Integrand size = 23, antiderivative size = 97 \[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\frac {21 (4+3 x)^{1+m}}{92 (5-2 x)^3}-\frac {(4+3 x)^{1+m}}{12 (1-m) (5-2 x)^2}+\frac {3 \left (332+537 m+189 m^2\right ) (4+3 x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {2}{23} (4+3 x)\right )}{1119364 \left (1-m^2\right )} \] Output:
21/92*(4+3*x)^(1+m)/(5-2*x)^3-1/12*(4+3*x)^(1+m)/(1-m)/(5-2*x)^2+3*(189*m^ 2+537*m+332)*(4+3*x)^(1+m)*hypergeom([3, 1+m],[2+m],8/23+6/23*x)/(-1119364 *m^2+1119364)
Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\frac {3 (4+3 x)^{1+m} \left (529 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {2}{23} (4+3 x)\right )-1104 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {2}{23} (4+3 x)\right )+567 \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,\frac {2}{23} (4+3 x)\right )\right )}{1119364 (1+m)} \] Input:
Integrate[((4 + 3*x)^m*(2 + 3*x + x^2))/(5 - 2*x)^4,x]
Output:
(3*(4 + 3*x)^(1 + m)*(529*Hypergeometric2F1[2, 1 + m, 2 + m, (2*(4 + 3*x)) /23] - 1104*Hypergeometric2F1[3, 1 + m, 2 + m, (2*(4 + 3*x))/23] + 567*Hyp ergeometric2F1[4, 1 + m, 2 + m, (2*(4 + 3*x))/23]))/(1119364*(1 + m))
Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1193, 27, 87, 78}
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.
| \(\displaystyle \int \frac {\left (x^2+3 x+2\right ) (3 x+4)^m}{(5-2 x)^4} \, dx\) |
| \(\Big \downarrow \) 1193 |
| \(\displaystyle \frac {1}{69} \int -\frac {3 (3 x+4)^m (63 m+46 x+127)}{4 (5-2 x)^3}dx+\frac {21 (3 x+4)^{m+1}}{92 (5-2 x)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {21 (3 x+4)^{m+1}}{92 (5-2 x)^3}-\frac {1}{92} \int \frac {(3 x+4)^m (63 m+46 x+127)}{(5-2 x)^3}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{92} \left (\frac {1}{46} \left (189 m^2+537 m+332\right ) \int \frac {(3 x+4)^m}{(5-2 x)^2}dx-\frac {(63 m+242) (3 x+4)^{m+1}}{46 (5-2 x)^2}\right )+\frac {21 (3 x+4)^{m+1}}{92 (5-2 x)^3}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {1}{92} \left (\frac {3 \left (189 m^2+537 m+332\right ) (3 x+4)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {2}{23} (3 x+4)\right )}{24334 (m+1)}-\frac {(63 m+242) (3 x+4)^{m+1}}{46 (5-2 x)^2}\right )+\frac {21 (3 x+4)^{m+1}}{92 (5-2 x)^3}\) |
Input:
Int[((4 + 3*x)^m*(2 + 3*x + x^2))/(5 - 2*x)^4,x]
Output:
(21*(4 + 3*x)^(1 + m))/(92*(5 - 2*x)^3) + (-1/46*((242 + 63*m)*(4 + 3*x)^( 1 + m))/(5 - 2*x)^2 + (3*(332 + 537*m + 189*m^2)*(4 + 3*x)^(1 + m)*Hyperge ometric2F1[2, 1 + m, 2 + m, (2*(4 + 3*x))/23])/(24334*(1 + m)))/92
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_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) ), x] + Simp[1/((m + 1)*(e*f - d*g)) Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a , b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] && !IntegerQ[n] && !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
\[\int \frac {\left (3 x +4\right )^{m} \left (x^{2}+3 x +2\right )}{\left (5-2 x \right )^{4}}d x\]
Input:
int((3*x+4)^m*(x^2+3*x+2)/(5-2*x)^4,x)
Output:
int((3*x+4)^m*(x^2+3*x+2)/(5-2*x)^4,x)
\[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\int { \frac {{\left (x^{2} + 3 \, x + 2\right )} {\left (3 \, x + 4\right )}^{m}}{{\left (2 \, x - 5\right )}^{4}} \,d x } \] Input:
integrate((4+3*x)^m*(x^2+3*x+2)/(5-2*x)^4,x, algorithm="fricas")
Output:
integral((x^2 + 3*x + 2)*(3*x + 4)^m/(16*x^4 - 160*x^3 + 600*x^2 - 1000*x + 625), x)
\[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\int \frac {\left (x + 1\right ) \left (x + 2\right ) \left (3 x + 4\right )^{m}}{\left (2 x - 5\right )^{4}}\, dx \] Input:
integrate((4+3*x)**m*(x**2+3*x+2)/(5-2*x)**4,x)
Output:
Integral((x + 1)*(x + 2)*(3*x + 4)**m/(2*x - 5)**4, x)
\[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\int { \frac {{\left (x^{2} + 3 \, x + 2\right )} {\left (3 \, x + 4\right )}^{m}}{{\left (2 \, x - 5\right )}^{4}} \,d x } \] Input:
integrate((4+3*x)^m*(x^2+3*x+2)/(5-2*x)^4,x, algorithm="maxima")
Output:
integrate((x^2 + 3*x + 2)*(3*x + 4)^m/(2*x - 5)^4, x)
\[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\int { \frac {{\left (x^{2} + 3 \, x + 2\right )} {\left (3 \, x + 4\right )}^{m}}{{\left (2 \, x - 5\right )}^{4}} \,d x } \] Input:
integrate((4+3*x)^m*(x^2+3*x+2)/(5-2*x)^4,x, algorithm="giac")
Output:
integrate((x^2 + 3*x + 2)*(3*x + 4)^m/(2*x - 5)^4, x)
Timed out. \[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\int \frac {{\left (3\,x+4\right )}^m\,\left (x^2+3\,x+2\right )}{{\left (2\,x-5\right )}^4} \,d x \] Input:
int(((3*x + 4)^m*(3*x + x^2 + 2))/(2*x - 5)^4,x)
Output:
int(((3*x + 4)^m*(3*x + x^2 + 2))/(2*x - 5)^4, x)
\[ \int \frac {(4+3 x)^m \left (2+3 x+x^2\right )}{(5-2 x)^4} \, dx=\text {too large to display} \] Input:
int((4+3*x)^m*(x^2+3*x+2)/(5-2*x)^4,x)
Output:
(90*(3*x + 4)**m*m**2*x**2 + 615*(3*x + 4)**m*m**2*x - 96*(3*x + 4)**m*m** 2 - 36*(3*x + 4)**m*m*x**2 + 1164*(3*x + 4)**m*m*x - 532*(3*x + 4)**m*m - 288*(3*x + 4)**m*x**2 + 288*(3*x + 4)**m*x - 432*(3*x + 4)**m + 521640*int (((3*x + 4)**m*x)/(240*m**3*x**5 - 2080*m**3*x**4 + 5800*m**3*x**3 - 3000* m**3*x**2 - 10625*m**3*x + 12500*m**3 - 336*m**2*x**5 + 2912*m**2*x**4 - 8 120*m**2*x**3 + 4200*m**2*x**2 + 14875*m**2*x - 17500*m**2 - 672*m*x**5 + 5824*m*x**4 - 16240*m*x**3 + 8400*m*x**2 + 29750*m*x - 35000*m + 768*x**5 - 6656*x**4 + 18560*x**3 - 9600*x**2 - 34000*x + 40000),x)*m**6*x**3 - 391 2300*int(((3*x + 4)**m*x)/(240*m**3*x**5 - 2080*m**3*x**4 + 5800*m**3*x**3 - 3000*m**3*x**2 - 10625*m**3*x + 12500*m**3 - 336*m**2*x**5 + 2912*m**2* x**4 - 8120*m**2*x**3 + 4200*m**2*x**2 + 14875*m**2*x - 17500*m**2 - 672*m *x**5 + 5824*m*x**4 - 16240*m*x**3 + 8400*m*x**2 + 29750*m*x - 35000*m + 7 68*x**5 - 6656*x**4 + 18560*x**3 - 9600*x**2 - 34000*x + 40000),x)*m**6*x* *2 + 9780750*int(((3*x + 4)**m*x)/(240*m**3*x**5 - 2080*m**3*x**4 + 5800*m **3*x**3 - 3000*m**3*x**2 - 10625*m**3*x + 12500*m**3 - 336*m**2*x**5 + 29 12*m**2*x**4 - 8120*m**2*x**3 + 4200*m**2*x**2 + 14875*m**2*x - 17500*m**2 - 672*m*x**5 + 5824*m*x**4 - 16240*m*x**3 + 8400*m*x**2 + 29750*m*x - 350 00*m + 768*x**5 - 6656*x**4 + 18560*x**3 - 9600*x**2 - 34000*x + 40000),x) *m**6*x - 8150625*int(((3*x + 4)**m*x)/(240*m**3*x**5 - 2080*m**3*x**4 + 5 800*m**3*x**3 - 3000*m**3*x**2 - 10625*m**3*x + 12500*m**3 - 336*m**2*x...