3.3069 \(\int \frac{(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx\)

Optimal. Leaf size=179 \[ \frac{\left (m^3+132 m^2+4358 m+32010\right ) (3 x+2)^{m-1} (2 x+1)^{1-m} \, _2F_1\left (2,1-m;2-m;\frac{23 (2 x+1)}{14 (3 x+2)}\right )}{2453889228 (1-m)}+\frac{\left (2 m^2+220 m+4359\right ) (3 x+2)^{m+1} (2 x+1)^{1-m}}{25039686 (5-4 x)^2}+\frac{(m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{77763 (5-4 x)^3}+\frac{(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4} \]

[Out]

((1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(322*(5 - 4*x)^4) + ((66 + m)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(7776
3*(5 - 4*x)^3) + ((4359 + 220*m + 2*m^2)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(25039686*(5 - 4*x)^2) + ((32010
+ 4358*m + 132*m^2 + m^3)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(-1 + m)*Hypergeometric2F1[2, 1 - m, 2 - m, (23*(1 + 2*
x))/(14*(2 + 3*x))])/(2453889228*(1 - m))

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Rubi [A]  time = 0.097856, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {129, 151, 12, 131} \[ \frac{\left (m^3+132 m^2+4358 m+32010\right ) (3 x+2)^{m-1} (2 x+1)^{1-m} \, _2F_1\left (2,1-m;2-m;\frac{23 (2 x+1)}{14 (3 x+2)}\right )}{2453889228 (1-m)}+\frac{\left (2 m^2+220 m+4359\right ) (3 x+2)^{m+1} (2 x+1)^{1-m}}{25039686 (5-4 x)^2}+\frac{(m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{77763 (5-4 x)^3}+\frac{(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(322*(5 - 4*x)^4) + ((66 + m)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(7776
3*(5 - 4*x)^3) + ((4359 + 220*m + 2*m^2)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(25039686*(5 - 4*x)^2) + ((32010
+ 4358*m + 132*m^2 + m^3)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(-1 + m)*Hypergeometric2F1[2, 1 - m, 2 - m, (23*(1 + 2*
x))/(14*(2 + 3*x))])/(2453889228*(1 - m))

Rule 129

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int \frac{(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx &=\frac{(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}-\frac{\int \frac{(-4 (51+m)-48 x) (1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^4} \, dx}{1288}\\ &=\frac{(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}+\frac{(66+m) (1+2 x)^{1-m} (2+3 x)^{1+m}}{77763 (5-4 x)^3}+\frac{\int \frac{(1+2 x)^{-m} (2+3 x)^m \left (8 \left (3369+205 m+2 m^2\right )+96 (66+m) x\right )}{(5-4 x)^3} \, dx}{1244208}\\ &=\frac{(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}+\frac{(66+m) (1+2 x)^{1-m} (2+3 x)^{1+m}}{77763 (5-4 x)^3}+\frac{\left (4359+220 m+2 m^2\right ) (1+2 x)^{1-m} (2+3 x)^{1+m}}{25039686 (5-4 x)^2}-\frac{\int -\frac{64 \left (32010+4358 m+132 m^2+m^3\right ) (1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^2} \, dx}{801269952}\\ &=\frac{(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}+\frac{(66+m) (1+2 x)^{1-m} (2+3 x)^{1+m}}{77763 (5-4 x)^3}+\frac{\left (4359+220 m+2 m^2\right ) (1+2 x)^{1-m} (2+3 x)^{1+m}}{25039686 (5-4 x)^2}-\frac{\left (-32010-4358 m-132 m^2-m^3\right ) \int \frac{(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^2} \, dx}{12519843}\\ &=\frac{(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}+\frac{(66+m) (1+2 x)^{1-m} (2+3 x)^{1+m}}{77763 (5-4 x)^3}+\frac{\left (4359+220 m+2 m^2\right ) (1+2 x)^{1-m} (2+3 x)^{1+m}}{25039686 (5-4 x)^2}+\frac{\left (32010+4358 m+132 m^2+m^3\right ) (1+2 x)^{1-m} (2+3 x)^{-1+m} \, _2F_1\left (2,1-m;2-m;\frac{23 (1+2 x)}{14 (2+3 x)}\right )}{2453889228 (1-m)}\\ \end{align*}

Mathematica [A]  time = 0.113299, size = 131, normalized size = 0.73 \[ \frac{(2 x+1)^{1-m} (3 x+2)^{m-1} \left (-\frac{\left (m^3+132 m^2+4358 m+32010\right ) \, _2F_1\left (2,1-m;2-m;\frac{46 x+23}{42 x+28}\right )}{m-1}+\frac{98 \left (2 m^2+220 m+4359\right ) (3 x+2)^2}{(5-4 x)^2}-\frac{31556 (m+66) (3 x+2)^2}{(4 x-5)^3}+\frac{7620774 (3 x+2)^2}{(5-4 x)^4}\right )}{2453889228} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*x)^(1 - m)*(2 + 3*x)^(-1 + m)*((7620774*(2 + 3*x)^2)/(5 - 4*x)^4 + (98*(4359 + 220*m + 2*m^2)*(2 + 3*x
)^2)/(5 - 4*x)^2 - (31556*(66 + m)*(2 + 3*x)^2)/(-5 + 4*x)^3 - ((32010 + 4358*m + 132*m^2 + m^3)*Hypergeometri
c2F1[2, 1 - m, 2 - m, (23 + 46*x)/(28 + 42*x)])/(-1 + m)))/2453889228

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Maple [F]  time = 0.073, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 2+3\,x \right ) ^{m}}{ \left ( 5-4\,x \right ) ^{5} \left ( 1+2\,x \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (3 \, x + 2\right )}^{m}}{{\left (1024 \, x^{5} - 6400 \, x^{4} + 16000 \, x^{3} - 20000 \, x^{2} + 12500 \, x - 3125\right )}{\left (2 \, x + 1\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*x + 2)^m/((1024*x^5 - 6400*x^4 + 16000*x^3 - 20000*x^2 + 12500*x - 3125)*(2*x + 1)^m), 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((2+3*x)**m/(5-4*x)**5/((1+2*x)**m),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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