3.2.39 \(\int x^2 (b+2 c x^3) (b x^3+c x^6)^p \, dx\) [139]

Optimal. Leaf size=24 \[ \frac {\left (b x^3+c x^6\right )^{1+p}}{3 (1+p)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1602} \begin {gather*} \frac {\left (b x^3+c x^6\right )^{p+1}}{3 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^3 + c*x^6)^(1 + p)/(3*(1 + p))

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^2 \left (b+2 c x^3\right ) \left (b x^3+c x^6\right )^p \, dx &=\frac {\left (b x^3+c x^6\right )^{1+p}}{3 (1+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 24, normalized size = 1.00 \begin {gather*} \frac {\left (x^3 \left (b+c x^3\right )\right )^{1+p}}{3 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b + c*x^3))^(1 + p)/(3*(1 + p))

________________________________________________________________________________________

Maple [A]
time = 0.15, size = 31, normalized size = 1.29

method	result	size
gosper	\(\frac {\left (c \,x^{3}+b \right ) x^{3} \left (c \,x^{6}+b \,x^{3}\right )^{p}}{3+3 p}\)	\(31\)
risch	\(\frac {x^{3} \left (c \,x^{3}+b \right ) \left (x^{3} \left (c \,x^{3}+b \right )\right )^{p}}{3+3 p}\)	\(31\)
norman	\(\frac {b \,x^{3} {\mathrm e}^{p \ln \left (c \,x^{6}+b \,x^{3}\right )}}{3+3 p}+\frac {c \,x^{6} {\mathrm e}^{p \ln \left (c \,x^{6}+b \,x^{3}\right )}}{3+3 p}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*x^6 + b*x^3)*e^(p*log(c*x^3 + b) + 3*p*log(x))/(p + 1)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (c x^{6} + b x^{3}\right )} {\left (c x^{6} + b x^{3}\right )}^{p}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 3.62, size = 22, normalized size = 0.92 \begin {gather*} \frac {{\left (c x^{6} + b x^{3}\right )}^{p + 1}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*x^6 + b*x^3)^(p + 1)/(p + 1)

________________________________________________________________________________________

Mupad [B]
time = 2.07, size = 31, normalized size = 1.29 \begin {gather*} \frac {x^3\,\left (c\,x^3+b\right )\,{\left (c\,x^6+b\,x^3\right )}^p}{3\,\left (p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________