3.31.89 \(\int (5-4 x)^3 (1+2 x)^{-3-m} (2+3 x)^m \, dx\) [3089]

Optimal. Leaf size=139 \[ -\frac {2}{3} (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {7 (1+2 x)^{-2-m} (2+3 x)^{1+m} \left (3 \left (186-m+2 m^2\right )+2 \left (677+102 m-8 m^2\right ) x\right )}{3 \left (2+3 m+m^2\right )}-\frac {2^{1-m} (63-2 m) (1+2 x)^{-m} \, _2F_1(-m,-m;1-m;-3 (1+2 x))}{3 m} \]

[Out]

-2/3*(5-4*x)^2*(1+2*x)^(-2-m)*(2+3*x)^(1+m)+7/3*(1+2*x)^(-2-m)*(2+3*x)^(1+m)*(6*m^2-3*m+558+2*(-8*m^2+102*m+67
7)*x)/(m^2+3*m+2)-1/3*2^(1-m)*(63-2*m)*hypergeom([-m, -m],[1-m],-3-6*x)/m/((1+2*x)^m)

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Rubi [A]
time = 0.07, antiderivative size = 139, 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, 150, 71} \begin {gather*} -\frac {2^{1-m} (63-2 m) (2 x+1)^{-m} \, _2F_1(-m,-m;1-m;-3 (2 x+1))}{3 m}+\frac {7 (3 x+2)^{m+1} \left (2 \left (-8 m^2+102 m+677\right ) x+3 \left (2 m^2-m+186\right )\right ) (2 x+1)^{-m-2}}{3 \left (m^2+3 m+2\right )}-\frac {2}{3} (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)^3*(1 + 2*x)^(-3 - m)*(2 + 3*x)^m,x]

[Out]

(-2*(5 - 4*x)^2*(1 + 2*x)^(-2 - m)*(2 + 3*x)^(1 + m))/3 + (7*(1 + 2*x)^(-2 - m)*(2 + 3*x)^(1 + m)*(3*(186 - m
+ 2*m^2) + 2*(677 + 102*m - 8*m^2)*x))/(3*(2 + 3*m + m^2)) - (2^(1 - m)*(63 - 2*m)*Hypergeometric2F1[-m, -m, 1
- m, -3*(1 + 2*x)])/(3*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]))

Rubi steps

\begin {align*} \int (5-4 x)^3 (1+2 x)^{-3-m} (2+3 x)^m \, dx &=-\frac {2}{3} (5-4 x)^2 (1+2 x)^{-2-m} (2+3 x)^{1+m}+\frac {1}{6} \int (5-4 x) (1+2 x)^{-3-m} (2+3 x)^m (-2 (7+10 m)-8 (63-2 m) x) \, dx\\ &=-\frac {2}{3} (5-4 x)^2 (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 (186-m+2 m^2\right )+2 \left (677+102 m-8 m^2\right ) x\right )}{3 \left (2+3 m+m^2\right )}+\frac {1}{3} (4 (63-2 m)) \int (1+2 x)^{-1-m} (2+3 x)^m \, dx\\ &=-\frac {2}{3} (5-4 x)^2 (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 (186-m+2 m^2\right )+2 \left (677+102 m-8 m^2\right ) x\right )}{3 \left (2+3 m+m^2\right )}-\frac {2^{1-m} (63-2 m) (1+2 x)^{-m} \, _2F_1(-m,-m;1-m;-3 (1+2 x))}{3 m}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 123, normalized size = 0.88 \begin {gather*} \frac {2^{-m} (1+2 x)^{-2-m} \left (-2^m m (2+3 x)^{1+m} \left (-3806-9638 x+64 x^2+8 (m+2 m x)^2+3 m \left (57-556 x+32 x^2\right )\right )+2 \left (-126-185 m-57 m^2+2 m^3\right ) (1+2 x)^2 \, _2F_1(-m,-m;1-m;-3-6 x)\right )}{3 m (1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*x)^(-2 - m)*(-(2^m*m*(2 + 3*x)^(1 + m)*(-3806 - 9638*x + 64*x^2 + 8*(m + 2*m*x)^2 + 3*m*(57 - 556*x +
32*x^2))) + 2*(-126 - 185*m - 57*m^2 + 2*m^3)*(1 + 2*x)^2*Hypergeometric2F1[-m, -m, 1 - m, -3 - 6*x]))/(3*2^m*
m*(1 + m)*(2 + m))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((5-4*x)^3*(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)^3*(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)^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)^(-3-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 - 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)**3*(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)^3*(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)^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+3}} \,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 + 3),x)

[Out]

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

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