\(\int (c x)^m (b x^2)^p \, dx\) [67]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {x (c x)^m \left (b x^2\right )^p}{1+m+2 p} \]

[Out]

x*(c*x)^m*(b*x^2)^p/(1+m+2*p)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, 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} \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {x \left (b x^2\right )^p (c x)^m}{m+2 p+1} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {x (c x)^m \left (b x^2\right )^p}{1+m+2 p} \]

[In]

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

[Out]

(x*(c*x)^m*(b*x^2)^p)/(1 + m + 2*p)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {x \left (c x \right )^{m} \left (b \,x^{2}\right )^{p}}{1+m +2 p}\)	\(23\)
parallelrisch	\(\frac {x \left (c x \right )^{m} \left (b \,x^{2}\right )^{p}}{1+m +2 p}\)	\(23\)
norman	\(\frac {x \,{\mathrm e}^{m \ln \left (c x \right )} {\mathrm e}^{p \ln \left (b \,x^{2}\right )}}{1+m +2 p}\)	\(27\)
risch	\(\frac {x^{2 p} b^{p} x^{m} c^{m} x \,{\mathrm e}^{\frac {i \pi \left (-\operatorname {csgn}\left (i c x \right )^{3} m +\operatorname {csgn}\left (i c x \right )^{2} \operatorname {csgn}\left (i c \right ) m +\operatorname {csgn}\left (i c x \right )^{2} \operatorname {csgn}\left (i x \right ) m -\operatorname {csgn}\left (i c x \right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x \right ) m -\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2} p +2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right ) p -\operatorname {csgn}\left (i x^{2}\right )^{3} p +\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i b \,x^{2}\right )^{2} p -\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i b \,x^{2}\right ) \operatorname {csgn}\left (i b \right ) p -\operatorname {csgn}\left (i b \,x^{2}\right )^{3} p +\operatorname {csgn}\left (i b \,x^{2}\right )^{2} \operatorname {csgn}\left (i b \right ) p \right )}{2}}}{1+m +2 p}\)	\(209\)

[In]

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

[Out]

x*(c*x)^m*(b*x^2)^p/(1+m+2*p)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {\left (c x\right )^{m} x e^{\left (2 \, p \log \left (c x\right ) + p \log \left (\frac {b}{c^{2}}\right )\right )}}{m + 2 \, p + 1} \]

[In]

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

[Out]

(c*x)^m*x*e^(2*p*log(c*x) + p*log(b/c^2))/(m + 2*p + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (19) = 38\).

Time = 0.47 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.09 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\begin {cases} \frac {x \left (b x^{2}\right )^{p} \left (c x\right )^{m}}{m + 2 p + 1} & \text {for}\: m \neq - 2 p - 1 \\x \left (b x^{2}\right )^{p} \left (c x\right )^{- 2 p - 1} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x)**m*(b*x**2)**p,x)

[Out]

Piecewise((x*(b*x**2)**p*(c*x)**m/(m + 2*p + 1), Ne(m, -2*p - 1)), (x*(b*x**2)**p*(c*x)**(-2*p - 1)*log(x), Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {b^{p} c^{m} x e^{\left (m \log \left (x\right ) + 2 \, p \log \left (x\right )\right )}}{m + 2 \, p + 1} \]

[In]

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

[Out]

b^p*c^m*x*e^(m*log(x) + 2*p*log(x))/(m + 2*p + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {x e^{\left (p \log \left (b\right ) + m \log \left (c\right ) + m \log \left (x\right ) + 2 \, p \log \left (x\right )\right )}}{m + 2 \, p + 1} \]

[In]

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

[Out]

x*e^(p*log(b) + m*log(c) + m*log(x) + 2*p*log(x))/(m + 2*p + 1)

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int (c x)^m \left (b x^2\right )^p \, dx=\frac {x\,{\left (c\,x\right )}^m\,{\left (b\,x^2\right )}^p}{m+2\,p+1} \]

[In]

int((c*x)^m*(b*x^2)^p,x)

[Out]

(x*(c*x)^m*(b*x^2)^p)/(m + 2*p + 1)