3.985 \(\int \frac{(1-x)^n (1+x)^{-n}}{x^4} \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 \left (2 n^2+1\right ) (1-x)^{n+1} (x+1)^{-n-1} \, _2F_1\left (2,n+1;n+2;\frac{1-x}{x+1}\right )}{3 (n+1)}+\frac{n (1-x)^{n+1} (x+1)^{1-n}}{3 x^2}-\frac{(1-x)^{n+1} (x+1)^{1-n}}{3 x^3} \]

[Out]

-((1 - x)^(1 + n)*(1 + x)^(1 - n))/(3*x^3) + (n*(1 - x)^(1 + n)*(1 + x)^(1 - n))/(3*x^2) - (2*(1 + 2*n^2)*(1 -
x)^(1 + n)*(1 + x)^(-1 - n)*Hypergeometric2F1[2, 1 + n, 2 + n, (1 - x)/(1 + x)])/(3*(1 + n))

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Rubi [A]  time = 0.0421368, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {129, 151, 12, 131} \[ -\frac{2 \left (2 n^2+1\right ) (1-x)^{n+1} (x+1)^{-n-1} \, _2F_1\left (2,n+1;n+2;\frac{1-x}{x+1}\right )}{3 (n+1)}+\frac{n (1-x)^{n+1} (x+1)^{1-n}}{3 x^2}-\frac{(1-x)^{n+1} (x+1)^{1-n}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^n/(x^4*(1 + x)^n),x]

[Out]

-((1 - x)^(1 + n)*(1 + x)^(1 - n))/(3*x^3) + (n*(1 - x)^(1 + n)*(1 + x)^(1 - n))/(3*x^2) - (2*(1 + 2*n^2)*(1 -
x)^(1 + n)*(1 + x)^(-1 - n)*Hypergeometric2F1[2, 1 + n, 2 + n, (1 - x)/(1 + x)])/(3*(1 + n))

Rule 129

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int \frac{(1-x)^n (1+x)^{-n}}{x^4} \, dx &=-\frac{(1-x)^{1+n} (1+x)^{1-n}}{3 x^3}-\frac{1}{3} \int \frac{(1-x)^n (2 n-x) (1+x)^{-n}}{x^3} \, dx\\ &=-\frac{(1-x)^{1+n} (1+x)^{1-n}}{3 x^3}+\frac{n (1-x)^{1+n} (1+x)^{1-n}}{3 x^2}+\frac{1}{6} \int \frac{\left (2+4 n^2\right ) (1-x)^n (1+x)^{-n}}{x^2} \, dx\\ &=-\frac{(1-x)^{1+n} (1+x)^{1-n}}{3 x^3}+\frac{n (1-x)^{1+n} (1+x)^{1-n}}{3 x^2}+\frac{1}{3} \left (1+2 n^2\right ) \int \frac{(1-x)^n (1+x)^{-n}}{x^2} \, dx\\ &=-\frac{(1-x)^{1+n} (1+x)^{1-n}}{3 x^3}+\frac{n (1-x)^{1+n} (1+x)^{1-n}}{3 x^2}-\frac{2 \left (1+2 n^2\right ) (1-x)^{1+n} (1+x)^{-1-n} \, _2F_1\left (2,1+n;2+n;\frac{1-x}{1+x}\right )}{3 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0313294, size = 77, normalized size = 0.73 \[ -\frac{(1-x)^{n+1} (x+1)^{-n-1} \left (2 \left (2 n^2+1\right ) x^3 \, _2F_1\left (2,n+1;n+2;\frac{1-x}{x+1}\right )-(n+1) (x+1)^2 (n x-1)\right )}{3 (n+1) x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^n/(x^4*(1 + x)^n),x]

[Out]

-((1 - x)^(1 + n)*(1 + x)^(-1 - n)*(-((1 + n)*(1 + x)^2*(-1 + n*x)) + 2*(1 + 2*n^2)*x^3*Hypergeometric2F1[2, 1
+ n, 2 + n, (1 - x)/(1 + x)]))/(3*(1 + n)*x^3)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 1-x \right ) ^{n}}{{x}^{4} \left ( 1+x \right ) ^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)^n/x^4/((1+x)^n),x)

[Out]

int((1-x)^n/x^4/((1+x)^n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n/x^4/((1+x)^n),x, algorithm="maxima")

[Out]

integrate((-x + 1)^n/((x + 1)^n*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n/x^4/((1+x)^n),x, algorithm="fricas")

[Out]

integral((-x + 1)^n/((x + 1)^n*x^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)**n/x**4/((1+x)**n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n/x^4/((1+x)^n),x, algorithm="giac")

[Out]

integrate((-x + 1)^n/((x + 1)^n*x^4), x)