3.991 \(\int x^m (3-2 a x)^{1+n} (6+4 a x)^n \, dx\)

Optimal. Leaf size=99 \[ \frac {2^n 3^{2 n+1} x^{m+1} \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};\frac {4 a^2 x^2}{9}\right )}{m+1}-\frac {a 2^{n+1} 9^n x^{m+2} \, _2F_1\left (\frac {m+2}{2},-n;\frac {m+4}{2};\frac {4 a^2 x^2}{9}\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {82, 125, 364} \[ \frac {2^n 3^{2 n+1} x^{m+1} \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};\frac {4 a^2 x^2}{9}\right )}{m+1}-\frac {a 2^{n+1} 9^n x^{m+2} \, _2F_1\left (\frac {m+2}{2},-n;\frac {m+4}{2};\frac {4 a^2 x^2}{9}\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(3 - 2*a*x)^(1 + n)*(6 + 4*a*x)^n,x]

[Out]

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

Rule 82

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[a, Int[(a + b*
x)^n*(c + d*x)^n*(f*x)^p, x], x] + Dist[b/f, Int[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b,
c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] &&  !IGtQ[m, 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)^{1+n} (6+4 a x)^n \, dx &=3 \int x^m (3-2 a x)^n (6+4 a x)^n \, dx-(2 a) \int x^{1+m} (3-2 a x)^n (6+4 a x)^n \, dx\\ &=3 \int x^m \left (18-8 a^2 x^2\right )^n \, dx-(2 a) \int x^{1+m} \left (18-8 a^2 x^2\right )^n \, dx\\ &=\frac {2^n 3^{1+2 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^{1+n} 9^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}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 115, normalized size = 1.16 \[ \frac {x^{m+1} \left (9-4 a^2 x^2\right )^n \left (\frac {1}{2}-\frac {2 a^2 x^2}{9}\right )^{-n} \left (3 (m+2) \, _2F_1\left (\frac {m+1}{2},-n;\frac {m+3}{2};\frac {4 a^2 x^2}{9}\right )-2 a (m+1) x \, _2F_1\left (\frac {m}{2}+1,-n;\frac {m}{2}+2;\frac {4 a^2 x^2}{9}\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(3 - 2*a*x)^(1 + n)*(6 + 4*a*x)^n,x]

[Out]

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

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n + 1} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-2*a*x+3)^(1+n)*(4*a*x+6)^n,x, algorithm="fricas")

[Out]

integral((4*a*x + 6)^n*(-2*a*x + 3)^(n + 1)*x^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n + 1} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-2*a*x+3)^(1+n)*(4*a*x+6)^n,x, algorithm="giac")

[Out]

integrate((4*a*x + 6)^n*(-2*a*x + 3)^(n + 1)*x^m, x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int x^{m} \left (-2 a x +3\right )^{n +1} \left (4 a x +6\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(-2*a*x+3)^(n+1)*(4*a*x+6)^n,x)

[Out]

int(x^m*(-2*a*x+3)^(n+1)*(4*a*x+6)^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n + 1} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(-2*a*x+3)^(1+n)*(4*a*x+6)^n,x, algorithm="maxima")

[Out]

integrate((4*a*x + 6)^n*(-2*a*x + 3)^(n + 1)*x^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,{\left (3-2\,a\,x\right )}^{n+1}\,{\left (4\,a\,x+6\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(3 - 2*a*x)^(n + 1)*(4*a*x + 6)^n,x)

[Out]

int(x^m*(3 - 2*a*x)^(n + 1)*(4*a*x + 6)^n, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(-2*a*x+3)**(1+n)*(4*a*x+6)**n,x)

[Out]

Timed out

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