Integrand size = 24, antiderivative size = 179 \[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\frac {(1+2 x)^{1-m} (2+3 x)^{1+m}}{322 (5-4 x)^4}+\frac {(66+m) (1+2 x)^{1-m} (2+3 x)^{1+m}}{77763 (5-4 x)^3}+\frac {\left (4359+220 m+2 m^2\right ) (1+2 x)^{1-m} (2+3 x)^{1+m}}{25039686 (5-4 x)^2}+\frac {\left (32010+4358 m+132 m^2+m^3\right ) (1+2 x)^{1-m} (2+3 x)^{-1+m} \operatorname {Hypergeometric2F1}\left (2,1-m,2-m,\frac {23 (1+2 x)}{14 (2+3 x)}\right )}{2453889228 (1-m)} \]
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)
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.73 \[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\frac {(1+2 x)^{1-m} (2+3 x)^{-1+m} \left (\frac {7620774 (2+3 x)^2}{(5-4 x)^4}+\frac {98 \left (4359+220 m+2 m^2\right ) (2+3 x)^2}{(5-4 x)^2}-\frac {31556 (66+m) (2+3 x)^2}{(-5+4 x)^3}-\frac {\left (32010+4358 m+132 m^2+m^3\right ) \operatorname {Hypergeometric2F1}\left (2,1-m,2-m,\frac {23+46 x}{28+42 x}\right )}{-1+m}\right )}{2453889228} \]
((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)*Hypergeometric2 F1[2, 1 - m, 2 - m, (23 + 46*x)/(28 + 42*x)])/(-1 + m)))/2453889228
Time = 0.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {114, 27, 168, 27, 168, 27, 141}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(2 x+1)^{-m} (3 x+2)^m}{(5-4 x)^5} \, dx\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {(2 x+1)^{1-m} (3 x+2)^{m+1}}{322 (5-4 x)^4}-\frac {\int -\frac {4 (2 x+1)^{-m} (3 x+2)^m (m+12 x+51)}{(5-4 x)^4}dx}{1288}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{322} \int \frac {(2 x+1)^{-m} (3 x+2)^m (m+12 x+51)}{(5-4 x)^4}dx+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{322} \left (\frac {2 (m+66) (2 x+1)^{1-m} (3 x+2)^{m+1}}{483 (5-4 x)^3}-\frac {1}{966} \int -\frac {2 (2 x+1)^{-m} (3 x+2)^m \left (2 m^2+205 m+12 (m+66) x+3369\right )}{(5-4 x)^3}dx\right )+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{322} \left (\frac {1}{483} \int \frac {(2 x+1)^{-m} (3 x+2)^m \left (2 m^2+205 m+12 (m+66) x+3369\right )}{(5-4 x)^3}dx+\frac {2 (m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{483 (5-4 x)^3}\right )+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{322} \left (\frac {1}{483} \left (\frac {\left (2 m^2+220 m+4359\right ) (2 x+1)^{1-m} (3 x+2)^{m+1}}{161 (5-4 x)^2}-\frac {1}{644} \int -\frac {8 \left (m^3+132 m^2+4358 m+32010\right ) (2 x+1)^{-m} (3 x+2)^m}{(5-4 x)^2}dx\right )+\frac {2 (m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{483 (5-4 x)^3}\right )+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{322} \left (\frac {1}{483} \left (\frac {2}{161} \left (m^3+132 m^2+4358 m+32010\right ) \int \frac {(2 x+1)^{-m} (3 x+2)^m}{(5-4 x)^2}dx+\frac {\left (2 m^2+220 m+4359\right ) (3 x+2)^{m+1} (2 x+1)^{1-m}}{161 (5-4 x)^2}\right )+\frac {2 (m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{483 (5-4 x)^3}\right )+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {1}{322} \left (\frac {1}{483} \left (\frac {\left (m^3+132 m^2+4358 m+32010\right ) (3 x+2)^{m-1} (2 x+1)^{1-m} \operatorname {Hypergeometric2F1}\left (2,1-m,2-m,\frac {23 (2 x+1)}{14 (3 x+2)}\right )}{15778 (1-m)}+\frac {\left (2 m^2+220 m+4359\right ) (3 x+2)^{m+1} (2 x+1)^{1-m}}{161 (5-4 x)^2}\right )+\frac {2 (m+66) (3 x+2)^{m+1} (2 x+1)^{1-m}}{483 (5-4 x)^3}\right )+\frac {(3 x+2)^{m+1} (2 x+1)^{1-m}}{322 (5-4 x)^4}\) |
((1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(322*(5 - 4*x)^4) + ((2*(66 + m)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(483*(5 - 4*x)^3) + (((4359 + 220*m + 2* m^2)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m))/(161*(5 - 4*x)^2) + ((32010 + 43 58*m + 132*m^2 + m^3)*(1 + 2*x)^(1 - m)*(2 + 3*x)^(-1 + m)*Hypergeometric2 F1[2, 1 - m, 2 - m, (23*(1 + 2*x))/(14*(2 + 3*x))])/(15778*(1 - m)))/483)/ 322
3.31.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S imp[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]
\[\int \frac {\left (2+3 x \right )^{m} \left (1+2 x \right )^{-m}}{\left (5-4 x \right )^{5}}d x\]
\[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m} {\left (4 \, x - 5\right )}^{5}} \,d x } \]
integral(-(3*x + 2)^m/((1024*x^5 - 6400*x^4 + 16000*x^3 - 20000*x^2 + 1250 0*x - 3125)*(2*x + 1)^m), x)
Timed out. \[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\text {Timed out} \]
\[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m} {\left (4 \, x - 5\right )}^{5}} \,d x } \]
\[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m} {\left (4 \, x - 5\right )}^{5}} \,d x } \]
Timed out. \[ \int \frac {(1+2 x)^{-m} (2+3 x)^m}{(5-4 x)^5} \, dx=-\int \frac {{\left (3\,x+2\right )}^m}{{\left (2\,x+1\right )}^m\,{\left (4\,x-5\right )}^5} \,d x \]