3.10.78 \(\int (1-x)^n x^3 (1+x)^{-n} \, dx\) [978]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 = {102, 152, 71} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - x)^n*x^3)/(1 + x)^n,x]

[Out]

-1/4*((1 - x)^(1 + n)*x^2*(1 + x)^(1 - n)) - ((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 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 102

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 152

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]

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.06, size = 18, normalized size = 0.17 \begin {gather*} \frac {1}{4} x^4 F_1(4;-n,n;5;x,-x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*AppellF1[4, -n, n, 5, x, -x])/4

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (1-x \right )^{n} x^{3} \left (1+x \right )^{-n}\, dx\]

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 = 0.00, 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^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 = 0.00, 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^3/((1+x)^n),x, algorithm="fricas")

[Out]

integral(x^3*(-x + 1)^n/(x + 1)^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (1 - x\right )^{n} \left (x + 1\right )^{- n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\left (1-x\right )}^n}{{\left (x+1\right )}^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(1 - x)^n)/(x + 1)^n,x)

[Out]

int((x^3*(1 - x)^n)/(x + 1)^n, x)

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