3.31.69 \(\int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx\) [3069]

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

[Out]

1/322*(1+2*x)^(1-m)*(2+3*x)^(1+m)/(5-4*x)^4+1/77763*(66+m)*(1+2*x)^(1-m)*(2+3*x)^(1+m)/(5-4*x)^3+1/25039686*(2
*m^2+220*m+4359)*(1+2*x)^(1-m)*(2+3*x)^(1+m)/(5-4*x)^2+1/2453889228*(m^3+132*m^2+4358*m+32010)*(1+2*x)^(1-m)*(
2+3*x)^(-1+m)*hypergeom([2, 1-m],[2-m],23/14*(1+2*x)/(2+3*x))/(1-m)

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Rubi [A]
time = 0.06, antiderivative size = 179, normalized size of antiderivative = 1.00, 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 = {105, 156, 12, 133} \begin {gather*} \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} \end {gather*}

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 12

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

Rule 105

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, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 156

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] && ILtQ[m, -1]

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*}

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Mathematica [A]
time = 0.16, size = 131, normalized size = 0.73 \begin {gather*} \frac {(1+2 x)^{1-m} (2+3 x)^{-1+m} \left (\frac {7620774 (2+3 x)^2}{(5-4 x)^4}+\frac {98 \left (4359+220 m+2 m^2\right ) (2+3 x)^2}{(5-4 x)^2}-\frac {31556 (66+m) (2+3 x)^2}{(-5+4 x)^3}-\frac {\left (32010+4358 m+132 m^2+m^3\right ) \, _2F_1\left (2,1-m;2-m;\frac {23+46 x}{28+42 x}\right )}{-1+m}\right )}{2453889228} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (2+3 x \right )^{m} \left (1+2 x \right )^{-m}}{\left (5-4 x \right )^{5}}\, dx\]

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 = 0.00, 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((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 = 0.00, 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((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)] 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((2+3*x)**m/(5-4*x)**5/((1+2*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((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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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