3.31.71 \(\int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx\) [3071]

Optimal. Leaf size=142 \[ -\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {2^{-1-m} \left (27783-8324 m+390 m^2-4 m^3\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{27 (1-m) m} \]

[Out]

-2/9*(5-4*x)^2*(2+3*x)^(1+m)/((1+2*x)^m)-1/27*(2+3*x)^(1+m)*(9261-512*m+4*m^2-4*(109-2*m)*m*x)/m/((1+2*x)^m)+1
/27*2^(-1-m)*(-4*m^3+390*m^2-8324*m+27783)*(1+2*x)^(1-m)*hypergeom([-m, 1-m],[2-m],-3-6*x)/(1-m)/m

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Rubi [A]
time = 0.12, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 = {102, 151, 71} \begin {gather*} \frac {2^{-m-1} \left (-4 m^3+390 m^2-8324 m+27783\right ) (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{27 (1-m) m}-\frac {(3 x+2)^{m+1} \left (4 m^2-4 (109-2 m) m x-512 m+9261\right ) (2 x+1)^{-m}}{27 m}-\frac {2}{9} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5 - 4*x)^2*(2 + 3*x)^(1 + m))/(9*(1 + 2*x)^m) - ((2 + 3*x)^(1 + m)*(9261 - 512*m + 4*m^2 - 4*(109 - 2*m)*
m*x))/(27*m*(1 + 2*x)^m) + (2^(-1 - m)*(27783 - 8324*m + 390*m^2 - 4*m^3)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[
1 - m, -m, 2 - m, -3*(1 + 2*x)])/(27*(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 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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rubi steps

\begin {align*} \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx &=-\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}+\frac {1}{18} \int (5-4 x) (1+2 x)^{-1-m} (2+3 x)^m (2 (223-10 m)-8 (109-2 m) x) \, dx\\ &=-\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {(288 (109-2 m) (1+m) (2+m)+6 (1+m) (-6 (-8 (223-10 m)-40 (109-2 m))-128 (109-2 m) m)[phone] m)+12 (-8 (223-10 m)-40 (109-2 m)) m-128 (109-2 m) (1-m) m)) \int (1+2 x)^{-m} (2+3 x)^m \, dx}{432 m}\\ &=-\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {2^{-1-m} \left (27783-8324 m+390 m^2-4 m^3\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{27 (1-m) m}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 106, normalized size = 0.75 \begin {gather*} \frac {2^{-1-m} (1+2 x)^{-m} \left (-\left ((-1+m) (4+6 x)^{1+m} \left (9261+m^2 (4+8 x)+m \left (-362-676 x+96 x^2\right )\right )\right )+\left (-27783+8324 m-390 m^2+4 m^3\right ) (1+2 x) \, _2F_1(1-m,-m;2-m;-3-6 x)\right )}{27 (-1+m) m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 - m)*(-((-1 + m)*(4 + 6*x)^(1 + m)*(9261 + m^2*(4 + 8*x) + m*(-362 - 676*x + 96*x^2))) + (-27783 + 8324
*m - 390*m^2 + 4*m^3)*(1 + 2*x)*Hypergeometric2F1[1 - m, -m, 2 - m, -3 - 6*x]))/(27*(-1 + m)*m*(1 + 2*x)^m)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (5-4 x \right )^{3} \left (1+2 x \right )^{-1-m} \left (2+3 x \right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-4*x)^3*(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)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-4*x)^3*(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)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^3}{{\left (2\,x+1\right )}^{m+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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