3.1235 \(\int (3-6 x)^m (2+4 x)^m \, dx\)

Optimal. Leaf size=20 \[ 6^m x \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};4 x^2\right ) \]

[Out]

6^m*x*Hypergeometric2F1[1/2, -m, 3/2, 4*x^2]

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Rubi [A]  time = 0.0057414, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {41, 245} \[ 6^m x \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};4 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(3 - 6*x)^m*(2 + 4*x)^m,x]

[Out]

6^m*x*Hypergeometric2F1[1/2, -m, 3/2, 4*x^2]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 245

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

Rubi steps

\begin{align*} \int (3-6 x)^m (2+4 x)^m \, dx &=\int \left (6-24 x^2\right )^m \, dx\\ &=6^m x \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};4 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0049051, size = 20, normalized size = 1. \[ 6^m x \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};4 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 - 6*x)^m*(2 + 4*x)^m,x]

[Out]

6^m*x*Hypergeometric2F1[1/2, -m, 3/2, 4*x^2]

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Maple [F]  time = 0.073, size = 0, normalized size = 0. \begin{align*} \int \left ( 3-6\,x \right ) ^{m} \left ( 2+4\,x \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3-6*x)^m*(2+4*x)^m,x)

[Out]

int((3-6*x)^m*(2+4*x)^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (4 \, x + 2\right )}^{m}{\left (-6 \, x + 3\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-6*x)^m*(4*x+2)^m,x, algorithm="maxima")

[Out]

integrate((4*x + 2)^m*(-6*x + 3)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (4 \, x + 2\right )}^{m}{\left (-6 \, x + 3\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-6*x)^m*(4*x+2)^m,x, algorithm="fricas")

[Out]

integral((4*x + 2)^m*(-6*x + 3)^m, x)

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Sympy [C]  time = 4.61746, size = 42, normalized size = 2.1 \begin{align*} \frac{24^{m} \left (x + \frac{1}{2}\right ) \left (x + \frac{1}{2}\right )^{m} \Gamma \left (m + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - m, m + 1 \\ m + 2 \end{matrix}\middle |{\left (x + \frac{1}{2}\right ) e^{2 i \pi }} \right )}}{\Gamma \left (m + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-6*x)**m*(4*x+2)**m,x)

[Out]

24**m*(x + 1/2)*(x + 1/2)**m*gamma(m + 1)*hyper((-m, m + 1), (m + 2,), (x + 1/2)*exp_polar(2*I*pi))/gamma(m +
2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (4 \, x + 2\right )}^{m}{\left (-6 \, x + 3\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-6*x)^m*(4*x+2)^m,x, algorithm="giac")

[Out]

integrate((4*x + 2)^m*(-6*x + 3)^m, x)