)/b