\(\int x^2 (a^2+2 a b x^3+b^2 x^6)^p \, dx\) [130]

Optimal result

Integrand size = 24, antiderivative size = 41 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b (1+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1366, 623} \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b (2 p+1)} \]

[In]

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

[Out]

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

Rule 623

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 1366

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^p \, 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 )^p}{3 b (1+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p}{3 b (1+2 p)} \]

[In]

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

[Out]

((a + b*x^3)*((a + b*x^3)^2)^p)/(3*b*(1 + 2*p))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\left (b \,x^{3}+a \right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{3 b \left (1+2 p \right )}\)	\(31\)
gosper	\(\frac {\left (b \,x^{3}+a \right ) \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{3 b \left (1+2 p \right )}\)	\(40\)
parallelrisch	\(\frac {x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a b +\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2}}{3 a \left (1+2 p \right ) b}\)	\(67\)
norman	\(\frac {x^{3} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 p +3}+\frac {a \,{\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{3 b \left (1+2 p \right )}\)	\(71\)

[In]

int(x^2*(b^2*x^6+2*a*b*x^3+a^2)^p,x,method=_RETURNVERBOSE)

[Out]

1/3*(b*x^3+a)/b/(1+2*p)*((b*x^3+a)^2)^p

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left (b x^{3} + a\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{3 \, {\left (2 \, b p + b\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\begin {cases} \frac {x^{3}}{3 \sqrt {a^{2}}} & \text {for}\: b = 0 \wedge p = - \frac {1}{2} \\\frac {x^{3} \left (a^{2}\right )^{p}}{3} & \text {for}\: b = 0 \\\int \frac {x^{2}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{6 b p + 3 b} + \frac {b x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{6 b p + 3 b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x**3/(3*sqrt(a**2)), Eq(b, 0) & Eq(p, -1/2)), (x**3*(a**2)**p/3, Eq(b, 0)), (Integral(x**2/sqrt((a
+ b*x**3)**2), x), Eq(p, -1/2)), (a*(a**2 + 2*a*b*x**3 + b**2*x**6)**p/(6*b*p + 3*b) + b*x**3*(a**2 + 2*a*b*x*
*3 + b**2*x**6)**p/(6*b*p + 3*b), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left (b x^{3} + a\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{3 \, b {\left (2 \, p + 1\right )}} \]

[In]

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

[Out]

1/3*(b*x^3 + a)*(b*x^3 + a)^(2*p)/(b*(2*p + 1))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b x^{3} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a}{3 \, {\left (2 \, b p + b\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int x^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\left (\frac {x^3}{3\,\left (2\,p+1\right )}+\frac {a}{3\,b\,\left (2\,p+1\right )}\right )\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p \]

[In]

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

[Out]

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