\(\int x^2 (b+2 c x^3) (-a+b x^3+c x^6)^p \, dx\) [135]

Optimal result

Integrand size = 28, antiderivative size = 27 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {\left (-a+b x^3+c x^6\right )^{1+p}}{3 (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1482, 643} \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {\left (-a+b x^3+c x^6\right )^{p+1}}{3 (p+1)} \]

[In]

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

[Out]

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

Rule 643

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

Rule 1482

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int (b+2 c x) \left (-a+b x+c x^2\right )^p \, dx,x,x^3\right ) \\ & = \frac {\left (-a+b x^3+c x^6\right )^{1+p}}{3 (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {\left (-a+b x^3+c x^6\right )^{1+p}}{3 (1+p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
gosper	\(\frac {\left (c \,x^{6}+b \,x^{3}-a \right )^{1+p}}{3+3 p}\)	\(26\)
risch	\(-\frac {\left (-c \,x^{6}-b \,x^{3}+a \right ) \left (c \,x^{6}+b \,x^{3}-a \right )^{p}}{3 \left (1+p \right )}\)	\(38\)
parallelrisch	\(\frac {x^{6} \left (c \,x^{6}+b \,x^{3}-a \right )^{p} c^{2}+x^{3} \left (c \,x^{6}+b \,x^{3}-a \right )^{p} b c -\left (c \,x^{6}+b \,x^{3}-a \right )^{p} a c}{3 c \left (1+p \right )}\)	\(77\)
norman	\(-\frac {a \,{\mathrm e}^{p \ln \left (c \,x^{6}+b \,x^{3}-a \right )}}{3 \left (1+p \right )}+\frac {b \,x^{3} {\mathrm e}^{p \ln \left (c \,x^{6}+b \,x^{3}-a \right )}}{3+3 p}+\frac {c \,x^{6} {\mathrm e}^{p \ln \left (c \,x^{6}+b \,x^{3}-a \right )}}{3+3 p}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {{\left (c x^{6} + b x^{3} - a\right )} {\left (c x^{6} + b x^{3} - a\right )}^{p}}{3 \, {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {{\left (c x^{6} + b x^{3} - a\right )} {\left (c x^{6} + b x^{3} - a\right )}^{p}}{3 \, {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx=\frac {{\left (c x^{6} + b x^{3} - a\right )}^{p + 1}}{3 \, {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.58 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int x^2 \left (b+2 c x^3\right ) \left (-a+b x^3+c x^6\right )^p \, dx={\left (c\,x^6+b\,x^3-a\right )}^p\,\left (\frac {b\,x^3}{3\,p+3}-\frac {a}{3\,p+3}+\frac {c\,x^6}{3\,p+3}\right ) \]

[In]

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

[Out]

(b*x^3 - a + c*x^6)^p*((b*x^3)/(3*p + 3) - a/(3*p + 3) + (c*x^6)/(3*p + 3))