3.61 \(\int x^2 (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\)

Optimal. Leaf size=36 \[ \frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{18 b} \]

[Out]

1/18*(b*x^3+a)*(b^2*x^6+2*a*b*x^3+a^2)^(5/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1352, 609} \[ \frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{18 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

((a + b*x^3)*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2))/(18*b)

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 1352

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^3\right )\\ &=\frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{18 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.02, size = 82, normalized size = 2.28 \[ \frac {x^3 \sqrt {\left (a+b x^3\right )^2} \left (6 a^5+15 a^4 b x^3+20 a^3 b^2 x^6+15 a^2 b^3 x^9+6 a b^4 x^{12}+b^5 x^{15}\right )}{18 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^3*Sqrt[(a + b*x^3)^2]*(6*a^5 + 15*a^4*b*x^3 + 20*a^3*b^2*x^6 + 15*a^2*b^3*x^9 + 6*a*b^4*x^12 + b^5*x^15))/(
18*(a + b*x^3))

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 57, normalized size = 1.58 \[ \frac {1}{18} \, b^{5} x^{18} + \frac {1}{3} \, a b^{4} x^{15} + \frac {5}{6} \, a^{2} b^{3} x^{12} + \frac {10}{9} \, a^{3} b^{2} x^{9} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{3} \, a^{5} x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/18*b^5*x^18 + 1/3*a*b^4*x^15 + 5/6*a^2*b^3*x^12 + 10/9*a^3*b^2*x^9 + 5/6*a^4*b*x^6 + 1/3*a^5*x^3

________________________________________________________________________________________

giac [B]  time = 0.37, size = 66, normalized size = 1.83 \[ \frac {1}{18} \, {\left (3 \, {\left (b x^{6} + 2 \, a x^{3}\right )} a^{4} + 3 \, {\left (b x^{6} + 2 \, a x^{3}\right )}^{2} a^{2} b + {\left (b x^{6} + 2 \, a x^{3}\right )}^{3} b^{2}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/18*(3*(b*x^6 + 2*a*x^3)*a^4 + 3*(b*x^6 + 2*a*x^3)^2*a^2*b + (b*x^6 + 2*a*x^3)^3*b^2)*sgn(b*x^3 + a)

________________________________________________________________________________________

maple [B]  time = 0.01, size = 79, normalized size = 2.19 \[ \frac {\left (b^{5} x^{15}+6 a \,b^{4} x^{12}+15 a^{2} b^{3} x^{9}+20 a^{3} b^{2} x^{6}+15 a^{4} b \,x^{3}+6 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}} x^{3}}{18 \left (b \,x^{3}+a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)

[Out]

1/18*x^3*(b^5*x^15+6*a*b^4*x^12+15*a^2*b^3*x^9+20*a^3*b^2*x^6+15*a^4*b*x^3+6*a^5)*((b*x^3+a)^2)^(5/2)/(b*x^3+a
)^5

________________________________________________________________________________________

maxima [A]  time = 0.66, size = 52, normalized size = 1.44 \[ \frac {1}{18} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} x^{3} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a}{18 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*x^3 + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a/b

________________________________________________________________________________________

mupad [B]  time = 1.24, size = 36, normalized size = 1.00 \[ \frac {\left (b^2\,x^3+a\,b\right )\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{18\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

((a*b + b^2*x^3)*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2))/(18*b^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**2*((a + b*x**3)**2)**(5/2), x)

________________________________________________________________________________________