\(\int x^{-1+n} (a+b x)^{-1-n} \, dx\) [753]

Optimal result

Integrand size = 17, antiderivative size = 19 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {x^n (a+b x)^{-n}}{a n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {37} \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {x^n (a+b x)^{-n}}{a n} \]

[In]

Int[x^(-1 + n)*(a + b*x)^(-1 - n),x]

[Out]

x^n/(a*n*(a + b*x)^n)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {x^n (a+b x)^{-n}}{a n} \]

[In]

Integrate[x^(-1 + n)*(a + b*x)^(-1 - n),x]

[Out]

x^n/(a*n*(a + b*x)^n)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {x^{n} \left (b x +a \right )^{-n}}{a n}\)	\(20\)
parallelrisch	\(\frac {x^{2} x^{-1+n} \left (b x +a \right )^{-1-n} b +x \,x^{-1+n} \left (b x +a \right )^{-1-n} a}{a n}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {{\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{-n - 1} x^{n - 1}}{a n} \]

[In]

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

[Out]

(b*x^2 + a*x)*(b*x + a)^(-n - 1)*x^(n - 1)/(a*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (12) = 24\).

Time = 1.53 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {a^{n} a^{- 2 n - 1} x^{n} \left (1 + \frac {b x}{a}\right )^{- n} \Gamma \left (n\right )}{\Gamma \left (n + 1\right )} \]

[In]

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

[Out]

a**n*a**(-2*n - 1)*x**n*gamma(n)/((1 + b*x/a)**n*gamma(n + 1))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {e^{\left (-n \log \left (b x + a\right ) + n \log \left (x\right )\right )}}{a n} \]

[In]

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

[Out]

e^(-n*log(b*x + a) + n*log(x))/(a*n)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} (a+b x)^{-1-n} \, dx=\frac {x^n}{a\,n\,{\left (a+b\,x\right )}^n} \]

[In]

int(x^(n - 1)/(a + b*x)^(n + 1),x)

[Out]

x^n/(a*n*(a + b*x)^n)