\(\int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx\) [1001]

Optimal result

Integrand size = 25, antiderivative size = 106 \[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=-\frac {2 a (1-a x)^{1-n} (1+a x)^{-1+n} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,\frac {1-a x}{1+a x}\right )}{1-n}+\frac {2^n a (1-a x)^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,-n,2-n,\frac {1}{2} (1-a x)\right )}{1-n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {130, 71, 133} \[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\frac {a 2^n (1-a x)^{1-n} \operatorname {Hypergeometric2F1}\left (1-n,-n,2-n,\frac {1}{2} (1-a x)\right )}{1-n}-\frac {2 a (1-a x)^{1-n} (a x+1)^{n-1} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,\frac {1-a x}{a x+1}\right )}{1-n} \]

[In]

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

[Out]

(-2*a*(1 - a*x)^(1 - n)*(1 + a*x)^(-1 + n)*Hypergeometric2F1[2, 1 - n, 2 - n, (1 - a*x)/(1 + a*x)])/(1 - n) +
(2^n*a*(1 - a*x)^(1 - n)*Hypergeometric2F1[1 - n, -n, 2 - n, (1 - a*x)/2])/(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 130

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.21 \[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=-\frac {\left (1-\frac {1}{a x}\right )^n \left (1+\frac {1}{a x}\right )^{-n} (1-a x)^{-n} (1+a x)^n \operatorname {AppellF1}\left (1,n,-n,2,\frac {1}{a x},-\frac {1}{a x}\right )}{x}-\frac {a (1-a x)^{-n} (1+a x)^{1+n} \left (1+\frac {1}{2} (-1-a x)\right )^n \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {1}{2} (1+a x)\right )}{1+n} \]

[In]

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

[Out]

-(((1 - 1/(a*x))^n*(1 + a*x)^n*AppellF1[1, n, -n, 2, 1/(a*x), -(1/(a*x))])/((1 + 1/(a*x))^n*x*(1 - a*x)^n)) -
(a*(1 + a*x)^(1 + n)*(1 + (-1 - a*x)/2)^n*Hypergeometric2F1[n, 1 + n, 2 + n, (1 + a*x)/2])/((1 + n)*(1 - a*x)^
n)

Maple [F]

\[\int \frac {\left (-a x +1\right )^{1-n} \left (a x +1\right )^{1+n}}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\int { \frac {{\left (a x + 1\right )}^{n + 1} {\left (-a x + 1\right )}^{-n + 1}}{x^{2}} \,d x } \]

[In]

integrate((-a*x+1)^(1-n)*(a*x+1)^(1+n)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\int \frac {\left (- a x + 1\right )^{1 - n} \left (a x + 1\right )^{n + 1}}{x^{2}}\, dx \]

[In]

integrate((-a*x+1)**(1-n)*(a*x+1)**(1+n)/x**2,x)

[Out]

Integral((-a*x + 1)**(1 - n)*(a*x + 1)**(n + 1)/x**2, x)

Maxima [F]

\[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\int { \frac {{\left (a x + 1\right )}^{n + 1} {\left (-a x + 1\right )}^{-n + 1}}{x^{2}} \,d x } \]

[In]

integrate((-a*x+1)^(1-n)*(a*x+1)^(1+n)/x^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\int { \frac {{\left (a x + 1\right )}^{n + 1} {\left (-a x + 1\right )}^{-n + 1}}{x^{2}} \,d x } \]

[In]

integrate((-a*x+1)^(1-n)*(a*x+1)^(1+n)/x^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-a x)^{1-n} (1+a x)^{1+n}}{x^2} \, dx=\int \frac {{\left (1-a\,x\right )}^{1-n}\,{\left (a\,x+1\right )}^{n+1}}{x^2} \,d x \]

[In]

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

[Out]

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