\(\int x (b x^n)^p \, dx\) [175]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int x \left (b x^n\right )^p \, dx=\frac {x^2 \left (b x^n\right )^p}{2+n p} \]

[Out]

x^2*(b*x^n)^p/(n*p+2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \[ \int x \left (b x^n\right )^p \, dx=\frac {x^2 \left (b x^n\right )^p}{n p+2} \]

[In]

Int[x*(b*x^n)^p,x]

[Out]

(x^2*(b*x^n)^p)/(2 + n*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^{-n p} \left (b x^n\right )^p\right ) \int x^{1+n p} \, dx \\ & = \frac {x^2 \left (b x^n\right )^p}{2+n p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int x \left (b x^n\right )^p \, dx=\frac {x^2 \left (b x^n\right )^p}{2+n p} \]

[In]

Integrate[x*(b*x^n)^p,x]

[Out]

(x^2*(b*x^n)^p)/(2 + n*p)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
gosper	\(\frac {x^{2} \left (b \,x^{n}\right )^{p}}{n p +2}\)	\(19\)
parallelrisch	\(\frac {x^{2} \left (b \,x^{n}\right )^{p}}{n p +2}\)	\(19\)

[In]

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

[Out]

x^2*(b*x^n)^p/(n*p+2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int x \left (b x^n\right )^p \, dx=\frac {x^{2} e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{n p + 2} \]

[In]

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

[Out]

x^2*e^(n*p*log(x) + p*log(b))/(n*p + 2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (14) = 28\).

Time = 0.57 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int x \left (b x^n\right )^p \, dx=\begin {cases} \frac {x^{2} \left (b x^{n}\right )^{p}}{n p + 2} & \text {for}\: n \neq - \frac {2}{p} \\x^{2} \left (b x^{- \frac {2}{p}}\right )^{p} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x*(b*x**n)**p,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int x \left (b x^n\right )^p \, dx=\frac {b^{p} x^{2} {\left (x^{n}\right )}^{p}}{n p + 2} \]

[In]

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

[Out]

b^p*x^2*(x^n)^p/(n*p + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int x \left (b x^n\right )^p \, dx=\frac {x^{2} e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{n p + 2} \]

[In]

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

[Out]

x^2*e^(n*p*log(x) + p*log(b))/(n*p + 2)

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int x \left (b x^n\right )^p \, dx=\frac {x^2\,{\left (b\,x^n\right )}^p}{n\,p+2} \]

[In]

int(x*(b*x^n)^p,x)

[Out]

(x^2*(b*x^n)^p)/(n*p + 2)