\(\int x^3 (b x^2)^p \, dx\) [69]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {x^4 \left (b x^2\right )^p}{2 (2+p)} \]

[Out]

1/2*x^4*(b*x^2)^p/(2+p)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {x^4 \left (b x^2\right )^p}{2 (p+2)} \]

[In]

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

[Out]

(x^4*(b*x^2)^p)/(2*(2 + p))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \left (x^{-2 p} \left (b x^2\right )^p\right ) \int x^{3+2 p} \, dx \\ & = \frac {x^4 \left (b x^2\right )^p}{2 (2+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {x^4 \left (b x^2\right )^p}{4+2 p} \]

[In]

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

[Out]

(x^4*(b*x^2)^p)/(4 + 2*p)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {\left (b \,x^{2}\right )^{p} x^{4}}{4+2 p}\)	\(18\)
risch	\(\frac {\left (b \,x^{2}\right )^{p} x^{4}}{4+2 p}\)	\(18\)
parallelrisch	\(\frac {\left (b \,x^{2}\right )^{p} x^{4}}{4+2 p}\)	\(18\)
norman	\(\frac {x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}\right )}}{4+2 p}\)	\(20\)

[In]

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

[Out]

1/2*x^4*(b*x^2)^p/(2+p)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {\left (b x^{2}\right )^{p} x^{4}}{2 \, {\left (p + 2\right )}} \]

[In]

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

[Out]

1/2*(b*x^2)^p*x^4/(p + 2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int x^3 \left (b x^2\right )^p \, dx=\begin {cases} \frac {x^{4} \left (b x^{2}\right )^{p}}{2 p + 4} & \text {for}\: p \neq -2 \\\frac {\log {\left (x \right )}}{b^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(b*x**2)**p,x)

[Out]

Piecewise((x**4*(b*x**2)**p/(2*p + 4), Ne(p, -2)), (log(x)/b**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {b^{p} {\left (x^{2}\right )}^{p} x^{4}}{2 \, {\left (p + 2\right )}} \]

[In]

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

[Out]

1/2*b^p*(x^2)^p*x^4/(p + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {\left (b x^{2}\right )^{p} x^{4}}{2 \, {\left (p + 2\right )}} \]

[In]

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

[Out]

1/2*(b*x^2)^p*x^4/(p + 2)

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int x^3 \left (b x^2\right )^p \, dx=\frac {x^4\,{\left (b\,x^2\right )}^p}{2\,\left (p+2\right )} \]

[In]

int(x^3*(b*x^2)^p,x)

[Out]

(x^4*(b*x^2)^p)/(2*(p + 2))