3.987 \(\int x^m (1-a x)^n (2+2 a x)^n \, dx\)

Optimal. Leaf size=39 \[ \frac {2^n x^{m+1} \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};a^2 x^2\right )}{m+1} \]

[Out]

2^n*x^(1+m)*hypergeom([-n, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {125, 364} \[ \frac {2^n x^{m+1} \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};a^2 x^2\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(1 - a*x)^n*(2 + 2*a*x)^n,x]

[Out]

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

Rule 125

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.00 \[ \frac {2^n x^{m+1} \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};a^2 x^2\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, a x + 2\right )}^{n} {\left (-a x + 1\right )}^{n} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (2 \, a x + 2\right )}^{n} {\left (-a x + 1\right )}^{n} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int x^{m} \left (-a x +1\right )^{n} \left (2 a x +2\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (2 \, a x + 2\right )}^{n} {\left (-a x + 1\right )}^{n} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int x^m\,{\left (1-a\,x\right )}^n\,{\left (2\,a\,x+2\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(1 - a*x)^n*(2*a*x + 2)^n,x)

[Out]

int(x^m*(1 - a*x)^n*(2*a*x + 2)^n, x)

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sympy [C]  time = 32.18, size = 209, normalized size = 5.36 \[ - \frac {2^{n} a^{- m} {G_{6, 6}^{5, 3}\left (\begin {matrix} - \frac {m}{2} - \frac {n}{2}, - \frac {m}{2} - \frac {n}{2} + \frac {1}{2}, 1 & \frac {1}{2} - \frac {m}{2}, - \frac {m}{2} - n, - \frac {m}{2} - n + \frac {1}{2} \\- \frac {m}{2} - n - \frac {1}{2}, - \frac {m}{2} - n, - \frac {m}{2} - \frac {n}{2}, - \frac {m}{2} - n + \frac {1}{2}, - \frac {m}{2} - \frac {n}{2} + \frac {1}{2} & 0 \end {matrix} \middle | {\frac {1}{a^{2} x^{2}}} \right )} e^{i \pi n}}{4 \pi a \Gamma \left (- n\right )} + \frac {2^{n} a^{- m} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {m}{2} - \frac {1}{2}, - \frac {m}{2}, \frac {1}{2} - \frac {m}{2}, - \frac {m}{2} - \frac {n}{2} - \frac {1}{2}, - \frac {m}{2} - \frac {n}{2}, 1 & \\- \frac {m}{2} - \frac {n}{2} - \frac {1}{2}, - \frac {m}{2} - \frac {n}{2} & - \frac {m}{2} - \frac {1}{2}, - \frac {m}{2}, - \frac {m}{2} - n - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{a^{2} x^{2}}} \right )} e^{- i \pi m}}{4 \pi a \Gamma \left (- n\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2**n*a**(-m)*meijerg(((-m/2 - n/2, -m/2 - n/2 + 1/2, 1), (1/2 - m/2, -m/2 - n, -m/2 - n + 1/2)), ((-m/2 - n -
1/2, -m/2 - n, -m/2 - n/2, -m/2 - n + 1/2, -m/2 - n/2 + 1/2), (0,)), 1/(a**2*x**2))*exp(I*pi*n)/(4*pi*a*gamma
(-n)) + 2**n*a**(-m)*meijerg(((-m/2 - 1/2, -m/2, 1/2 - m/2, -m/2 - n/2 - 1/2, -m/2 - n/2, 1), ()), ((-m/2 - n/
2 - 1/2, -m/2 - n/2), (-m/2 - 1/2, -m/2, -m/2 - n - 1/2, 0)), exp_polar(-2*I*pi)/(a**2*x**2))*exp(-I*pi*m)/(4*
pi*a*gamma(-n))

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