3.10.85 \(\int \frac {(1-x)^n (1+x)^{-n}}{x^4} \, dx\) [985]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 = {105, 156, 12, 133} \begin {gather*} -\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 {(1-x)^{n+1} (x+1)^{1-n}}{3 x^3}+\frac {n (1-x)^{n+1} (x+1)^{1-n}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x)^n/(x^4*(1 + x)^n),x]

[Out]

-1/3*((1 - x)^(1 + n)*(1 + x)^(1 - n))/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 12

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

Rule 105

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, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 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]

Rule 156

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] && ILtQ[m, -1]

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*}

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Mathematica [A]
time = 0.05, size = 77, normalized size = 0.73 \begin {gather*} -\frac {(1-x)^{1+n} (1+x)^{-1-n} \left (-\left ((1+n) (1+x)^2 (-1+n x)\right )+2 \left (1+2 n^2\right ) x^3 \, _2F_1\left (2,1+n;2+n;\frac {1-x}{1+x}\right )\right )}{3 (1+n) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((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)]))/((1 + n)*x^3)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (1-x \right )^{n} \left (1+x \right )^{-n}}{x^{4}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1 - x\right )^{n} \left (x + 1\right )^{- n}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-x\right )}^n}{x^4\,{\left (x+1\right )}^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x)^n/(x^4*(x + 1)^n),x)

[Out]

int((1 - x)^n/(x^4*(x + 1)^n), x)

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