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

Optimal. Leaf size=161 \[ -\frac{512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac{64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b} \]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(15*b) - (512*a^5*(a*x^2 + b*x^3)^(5/2))/(45045*b^6*x^5) + (256*a^4*(a*x^2 + b*x^3)^
(5/2))/(9009*b^5*x^4) - (64*a^3*(a*x^2 + b*x^3)^(5/2))/(1287*b^4*x^3) + (32*a^2*(a*x^2 + b*x^3)^(5/2))/(429*b^
3*x^2) - (4*a*(a*x^2 + b*x^3)^(5/2))/(39*b^2*x)

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Rubi [A]  time = 0.23122, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \[ -\frac{512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac{64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(15*b) - (512*a^5*(a*x^2 + b*x^3)^(5/2))/(45045*b^6*x^5) + (256*a^4*(a*x^2 + b*x^3)^
(5/2))/(9009*b^5*x^4) - (64*a^3*(a*x^2 + b*x^3)^(5/2))/(1287*b^4*x^3) + (32*a^2*(a*x^2 + b*x^3)^(5/2))/(429*b^
3*x^2) - (4*a*(a*x^2 + b*x^3)^(5/2))/(39*b^2*x)

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 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 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 x^2 \left (a x^2+b x^3\right )^{3/2} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac{(2 a) \int x \left (a x^2+b x^3\right )^{3/2} \, dx}{3 b}\\ &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac{\left (16 a^2\right ) \int \left (a x^2+b x^3\right )^{3/2} \, dx}{39 b^2}\\ &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}-\frac{\left (32 a^3\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x} \, dx}{143 b^3}\\ &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac{64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac{\left (128 a^4\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{1287 b^4}\\ &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}+\frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac{64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}-\frac{\left (256 a^5\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{9009 b^5}\\ &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac{512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac{64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac{4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0414666, size = 80, normalized size = 0.5 \[ \frac{2 x (a+b x)^3 \left (-1120 a^3 b^2 x^2+1680 a^2 b^3 x^3+640 a^4 b x-256 a^5-2310 a b^4 x^4+3003 b^5 x^5\right )}{45045 b^6 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*x)^3*(-256*a^5 + 640*a^4*b*x - 1120*a^3*b^2*x^2 + 1680*a^2*b^3*x^3 - 2310*a*b^4*x^4 + 3003*b^5*x^5
))/(45045*b^6*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.005, size = 79, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -3003\,{x}^{5}{b}^{5}+2310\,a{x}^{4}{b}^{4}-1680\,{x}^{3}{a}^{2}{b}^{3}+1120\,{x}^{2}{a}^{3}{b}^{2}-640\,{a}^{4}xb+256\,{a}^{5} \right ) }{45045\,{b}^{6}{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(x^2*(b*x^3+a*x^2)^(3/2),x)

[Out]

-2/45045*(b*x+a)*(-3003*b^5*x^5+2310*a*b^4*x^4-1680*a^2*b^3*x^3+1120*a^3*b^2*x^2-640*a^4*b*x+256*a^5)*(b*x^3+a
*x^2)^(3/2)/b^6/x^3

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Maxima [A]  time = 1.00878, size = 116, normalized size = 0.72 \begin{align*} \frac{2 \,{\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt{b x + a}}{45045 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*b^7*x^7 + 3696*a*b^6*x^6 + 63*a^2*b^5*x^5 - 70*a^3*b^4*x^4 + 80*a^4*b^3*x^3 - 96*a^5*b^2*x^2 + 1
28*a^6*b*x - 256*a^7)*sqrt(b*x + a)/b^6

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Fricas [A]  time = 0.868384, size = 217, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt{b x^{3} + a x^{2}}}{45045 \, b^{6} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*b^7*x^7 + 3696*a*b^6*x^6 + 63*a^2*b^5*x^5 - 70*a^3*b^4*x^4 + 80*a^4*b^3*x^3 - 96*a^5*b^2*x^2 + 1
28*a^6*b*x - 256*a^7)*sqrt(b*x^3 + a*x^2)/(b^6*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13546, size = 242, normalized size = 1.5 \begin{align*} \frac{512 \, a^{\frac{15}{2}} \mathrm{sgn}\left (x\right )}{45045 \, b^{6}} + \frac{2 \,{\left (\frac{5 \,{\left (693 \,{\left (b x + a\right )}^{\frac{13}{2}} - 4095 \,{\left (b x + a\right )}^{\frac{11}{2}} a + 10010 \,{\left (b x + a\right )}^{\frac{9}{2}} a^{2} - 12870 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{3} + 9009 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{4} - 3003 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{5}\right )} a \mathrm{sgn}\left (x\right )}{b^{5}} + \frac{{\left (3003 \,{\left (b x + a\right )}^{\frac{15}{2}} - 20790 \,{\left (b x + a\right )}^{\frac{13}{2}} a + 61425 \,{\left (b x + a\right )}^{\frac{11}{2}} a^{2} - 100100 \,{\left (b x + a\right )}^{\frac{9}{2}} a^{3} + 96525 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{4} - 54054 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{5} + 15015 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{6}\right )} \mathrm{sgn}\left (x\right )}{b^{5}}\right )}}{45045 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

512/45045*a^(15/2)*sgn(x)/b^6 + 2/45045*(5*(693*(b*x + a)^(13/2) - 4095*(b*x + a)^(11/2)*a + 10010*(b*x + a)^(
9/2)*a^2 - 12870*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 3003*(b*x + a)^(3/2)*a^5)*a*sgn(x)/b^5 + (30
03*(b*x + a)^(15/2) - 20790*(b*x + a)^(13/2)*a + 61425*(b*x + a)^(11/2)*a^2 - 100100*(b*x + a)^(9/2)*a^3 + 965
25*(b*x + a)^(7/2)*a^4 - 54054*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2)*a^6)*sgn(x)/b^5)/b