\(\int x^m (b x^n)^p \, dx\) [2712]

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x^{1+m} \left (b x^n\right )^p}{1+m+n p} \]

[Out]

x^(1+m)*(b*x^n)^p/(n*p+m+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, 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} \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x^{m+1} \left (b x^n\right )^p}{m+n p+1} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x^{1+m} \left (b x^n\right )^p}{1+m+n p} \]

[In]

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

[Out]

(x^(1 + m)*(b*x^n)^p)/(1 + m + n*p)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {x \,x^{m} \left (b \,x^{n}\right )^{p}}{n p +m +1}\)	\(21\)
gosper	\(\frac {x^{1+m} \left (b \,x^{n}\right )^{p}}{n p +m +1}\)	\(22\)

[In]

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

[Out]

x/(n*p+m+1)*x^m*(b*x^n)^p

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x x^{m} e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

x*x^m*e^(n*p*log(x) + p*log(b))/(n*p + m + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 0.97 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int x^m \left (b x^n\right )^p \, dx=\begin {cases} \frac {x x^{m} \left (b x^{n}\right )^{p}}{m + n p + 1} & \text {for}\: m \neq - n p - 1 \\x x^{- n p - 1} \left (b x^{n}\right )^{p} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*x**m*(b*x**n)**p/(m + n*p + 1), Ne(m, -n*p - 1)), (x*x**(-n*p - 1)*(b*x**n)**p*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {b^{p} x e^{\left (m \log \left (x\right ) + p \log \left (x^{n}\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

b^p*x*e^(m*log(x) + p*log(x^n))/(n*p + m + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x x^{m} e^{\left (n p \log \left (x\right ) + p \log \left (b\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

x*x^m*e^(n*p*log(x) + p*log(b))/(n*p + m + 1)

Mupad [B] (verification not implemented)

Time = 6.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int x^m \left (b x^n\right )^p \, dx=\frac {x^{m+1}\,{\left (b\,x^n\right )}^p}{m+n\,p+1} \]

[In]

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

[Out]

(x^(m + 1)*(b*x^n)^p)/(m + n*p + 1)