3.3072 \(\int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx\)

Optimal. Leaf size=121 \[ \frac{2^{-m-1} \left (2 m^2-86 m+441\right ) (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{3 (1-m) m}-\frac{1}{3} (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m}-\frac{7 (21-m) (3 x+2)^{m+1} (2 x+1)^{-m}}{3 m} \]

[Out]

(-7*(21 - m)*(2 + 3*x)^(1 + m))/(3*m*(1 + 2*x)^m) - ((5 - 4*x)*(2 + 3*x)^(1 + m))/(3*(1 + 2*x)^m) + (2^(-1 - m
)*(441 - 86*m + 2*m^2)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(3*(1 - m)*m)

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Rubi [A]  time = 0.098925, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {90, 79, 69} \[ \frac{2^{-m-1} \left (2 m^2-86 m+441\right ) (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{3 (1-m) m}-\frac{1}{3} (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m}-\frac{7 (21-m) (3 x+2)^{m+1} (2 x+1)^{-m}}{3 m} \]

Antiderivative was successfully verified.

[In]

Int[(5 - 4*x)^2*(1 + 2*x)^(-1 - m)*(2 + 3*x)^m,x]

[Out]

(-7*(21 - m)*(2 + 3*x)^(1 + m))/(3*m*(1 + 2*x)^m) - ((5 - 4*x)*(2 + 3*x)^(1 + m))/(3*(1 + 2*x)^m) + (2^(-1 - m
)*(441 - 86*m + 2*m^2)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(3*(1 - m)*m)

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

\begin{align*} \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx &=-\frac{1}{3} (5-4 x) (1+2 x)^{-m} (2+3 x)^{1+m}+\frac{1}{12} \int (1+2 x)^{-1-m} (2+3 x)^m (4 (82-5 m)-8 (65-2 m) x) \, dx\\ &=-\frac{7 (21-m) (1+2 x)^{-m} (2+3 x)^{1+m}}{3 m}-\frac{1}{3} (5-4 x) (1+2 x)^{-m} (2+3 x)^{1+m}+\frac{(24 (82-5 m)+8 (65-2 m) (-4 m+3 (1+m))) \int (1+2 x)^{-m} (2+3 x)^m \, dx}{24 m}\\ &=-\frac{7 (21-m) (1+2 x)^{-m} (2+3 x)^{1+m}}{3 m}-\frac{1}{3} (5-4 x) (1+2 x)^{-m} (2+3 x)^{1+m}+\frac{2^{-1-m} \left (441-86 m+2 m^2\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{3 (1-m) m}\\ \end{align*}

Mathematica [A]  time = 0.0890314, size = 96, normalized size = 0.79 \[ \frac{1}{6} (2 x+1)^{-m} \left (-\frac{2^{-m} \left (2 m^2-86 m+441\right ) (2 x+1) \, _2F_1(1-m,-m;2-m;-6 x-3)}{(m-1) m}+2 (4 x-5) (3 x+2)^{m+1}+\frac{14 (m-21) (3 x+2)^{m+1}}{m}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - 4*x)^2*(1 + 2*x)^(-1 - m)*(2 + 3*x)^m,x]

[Out]

((14*(-21 + m)*(2 + 3*x)^(1 + m))/m + 2*(2 + 3*x)^(1 + m)*(-5 + 4*x) - ((441 - 86*m + 2*m^2)*(1 + 2*x)*Hyperge
ometric2F1[1 - m, -m, 2 - m, -3 - 6*x])/(2^m*(-1 + m)*m))/(6*(1 + 2*x)^m)

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int \left ( 5-4\,x \right ) ^{2} \left ( 1+2\,x \right ) ^{-1-m} \left ( 2+3\,x \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^m,x)

[Out]

int((5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 1}{\left (4 \, x - 5\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^m,x, algorithm="maxima")

[Out]

integrate((3*x + 2)^m*(2*x + 1)^(-m - 1)*(4*x - 5)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (16 \, x^{2} - 40 \, x + 25\right )}{\left (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^m,x, algorithm="fricas")

[Out]

integral((16*x^2 - 40*x + 25)*(3*x + 2)^m*(2*x + 1)^(-m - 1), 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((5-4*x)**2*(1+2*x)**(-1-m)*(2+3*x)**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 1}{\left (4 \, x - 5\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^m,x, algorithm="giac")

[Out]

integrate((3*x + 2)^m*(2*x + 1)^(-m - 1)*(4*x - 5)^2, x)