\(\int (1-x)^n x (1+x)^{-n} \, dx\) [980]

Optimal result

Integrand size = 16, antiderivative size = 61 \[ \int (1-x)^n x (1+x)^{-n} \, dx=-\frac {1}{2} (1-x)^{1+n} (1+x)^{1-n}+\frac {2^{-n} n (1-x)^{1+n} \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {1-x}{2}\right )}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {81, 71} \[ \int (1-x)^n x (1+x)^{-n} \, dx=\frac {2^{-n} n (1-x)^{n+1} \operatorname {Hypergeometric2F1}\left (n,n+1,n+2,\frac {1-x}{2}\right )}{n+1}-\frac {1}{2} (1-x)^{n+1} (x+1)^{1-n} \]

[In]

Int[((1 - x)^n*x)/(1 + x)^n,x]

[Out]

-1/2*((1 - x)^(1 + n)*(1 + x)^(1 - n)) + (n*(1 - x)^(1 + n)*Hypergeometric2F1[n, 1 + n, 2 + n, (1 - x)/2])/(2^
n*(1 + 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 81

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int (1-x)^n x (1+x)^{-n} \, dx=\frac {1}{2} (1-x)^{1+n} \left (-(1+x)^{1-n}+\frac {2^{1-n} n \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {1-x}{2}\right )}{1+n}\right ) \]

[In]

Integrate[((1 - x)^n*x)/(1 + x)^n,x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x*(-x + 1)^n/(x + 1)^n, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x*(1 - x)**n/(x + 1)**n, x)

Maxima [F]

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

[In]

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

[Out]

integrate(x*(-x + 1)^n/(x + 1)^n, x)

Giac [F]

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

[In]

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

[Out]

integrate(x*(-x + 1)^n/(x + 1)^n, x)

Mupad [F(-1)]

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

[In]

int((x*(1 - x)^n)/(x + 1)^n,x)

[Out]

int((x*(1 - x)^n)/(x + 1)^n, x)