3.990 \(\int x^m (3-2 a x)^{2+n} (6+4 a x)^n \, dx\)

Optimal. Leaf size=151 \[ \frac{2^n 9^{n+1} x^{m+1} \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{4 a^2 x^2}{9}\right )}{m+1}-\frac{a 2^{n+2} 3^{2 n+1} x^{m+2} \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{4 a^2 x^2}{9}\right )}{m+2}+\frac{a^2 2^{n+2} 9^n x^{m+3} \, _2F_1\left (\frac{m+3}{2},-n;\frac{m+5}{2};\frac{4 a^2 x^2}{9}\right )}{m+3} \]

[Out]

(2^n*9^(1 + n)*x^(1 + m)*Hypergeometric2F1[(1 + m)/2, -n, (3 + m)/2, (4*a^2*x^2)/9])/(1 + m) - (2^(2 + n)*3^(1
+ 2*n)*a*x^(2 + m)*Hypergeometric2F1[(2 + m)/2, -n, (4 + m)/2, (4*a^2*x^2)/9])/(2 + m) + (2^(2 + n)*9^n*a^2*x
^(3 + m)*Hypergeometric2F1[(3 + m)/2, -n, (5 + m)/2, (4*a^2*x^2)/9])/(3 + m)

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Rubi [A]  time = 0.096674, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {127, 125, 364} \[ \frac{2^n 9^{n+1} x^{m+1} \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{4 a^2 x^2}{9}\right )}{m+1}-\frac{a 2^{n+2} 3^{2 n+1} x^{m+2} \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{4 a^2 x^2}{9}\right )}{m+2}+\frac{a^2 2^{n+2} 9^n x^{m+3} \, _2F_1\left (\frac{m+3}{2},-n;\frac{m+5}{2};\frac{4 a^2 x^2}{9}\right )}{m+3} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(3 - 2*a*x)^(2 + n)*(6 + 4*a*x)^n,x]

[Out]

(2^n*9^(1 + n)*x^(1 + m)*Hypergeometric2F1[(1 + m)/2, -n, (3 + m)/2, (4*a^2*x^2)/9])/(1 + m) - (2^(2 + n)*3^(1
+ 2*n)*a*x^(2 + m)*Hypergeometric2F1[(2 + m)/2, -n, (4 + m)/2, (4*a^2*x^2)/9])/(2 + m) + (2^(2 + n)*9^n*a^2*x
^(3 + m)*Hypergeometric2F1[(3 + m)/2, -n, (5 + m)/2, (4*a^2*x^2)/9])/(3 + m)

Rule 127

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

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 (3-2 a x)^{2+n} (6+4 a x)^n \, dx &=\int \left (9 x^m (3-2 a x)^n (6+4 a x)^n-12 a x^{1+m} (3-2 a x)^n (6+4 a x)^n+4 a^2 x^{2+m} (3-2 a x)^n (6+4 a x)^n\right ) \, dx\\ &=9 \int x^m (3-2 a x)^n (6+4 a x)^n \, dx-(12 a) \int x^{1+m} (3-2 a x)^n (6+4 a x)^n \, dx+\left (4 a^2\right ) \int x^{2+m} (3-2 a x)^n (6+4 a x)^n \, dx\\ &=9 \int x^m \left (18-8 a^2 x^2\right )^n \, dx-(12 a) \int x^{1+m} \left (18-8 a^2 x^2\right )^n \, dx+\left (4 a^2\right ) \int x^{2+m} \left (18-8 a^2 x^2\right )^n \, dx\\ &=\frac{2^n 9^{1+n} x^{1+m} \, _2F_1\left (\frac{1+m}{2},-n;\frac{3+m}{2};\frac{4 a^2 x^2}{9}\right )}{1+m}-\frac{2^{2+n} 3^{1+2 n} a x^{2+m} \, _2F_1\left (\frac{2+m}{2},-n;\frac{4+m}{2};\frac{4 a^2 x^2}{9}\right )}{2+m}+\frac{2^{2+n} 9^n a^2 x^{3+m} \, _2F_1\left (\frac{3+m}{2},-n;\frac{5+m}{2};\frac{4 a^2 x^2}{9}\right )}{3+m}\\ \end{align*}

Mathematica [A]  time = 0.133098, size = 166, normalized size = 1.1 \[ \frac{x^{m+1} \left (9-4 a^2 x^2\right )^n \left (\frac{1}{2}-\frac{2 a^2 x^2}{9}\right )^{-n} \left (9 \left (m^2+5 m+6\right ) \, _2F_1\left (\frac{m+1}{2},-n;\frac{m+3}{2};\frac{4 a^2 x^2}{9}\right )-4 a (m+1) x \left (3 (m+3) \, _2F_1\left (\frac{m+2}{2},-n;\frac{m+4}{2};\frac{4 a^2 x^2}{9}\right )-a (m+2) x \, _2F_1\left (\frac{m+3}{2},-n;\frac{m+5}{2};\frac{4 a^2 x^2}{9}\right )\right )\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(3 - 2*a*x)^(2 + n)*(6 + 4*a*x)^n,x]

[Out]

(x^(1 + m)*(9 - 4*a^2*x^2)^n*(9*(6 + 5*m + m^2)*Hypergeometric2F1[(1 + m)/2, -n, (3 + m)/2, (4*a^2*x^2)/9] - 4
*a*(1 + m)*x*(3*(3 + m)*Hypergeometric2F1[(2 + m)/2, -n, (4 + m)/2, (4*a^2*x^2)/9] - a*(2 + m)*x*Hypergeometri
c2F1[(3 + m)/2, -n, (5 + m)/2, (4*a^2*x^2)/9])))/((1 + m)*(2 + m)*(3 + m)*(1/2 - (2*a^2*x^2)/9)^n)

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Maple [F]  time = 0.148, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( -2\,ax+3 \right ) ^{2+n} \left ( 4\,ax+6 \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((4*a*x + 6)^n*(-2*a*x + 3)^(n + 2)*x^m, 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(x**m*(-2*a*x+3)**(2+n)*(4*a*x+6)**n,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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