3.3080 \(\int (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^m \, dx\)

Optimal. Leaf size=132 \[ \frac{2^{-m} \left (2 m^2-128 m+1323\right ) (2 x+1)^{-m} \, _2F_1(-m,-m;1-m;-3 (2 x+1))}{9 m}-\frac{(3 x+2)^{m+1} \left (4 m^2-8 (43-m) (m+1) x-315 m+2768\right ) (2 x+1)^{-m-1}}{9 (m+1)}-\frac{1}{3} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m-1} \]

[Out]

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

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Rubi [A]  time = 0.136775, antiderivative size = 132, 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 = {100, 143, 69} \[ \frac{2^{-m} \left (2 m^2-128 m+1323\right ) (2 x+1)^{-m} \, _2F_1(-m,-m;1-m;-3 (2 x+1))}{9 m}-\frac{(3 x+2)^{m+1} \left (4 m^2-8 (43-m) (m+1) x-315 m+2768\right ) (2 x+1)^{-m-1}}{9 (m+1)}-\frac{1}{3} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m-1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5 - 4*x)^2*(1 + 2*x)^(-1 - m)*(2 + 3*x)^(1 + m))/3 - ((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 100

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 143

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[n, 1] &&  !SumSimplerQ[m, 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)^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}{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]  time = 0.0997457, size = 114, normalized size = 0.86 \[ \frac{2^{-m} (2 x+1)^{-m-1} \left (\left (2 m^3-126 m^2+1195 m+1323\right ) (2 x+1) \, _2F_1(-m,-m;1-m;-6 x-3)-2^m m (3 x+2)^{m+1} \left (m^2 (8 x+4)+24 m \left (2 x^2-19 x-10\right )+48 x^2-464 x+2843\right )\right )}{9 m (m+1)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( 5-4\,x \right ) ^{3} \left ( 1+2\,x \right ) ^{-2-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)^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)

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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 - 2}{\left (4 \, x - 5\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (64 \, x^{3} - 240 \, x^{2} + 300 \, x - 125\right )}{\left (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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)**3*(1+2*x)**(-2-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 - 2}{\left (4 \, x - 5\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)