\(\int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx\) [3080]

Optimal result

Integrand size = 26, antiderivative size = 132 \[ \int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx=-\frac {1}{3} (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-1-m} (2+3 x)^{1+m} \left (2768-315 m+4 m^2-8 (43-m) (1+m) x\right )}{9 (1+m)}+\frac {2^{-m} \left (1323-128 m+2 m^2\right ) (1+2 x)^{-m} \operatorname {Hypergeometric2F1}(-m,-m,1-m,-3 (1+2 x))}{9 m} \]

[Out]

-1/3*(5-4*x)^2*(1+2*x)^(-1-m)*(2+3*x)^(1+m)-1/9*(1+2*x)^(-1-m)*(2+3*x)^(1+m)*(2768-315*m+4*m^2-8*(43-m)*(1+m)*
x)/(1+m)+1/9*(2*m^2-128*m+1323)*hypergeom([-m, -m],[1-m],-3-6*x)/(2^m)/m/((1+2*x)^m)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 132, 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 = {102, 148, 71} \[ \int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx=\frac {2^{-m} \left (2 m^2-128 m+1323\right ) (2 x+1)^{-m} \operatorname {Hypergeometric2F1}(-m,-m,1-m,-3 (2 x+1))}{9 m}-\frac {(3 x+2)^{m+1} \left (4 m^2-8 (43-m) (m+1) x-315 m+2768\right ) (2 x+1)^{-m-1}}{9 (m+1)}-\frac {1}{3} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m-1} \]

[In]

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

[Out]

-1/3*((5 - 4*x)^2*(1 + 2*x)^(-1 - m)*(2 + 3*x)^(1 + m)) - ((1 + 2*x)^(-1 - m)*(2 + 3*x)^(1 + m)*(2768 - 315*m
+ 4*m^2 - 8*(43 - m)*(1 + m)*x))/(9*(1 + m)) + ((1323 - 128*m + 2*m^2)*Hypergeometric2F1[-m, -m, 1 - m, -3*(1
+ 2*x)])/(9*2^m*m*(1 + 2*x)^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 102

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

Rule 148

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[(b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^{1+m}+\frac {1}{12} \int (5-4 x) (1+2 x)^{-2-m} (2+3 x)^m (4 (54-5 m)-16 (43-m) x) \, dx \\ & = -\frac {1}{3} (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-1-m} (2+3 x)^{1+m} \left (2768-315 m+4 m^2-8 (43-m) (1+m) x\right )}{9 (1+m)}-\frac {1}{9} \left (2 \left (1323-128 m+2 m^2\right )\right ) \int (1+2 x)^{-1-m} (2+3 x)^m \, dx \\ & = -\frac {1}{3} (5-4 x)^2 (1+2 x)^{-1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-1-m} (2+3 x)^{1+m} \left (2768-315 m+4 m^2-8 (43-m) (1+m) x\right )}{9 (1+m)}+\frac {2^{-m} \left (1323-128 m+2 m^2\right ) (1+2 x)^{-m} \, _2F_1(-m,-m;1-m;-3 (1+2 x))}{9 m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx=\frac {2^{-m} (1+2 x)^{-1-m} \left (-2^m m (2+3 x)^{1+m} \left (2843-464 x+48 x^2+m^2 (4+8 x)+24 m \left (-10-19 x+2 x^2\right )\right )+\left (1323+1195 m-126 m^2+2 m^3\right ) (1+2 x) \operatorname {Hypergeometric2F1}(-m,-m,1-m,-3-6 x)\right )}{9 m (1+m)} \]

[In]

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

[Out]

((1 + 2*x)^(-1 - m)*(-(2^m*m*(2 + 3*x)^(1 + m)*(2843 - 464*x + 48*x^2 + m^2*(4 + 8*x) + 24*m*(-10 - 19*x + 2*x
^2))) + (1323 + 1195*m - 126*m^2 + 2*m^3)*(1 + 2*x)*Hypergeometric2F1[-m, -m, 1 - m, -3 - 6*x]))/(9*2^m*m*(1 +
m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(64*x^3 - 240*x^2 + 300*x - 125)*(3*x + 2)^m*(2*x + 1)^(-m - 2), x)

Sympy [F]

\[ \int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx=- \int \left (- 125 \left (2 x + 1\right )^{- m - 2} \left (3 x + 2\right )^{m}\right )\, dx - \int 300 x \left (2 x + 1\right )^{- m - 2} \left (3 x + 2\right )^{m}\, dx - \int \left (- 240 x^{2} \left (2 x + 1\right )^{- m - 2} \left (3 x + 2\right )^{m}\right )\, dx - \int 64 x^{3} \left (2 x + 1\right )^{- m - 2} \left (3 x + 2\right )^{m}\, dx \]

[In]

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

[Out]

-Integral(-125*(2*x + 1)**(-m - 2)*(3*x + 2)**m, x) - Integral(300*x*(2*x + 1)**(-m - 2)*(3*x + 2)**m, x) - In
tegral(-240*x**2*(2*x + 1)**(-m - 2)*(3*x + 2)**m, x) - Integral(64*x**3*(2*x + 1)**(-m - 2)*(3*x + 2)**m, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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