\(\int (c x)^m (b x^n)^p \, dx\) [173]

Optimal result

Integrand size = 13, antiderivative size = 26 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {(c x)^{1+m} \left (b x^n\right )^p}{c (1+m+n p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {15, 20, 30} \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {x (c x)^m \left (b x^n\right )^p}{m+n p+1} \]

[In]

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

[Out]

(x*(c*x)^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 20

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {x (c x)^m \left (b x^n\right )^p}{1+m+n p} \]

[In]

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

[Out]

(x*(c*x)^m*(b*x^n)^p)/(1 + m + n*p)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {x \left (c x \right )^{m} \left (b \,x^{n}\right )^{p}}{n p +m +1}\)	\(23\)
parallelrisch	\(\frac {x \left (c x \right )^{m} \left (b \,x^{n}\right )^{p}}{n p +m +1}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {x e^{\left (n p \log \left (x\right ) + p \log \left (b\right ) + m \log \left (c\right ) + m \log \left (x\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).

Time = 0.98 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\begin {cases} \frac {x \left (b x^{n}\right )^{p} \left (c x\right )^{m}}{m + n p + 1} & \text {for}\: m \neq - n p - 1 \\x \left (b x^{n}\right )^{p} \left (c x\right )^{- n p - 1} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {b^{p} c^{m} x e^{\left (m \log \left (x\right ) + p \log \left (x^{n}\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {x e^{\left (n p \log \left (x\right ) + p \log \left (b\right ) + m \log \left (c\right ) + m \log \left (x\right )\right )}}{n p + m + 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int (c x)^m \left (b x^n\right )^p \, dx=\frac {x\,{\left (b\,x^n\right )}^p\,{\left (c\,x\right )}^m}{m+n\,p+1} \]

[In]

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

[Out]

(x*(b*x^n)^p*(c*x)^m)/(m + n*p + 1)