3.3089 \(\int (5-4 x)^3 (1+2 x)^{-3-m} (2+3 x)^m \, dx\)

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

[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)

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Rubi [A]  time = 0.104301, antiderivative size = 139, 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, 145, 69} \[ -\frac{2^{1-m} (63-2 m) (2 x+1)^{-m} \, _2F_1(-m,-m;1-m;-3 (2 x+1))}{3 m}+\frac{7 (3 x+2)^{m+1} \left (2 \left (-8 m^2+102 m+677\right ) x+3 \left (2 m^2-m+186\right )\right ) (2 x+1)^{-m-2}}{3 \left (m^2+3 m+2\right )}-\frac{2}{3} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m-2} \]

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 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 145

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)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), 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 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)^{-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*}

Mathematica [A]  time = 0.101234, size = 123, normalized size = 0.88 \[ \frac{2^{-m} (2 x+1)^{-m-2} \left (2 \left (2 m^3-57 m^2-185 m-126\right ) (2 x+1)^2 \, _2F_1(-m,-m;1-m;-6 x-3)-2^m m (3 x+2)^{m+1} \left (3 m \left (32 x^2-556 x+57\right )+8 (2 m x+m)^2+64 x^2-9638 x-3806\right )\right )}{3 m (m+1) (m+2)} \]

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 = 0.055, size = 0, normalized size = 0. \begin{align*} \int \left ( 5-4\,x \right ) ^{3} \left ( 1+2\,x \right ) ^{-3-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)^(-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., size = 0, normalized size = 0. \begin{align*} -\int{\left (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 3}{\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)^(-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., 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 - 3}, x\right ) \end{align*}

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)]  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)**(-3-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 - 3}{\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)^(-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)