\(\int x^{-1+k} (a+b x^k)^n \, dx\) [89]

Optimal result

Integrand size = 15, antiderivative size = 23 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {\left (a+b x^k\right )^{1+n}}{b k (1+n)} \]

[Out]

(a+b*x^k)^(1+n)/b/k/(1+n)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {\left (a+b x^k\right )^{n+1}}{b k (n+1)} \]

[In]

Int[x^(-1 + k)*(a + b*x^k)^n,x]

[Out]

(a + b*x^k)^(1 + n)/(b*k*(1 + n))

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x^k\right )^{1+n}}{b k (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {\left (a+b x^k\right )^{1+n}}{b k (1+n)} \]

[In]

Integrate[x^(-1 + k)*(a + b*x^k)^n,x]

[Out]

(a + b*x^k)^(1 + n)/(b*k*(1 + n))

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26

method	result	size
risch	\(\frac {\left (a +b \,x^{k}\right ) \left (a +b \,x^{k}\right )^{n}}{b \left (1+n \right ) k}\)	\(29\)

[In]

int(x^(-1+k)*(a+b*x^k)^n,x,method=_RETURNVERBOSE)

[Out]

(a+b*x^k)/b/(1+n)/k*(a+b*x^k)^n

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {{\left (b x^{k} + a\right )} {\left (b x^{k} + a\right )}^{n}}{b k n + b k} \]

[In]

integrate(x^(-1+k)*(a+b*x^k)^n,x, algorithm="fricas")

[Out]

(b*x^k + a)*(b*x^k + a)^n/(b*k*n + b*k)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (15) = 30\).

Time = 12.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.39 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge k = 0 \wedge n = -1 \\\frac {a^{n} x x^{k - 1}}{k} & \text {for}\: b = 0 \\\left (a + b\right )^{n} \log {\left (x \right )} & \text {for}\: k = 0 \\\frac {\log {\left (\frac {a}{b} + x^{k} \right )}}{b k} & \text {for}\: n = -1 \\\frac {a \left (a + b x^{k}\right )^{n}}{b k n + b k} + \frac {b x^{k} \left (a + b x^{k}\right )^{n}}{b k n + b k} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1+k)*(a+b*x**k)**n,x)

[Out]

Piecewise((log(x)/a, Eq(b, 0) & Eq(k, 0) & Eq(n, -1)), (a**n*x*x**(k - 1)/k, Eq(b, 0)), ((a + b)**n*log(x), Eq
(k, 0)), (log(a/b + x**k)/(b*k), Eq(n, -1)), (a*(a + b*x**k)**n/(b*k*n + b*k) + b*x**k*(a + b*x**k)**n/(b*k*n
+ b*k), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {{\left (b x^{k} + a\right )}^{n + 1}}{b k {\left (n + 1\right )}} \]

[In]

integrate(x^(-1+k)*(a+b*x^k)^n,x, algorithm="maxima")

[Out]

(b*x^k + a)^(n + 1)/(b*k*(n + 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {{\left (b x^{k} + a\right )}^{n + 1}}{b k {\left (n + 1\right )}} \]

[In]

integrate(x^(-1+k)*(a+b*x^k)^n,x, algorithm="giac")

[Out]

(b*x^k + a)^(n + 1)/(b*k*(n + 1))

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int x^{-1+k} \left (a+b x^k\right )^n \, dx=\frac {{\left (a+b\,x^k\right )}^{n+1}}{b\,k\,\left (n+1\right )} \]

[In]

int(x^(k - 1)*(a + b*x^k)^n,x)

[Out]

(a + b*x^k)^(n + 1)/(b*k*(n + 1))