\(\int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx\) [3072]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=-\frac {7 (21-m) (1+2 x)^{-m} (2+3 x)^{1+m}}{3 m}-\frac {1}{3} (5-4 x) (1+2 x)^{-m} (2+3 x)^{1+m}+\frac {2^{-1-m} \left (441-86 m+2 m^2\right ) (1+2 x)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (1+2 x))}{3 (1-m) m} \]

[Out]

-7/3*(21-m)*(2+3*x)^(1+m)/m/((1+2*x)^m)-1/3*(5-4*x)*(2+3*x)^(1+m)/((1+2*x)^m)+1/3*2^(-1-m)*(2*m^2-86*m+441)*(1
+2*x)^(1-m)*hypergeom([-m, 1-m],[2-m],-3-6*x)/(1-m)/m

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {92, 80, 71} \[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\frac {2^{-m-1} \left (2 m^2-86 m+441\right ) (2 x+1)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (2 x+1))}{3 (1-m) m}-\frac {1}{3} (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m}-\frac {7 (21-m) (3 x+2)^{m+1} (2 x+1)^{-m}}{3 m} \]

[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 71

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

Rule 80

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 92

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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\frac {1}{6} (1+2 x)^{-m} \left (\frac {14 (-21+m) (2+3 x)^{1+m}}{m}+2 (2+3 x)^{1+m} (-5+4 x)-\frac {2^{-m} \left (441-86 m+2 m^2\right ) (1+2 x) \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3-6 x)}{(-1+m) m}\right ) \]

[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)

Maple [F]

\[\int \left (5-4 x \right )^{2} \left (1+2 x \right )^{-1-m} \left (2+3 x \right )^{m}d x\]

[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)

Fricas [F]

\[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { {\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{2} \,d x } \]

[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)

Sympy [F]

\[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int \left (2 x + 1\right )^{- m - 1} \left (3 x + 2\right )^{m} \left (4 x - 5\right )^{2}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { {\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{2} \,d x } \]

[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)

Giac [F]

\[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { {\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^2}{{\left (2\,x+1\right )}^{m+1}} \,d x \]

[In]

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

[Out]

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