3.31.88 \(\int (5-4 x)^4 (1+2 x)^{-3-m} (2+3 x)^m \, dx\) [3088]

Optimal. Leaf size=188 \[ -\frac {1}{9} (107-2 m) (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}-\frac {1}{3} (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {7 (1+2 x)^{-2-m} (2+3 x)^{1+m} \left (3 \left (4638+485 m+108 m^2-2 m^3\right )+2 \left (15209+1882 m-530 m^2+8 m^3\right ) x\right )}{9 \left (2+3 m+m^2\right )}-\frac {2^{2-m} \left (1323-85 m+m^2\right ) (1+2 x)^{-m} \, _2F_1(-m,-m;1-m;-3 (1+2 x))}{9 m} \]

[Out]

-1/9*(107-2*m)*(5-4*x)^2*(1+2*x)^(-2-m)*(2+3*x)^(1+m)-1/3*(5-4*x)^3*(1+2*x)^(-2-m)*(2+3*x)^(1+m)+7/9*(1+2*x)^(
-2-m)*(2+3*x)^(1+m)*(-6*m^3+324*m^2+1455*m+13914+2*(8*m^3-530*m^2+1882*m+15209)*x)/(m^2+3*m+2)-1/9*2^(2-m)*(m^
2-85*m+1323)*hypergeom([-m, -m],[1-m],-3-6*x)/m/((1+2*x)^m)

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Rubi [A]
time = 0.13, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {102, 158, 150, 71} \begin {gather*} -\frac {2^{2-m} \left (m^2-85 m+1323\right ) (2 x+1)^{-m} \, _2F_1(-m,-m;1-m;-3 (2 x+1))}{9 m}+\frac {7 (3 x+2)^{m+1} \left (2 \left (8 m^3-530 m^2+1882 m+15209\right ) x+3 \left (-2 m^3+108 m^2+485 m+4638\right )\right ) (2 x+1)^{-m-2}}{9 \left (m^2+3 m+2\right )}-\frac {1}{3} (5-4 x)^3 (3 x+2)^{m+1} (2 x+1)^{-m-2}-\frac {1}{9} (107-2 m) (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m-2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*((107 - 2*m)*(5 - 4*x)^2*(1 + 2*x)^(-2 - m)*(2 + 3*x)^(1 + m)) - ((5 - 4*x)^3*(1 + 2*x)^(-2 - m)*(2 + 3*x
)^(1 + m))/3 + (7*(1 + 2*x)^(-2 - m)*(2 + 3*x)^(1 + m)*(3*(4638 + 485*m + 108*m^2 - 2*m^3) + 2*(15209 + 1882*m
- 530*m^2 + 8*m^3)*x))/(9*(2 + 3*m + m^2)) - (2^(2 - m)*(1323 - 85*m + m^2)*Hypergeometric2F1[-m, -m, 1 - m,
-3*(1 + 2*x)])/(9*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 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(
n + 1), x] + Dist[f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 158

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

Rubi steps

\begin {align*} \int (5-4 x)^4 (1+2 x)^{-3-m} (2+3 x)^m \, dx &=-\frac {1}{3} (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {1}{12} \int (5-4 x)^2 (1+2 x)^{-3-m} (2+3 x)^m (4 (26-5 m)-8 (107-2 m) x) \, dx\\ &=-\frac {1}{9} (107-2 m) (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}-\frac {1}{3} (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {1}{72} \int (5-4 x) (1+2 x)^{-3-m} (2+3 x)^m \left (-8 \left (3997+528 m-10 m^2\right )-64 \left (1323-85 m+m^2\right ) x\right ) \, dx\\ &=-\frac {1}{9} (107-2 m) (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}-\frac {1}{3} (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {7 (1+2 x)^{-2-m} (2+3 x)^{1+m} \left (3 \left (4638+485 m+108 m^2-2 m^3\right )+2 \left (15209+1882 m-530 m^2+8 m^3\right ) x\right )}{9 \left (2+3 m+m^2\right )}+\frac {1}{9} \left (8 \left (1323-85 m+m^2\right )\right ) \int (1+2 x)^{-1-m} (2+3 x)^m \, dx\\ &=-\frac {1}{9} (107-2 m) (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}-\frac {1}{3} (5-4 x)^3 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {7 (1+2 x)^{-2-m} (2+3 x)^{1+m} \left (3 \left (4638+485 m+108 m^2-2 m^3\right )+2 \left (15209+1882 m-530 m^2+8 m^3\right ) x\right )}{9 \left (2+3 m+m^2\right )}-\frac {2^{2-m} \left (1323-85 m+m^2\right ) (1+2 x)^{-m} \, _2F_1(-m,-m;1-m;-3 (1+2 x))}{9 m}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.87, size = 214, normalized size = 1.14 \begin {gather*} \frac {2744 (1+2 x)^{-1-m} (2+3 x)^{1+m}}{1+m}+\left (\frac {23}{14}\right )^m (5-4 x)^2 F_1\left (2;-m,m;3;\frac {3}{23} (5-4 x),\frac {1}{7} (5-4 x)\right )+\frac {21\ 2^{2-m} (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3-6 x)}{-1+m}-\frac {49\ 2^{3+m} (-1-2 x)^m (2+4 x)^{-m} (6+9 x)^{1+m} \, _2F_1(1+m,1+m;2+m;4+6 x)}{1+m}-\frac {2401\ 3^{2+m} (-1-2 x)^m (1+2 x)^{-m} (2+3 x)^{1+m} \, _2F_1(1+m,3+m;2+m;4+6 x)}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2744*(1 + 2*x)^(-1 - m)*(2 + 3*x)^(1 + m))/(1 + m) + (23/14)^m*(5 - 4*x)^2*AppellF1[2, -m, m, 3, (3*(5 - 4*x)
)/23, (5 - 4*x)/7] + (21*2^(2 - m)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3 - 6*x])/(-1 + m) -
(49*2^(3 + m)*(-1 - 2*x)^m*(6 + 9*x)^(1 + m)*Hypergeometric2F1[1 + m, 1 + m, 2 + m, 4 + 6*x])/((1 + m)*(2 + 4
*x)^m) - (2401*3^(2 + m)*(-1 - 2*x)^m*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1 + m, 3 + m, 2 + m, 4 + 6*x])/((1 +
m)*(1 + 2*x)^m)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((256*x^4 - 1280*x^3 + 2400*x^2 - 2000*x + 625)*(3*x + 2)^m*(2*x + 1)^(-m - 3), 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)**4*(1+2*x)**(-3-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)^4*(1+2*x)^(-3-m)*(2+3*x)^m,x, algorithm="giac")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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