\(\int \frac {(1-x)^n (1+x)^{-n}}{x} \, dx\) [982]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 68 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x} \, dx=-\frac {(1-x)^n (1+x)^{-n} \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1-x}{1+x}\right )}{n}+\frac {2^{-n} (1-x)^n \operatorname {Hypergeometric2F1}\left (n,n,1+n,\frac {1-x}{2}\right )}{n} \]

[Out]

-(1-x)^n*hypergeom([1, n],[1+n],(1-x)/(1+x))/n/((1+x)^n)+(1-x)^n*hypergeom([n, n],[1+n],1/2-1/2*x)/(2^n)/n

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {132, 71, 133} \[ \int \frac {(1-x)^n (1+x)^{-n}}{x} \, dx=\frac {2^{-n} (1-x)^n \operatorname {Hypergeometric2F1}\left (n,n,n+1,\frac {1-x}{2}\right )}{n}-\frac {(1-x)^n (x+1)^{-n} \operatorname {Hypergeometric2F1}\left (1,n,n+1,\frac {1-x}{x+1}\right )}{n} \]

[In]

Int[(1 - x)^n/(x*(1 + x)^n),x]

[Out]

-(((1 - x)^n*Hypergeometric2F1[1, n, 1 + n, (1 - x)/(1 + x)])/(n*(1 + x)^n)) + ((1 - x)^n*Hypergeometric2F1[n,
 n, 1 + n, (1 - x)/2])/(2^n*n)

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
 - a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
 -1]))

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*c - a
*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
 + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\int (1-x)^{-1+n} (1+x)^{-n} \, dx+\int \frac {(1-x)^{-1+n} (1+x)^{-n}}{x} \, dx \\ & = -\frac {(1-x)^n (1+x)^{-n} \, _2F_1\left (1,n;1+n;\frac {1-x}{1+x}\right )}{n}+\frac {2^{-n} (1-x)^n \, _2F_1\left (n,n;1+n;\frac {1-x}{2}\right )}{n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {(1-x)^n (1+x)^{-n}}{x} \, dx=\frac {2^{-n} (1-x)^n (1+x)^{-n} \left (-2^n \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1-x}{1+x}\right )+(1+x)^n \operatorname {Hypergeometric2F1}\left (n,n,1+n,\frac {1-x}{2}\right )\right )}{n} \]

[In]

Integrate[(1 - x)^n/(x*(1 + x)^n),x]

[Out]

((1 - x)^n*(-(2^n*Hypergeometric2F1[1, n, 1 + n, (1 - x)/(1 + x)]) + (1 + x)^n*Hypergeometric2F1[n, n, 1 + n,
(1 - x)/2]))/(2^n*n*(1 + x)^n)

Maple [F]

\[\int \frac {\left (1-x \right )^{n} \left (1+x \right )^{-n}}{x}d x\]

[In]

int((1-x)^n/x/((1+x)^n),x)

[Out]

int((1-x)^n/x/((1+x)^n),x)

Fricas [F]

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

[In]

integrate((1-x)^n/x/((1+x)^n),x, algorithm="fricas")

[Out]

integral((-x + 1)^n/((x + 1)^n*x), x)

Sympy [F]

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

[In]

integrate((1-x)**n/x/((1+x)**n),x)

[Out]

Integral((1 - x)**n/(x*(x + 1)**n), x)

Maxima [F]

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

[In]

integrate((1-x)^n/x/((1+x)^n),x, algorithm="maxima")

[Out]

integrate((-x + 1)^n/((x + 1)^n*x), x)

Giac [F]

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

[In]

integrate((1-x)^n/x/((1+x)^n),x, algorithm="giac")

[Out]

integrate((-x + 1)^n/((x + 1)^n*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-x)^n (1+x)^{-n}}{x} \, dx=\int \frac {{\left (1-x\right )}^n}{x\,{\left (x+1\right )}^n} \,d x \]

[In]

int((1 - x)^n/(x*(x + 1)^n),x)

[Out]

int((1 - x)^n/(x*(x + 1)^n), x)