3.978 \(\int (1-x)^n x^3 (1+x)^{-n} \, dx\)

Optimal. Leaf size=105 \[ \frac{2^{-n} n \left (n^2+2\right ) (1-x)^{n+1} \, _2F_1\left (n,n+1;n+2;\frac{1-x}{2}\right )}{3 (n+1)}-\frac{1}{12} (1-x)^{n+1} \left (2 n^2-2 n x+3\right ) (x+1)^{1-n}-\frac{1}{4} x^2 (1-x)^{n+1} (x+1)^{1-n} \]

[Out]

-((1 - x)^(1 + n)*x^2*(1 + x)^(1 - n))/4 - ((1 - x)^(1 + n)*(1 + x)^(1 - n)*(3 + 2*n^2 - 2*n*x))/12 + (n*(2 +
n^2)*(1 - x)^(1 + n)*Hypergeometric2F1[n, 1 + n, 2 + n, (1 - x)/2])/(3*2^n*(1 + n))

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Rubi [A]  time = 0.0798303, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {100, 147, 69} \[ \frac{2^{-n} n \left (n^2+2\right ) (1-x)^{n+1} \, _2F_1\left (n,n+1;n+2;\frac{1-x}{2}\right )}{3 (n+1)}-\frac{1}{12} (1-x)^{n+1} \left (2 n^2-2 n x+3\right ) (x+1)^{1-n}-\frac{1}{4} x^2 (1-x)^{n+1} (x+1)^{1-n} \]

Antiderivative was successfully verified.

[In]

Int[((1 - x)^n*x^3)/(1 + x)^n,x]

[Out]

-((1 - x)^(1 + n)*x^2*(1 + x)^(1 - n))/4 - ((1 - x)^(1 + n)*(1 + x)^(1 - n)*(3 + 2*n^2 - 2*n*x))/12 + (n*(2 +
n^2)*(1 - x)^(1 + n)*Hypergeometric2F1[n, 1 + n, 2 + n, (1 - x)/2])/(3*2^n*(1 + n))

Rule 100

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

\begin{align*} \int (1-x)^n x^3 (1+x)^{-n} \, dx &=-\frac{1}{4} (1-x)^{1+n} x^2 (1+x)^{1-n}-\frac{1}{4} \int (1-x)^n x (1+x)^{-n} (-2+2 n x) \, dx\\ &=-\frac{1}{4} (1-x)^{1+n} x^2 (1+x)^{1-n}-\frac{1}{12} (1-x)^{1+n} (1+x)^{1-n} \left (3+2 n^2-2 n x\right )-\frac{1}{3} \left (n \left (2+n^2\right )\right ) \int (1-x)^n (1+x)^{-n} \, dx\\ &=-\frac{1}{4} (1-x)^{1+n} x^2 (1+x)^{1-n}-\frac{1}{12} (1-x)^{1+n} (1+x)^{1-n} \left (3+2 n^2-2 n x\right )+\frac{2^{-n} n \left (2+n^2\right ) (1-x)^{1+n} \, _2F_1\left (n,1+n;2+n;\frac{1-x}{2}\right )}{3 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.116, size = 128, normalized size = 1.22 \[ \frac{2^{-n-2} (x-1) (1-x)^n (x+1)^{-n} \left ((n+1) \left (2^n x^2 (x+1)-2 (x+1)^n \, _2F_1\left (n,n+1;n+2;\frac{1-x}{2}\right )\right )-8 n (x+1)^n \, _2F_1\left (n-2,n+1;n+2;\frac{1-x}{2}\right )+4 (2 n+1) (x+1)^n \, _2F_1\left (n-1,n+1;n+2;\frac{1-x}{2}\right )\right )}{n+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-2 - n)*(1 - x)^n*(-1 + x)*(-8*n*(1 + x)^n*Hypergeometric2F1[-2 + n, 1 + n, 2 + n, (1 - x)/2] + 4*(1 + 2*n
)*(1 + x)^n*Hypergeometric2F1[-1 + n, 1 + n, 2 + n, (1 - x)/2] + (1 + n)*(2^n*x^2*(1 + x) - 2*(1 + x)^n*Hyperg
eometric2F1[n, 1 + n, 2 + n, (1 - x)/2])))/((1 + n)*(1 + x)^n)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3} \left ( 1-x \right ) ^{n}}{ \left ( 1+x \right ) ^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)^n*x^3/((1+x)^n),x)

[Out]

int((1-x)^n*x^3/((1+x)^n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n*x^3/((1+x)^n),x, algorithm="maxima")

[Out]

integrate(x^3*(-x + 1)^n/(x + 1)^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3}{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n*x^3/((1+x)^n),x, algorithm="fricas")

[Out]

integral(x^3*(-x + 1)^n/(x + 1)^n, 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**3/((1+x)**n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}{\left (-x + 1\right )}^{n}}{{\left (x + 1\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^n*x^3/((1+x)^n),x, algorithm="giac")

[Out]

integrate(x^3*(-x + 1)^n/(x + 1)^n, x)