\(\int (x^m (a+b x^{1+m p}))^p \, dx\) [445]

Optimal result

Integrand size = 17, antiderivative size = 44 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (1+p)} \left (a x^m+b x^{1+m+m p}\right )^{1+p}}{b (1+p) (1+m p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2004, 2025} \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (p+1)} \left (a x^m+b x^{m p+m+1}\right )^{p+1}}{b (p+1) (m p+1)} \]

[In]

Int[(x^m*(a + b*x^(1 + m*p)))^p,x]

[Out]

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

Rule 2004

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx \\ & = \frac {x^{-m (1+p)} \left (a x^m+b x^{1+m+m p}\right )^{1+p}}{b (1+p) (1+m p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (1+p)} \left (x^m \left (a+b x^{1+m p}\right )\right )^{1+p}}{b (1+p) (1+m p)} \]

[In]

Integrate[(x^m*(a + b*x^(1 + m*p)))^p,x]

[Out]

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

Maple [F]

\[\int \left (x^{m} \left (a +b \,x^{m p +1}\right )\right )^{p}d x\]

[In]

int((x^m*(a+b*x^(m*p+1)))^p,x)

[Out]

int((x^m*(a+b*x^(m*p+1)))^p,x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {{\left (b x x^{m p + 1} + a x\right )} {\left (b x^{m p + 1} x^{m} + a x^{m}\right )}^{p}}{{\left (b m p^{2} + {\left (b m + b\right )} p + b\right )} x^{m p + 1}} \]

[In]

integrate((x^m*(a+b*x^(m*p+1)))^p,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int \left (x^{m} \left (a + b x^{m p + 1}\right )\right )^{p}\, dx \]

[In]

integrate((x**m*(a+b*x**(m*p+1)))**p,x)

[Out]

Integral((x**m*(a + b*x**(m*p + 1)))**p, x)

Maxima [F]

\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int { \left ({\left (b x^{m p + 1} + a\right )} x^{m}\right )^{p} \,d x } \]

[In]

integrate((x^m*(a+b*x^(m*p+1)))^p,x, algorithm="maxima")

[Out]

integrate(((b*x^(m*p + 1) + a)*x^m)^p, x)

Giac [F]

\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int { \left ({\left (b x^{m p + 1} + a\right )} x^{m}\right )^{p} \,d x } \]

[In]

integrate((x^m*(a+b*x^(m*p+1)))^p,x, algorithm="giac")

[Out]

integrate(((b*x^(m*p + 1) + a)*x^m)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int {\left (x^m\,\left (a+b\,x^{m\,p+1}\right )\right )}^p \,d x \]

[In]

int((x^m*(a + b*x^(m*p + 1)))^p,x)

[Out]

int((x^m*(a + b*x^(m*p + 1)))^p, x)