Integrand size = 38, antiderivative size = 377 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\frac {4 (d+e x)^{1+m}}{25 e (1+m)}-\frac {(1367 d-293 e+(423 d-1367 e) x) (d+e x)^{1+m}}{700 \left (5 d^2-2 d e+3 e^2\right ) \left (3+2 x+5 x^2\right )}+\frac {\left (80360 d^2-32144 d e+48216 e^2+i \sqrt {14} \left (6565 d^2-2 d e (1313-3206 m)+e^2 (3939-98 m)\right )-5922 d e m+19138 e^2 m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-e+i \sqrt {14} e}\right )}{19600 \left (5 d+i \left (i+\sqrt {14}\right ) e\right ) \left (5 d^2-2 d e+3 e^2\right ) (1+m)}+\frac {\left (80360 d^2-32144 d e+48216 e^2-i \sqrt {14} \left (6565 d^2-2 d e (1313-3206 m)+e^2 (3939-98 m)\right )-5922 d e m+19138 e^2 m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{19600 \left (5 d-\left (1+i \sqrt {14}\right ) e\right ) \left (5 d^2-2 d e+3 e^2\right ) (1+m)} \]
4/25*(e*x+d)^(1+m)/e/(1+m)-1/700*(1367*d-293*e+(423*d-1367*e)*x)*(e*x+d)^( 1+m)/(5*d^2-2*d*e+3*e^2)/(5*x^2+2*x+3)+1/19600*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e*(1+I*14^(1/2))))*(80360*d^2-32144*d*e+48216*e ^2-5922*d*e*m+19138*e^2*m-I*(6565*d^2-2*d*e*(1313-3206*m)+e^2*(3939-98*m)) *14^(1/2))/(5*d^2-2*d*e+3*e^2)/(1+m)/(5*d-e*(1+I*14^(1/2)))+1/19600*(e*x+d )^(1+m)*hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e+I*14^(1/2)*e))*(80360*d^ 2-32144*d*e+48216*e^2-5922*d*e*m+19138*e^2*m+I*(6565*d^2-2*d*e*(1313-3206* m)+e^2*(3939-98*m))*14^(1/2))/(5*d^2-2*d*e+3*e^2)/(1+m)/(5*d+I*e*(I+14^(1/ 2)))
Time = 1.35 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {3136}{e+e m}-\frac {28 (d (1367+423 x)-e (293+1367 x))}{\left (5 d^2-2 d e+3 e^2\right ) \left (3+2 x+5 x^2\right )}+\frac {56 \left (287 i+31 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+\left (-1-i \sqrt {14}\right ) e}\right )}{\left (5 i d+\left (-i+\sqrt {14}\right ) e\right ) (1+m)}+\frac {56 \left (-287 i+31 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+i \left (i+\sqrt {14}\right ) e}\right )}{\left (-5 i d+\left (i+\sqrt {14}\right ) e\right ) (1+m)}-\frac {\sqrt {14} \left (\frac {\left (2115 d^2+d e \left (-846+\left (-6412+423 i \sqrt {14}\right ) m\right )+e^2 \left (1269+\left (98-1367 i \sqrt {14}\right ) m\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+\left (-1-i \sqrt {14}\right ) e}\right )}{5 i d+\left (-i+\sqrt {14}\right ) e}-\frac {\left (2115 d^2-d e \left (846+\left (6412+423 i \sqrt {14}\right ) m\right )+e^2 \left (1269+\left (98+1367 i \sqrt {14}\right ) m\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+i \left (i+\sqrt {14}\right ) e}\right )}{5 i d-\left (i+\sqrt {14}\right ) e}\right )}{\left (5 d^2-2 d e+3 e^2\right ) (1+m)}\right )}{19600} \]
((d + e*x)^(1 + m)*(3136/(e + e*m) - (28*(d*(1367 + 423*x) - e*(293 + 1367 *x)))/((5*d^2 - 2*d*e + 3*e^2)*(3 + 2*x + 5*x^2)) + (56*(287*I + 31*Sqrt[1 4])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d + (-1 - I*Sqrt[1 4])*e)])/(((5*I)*d + (-I + Sqrt[14])*e)*(1 + m)) + (56*(-287*I + 31*Sqrt[1 4])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d + I*(I + Sqrt[14 ])*e)])/(((-5*I)*d + (I + Sqrt[14])*e)*(1 + m)) - (Sqrt[14]*(((2115*d^2 + d*e*(-846 + (-6412 + (423*I)*Sqrt[14])*m) + e^2*(1269 + (98 - (1367*I)*Sqr t[14])*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d + (-1 - I *Sqrt[14])*e)])/((5*I)*d + (-I + Sqrt[14])*e) - ((2115*d^2 - d*e*(846 + (6 412 + (423*I)*Sqrt[14])*m) + e^2*(1269 + (98 + (1367*I)*Sqrt[14])*m))*Hype rgeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d + I*(I + Sqrt[14])*e)])/ ((5*I)*d - (I + Sqrt[14])*e)))/((5*d^2 - 2*d*e + 3*e^2)*(1 + m))))/19600
Time = 0.86 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2179, 27, 2159, 2009}
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.
| \(\displaystyle \int \frac {\left (4 x^4-5 x^3+3 x^2+x+2\right ) (d+e x)^m}{\left (5 x^2+2 x+3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2179 |
| \(\displaystyle \frac {\int \frac {2 (d+e x)^m \left (1845 d^2-e (738-1367 m) d+560 \left (5 d^2-2 e d+3 e^2\right ) x^2+e^2 (1107-293 m)-\left (4620 d^2-3 e (141 m+616) d+e^2 (1367 m+2772)\right ) x\right )}{25 \left (5 x^2+2 x+3\right )}dx}{56 \left (5 d^2-2 d e+3 e^2\right )}-\frac {(x (423 d-1367 e)+1367 d-293 e) (d+e x)^{m+1}}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d+e x)^m \left (1845 d^2-e (738-1367 m) d+560 \left (5 d^2-2 e d+3 e^2\right ) x^2+e^2 (1107-293 m)-\left (4620 d^2-3 e (141 m+616) d+e^2 (1367 m+2772)\right ) x\right )}{5 x^2+2 x+3}dx}{700 \left (5 d^2-2 d e+3 e^2\right )}-\frac {(x (423 d-1367 e)+1367 d-293 e) (d+e x)^{m+1}}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle -\frac {(x (423 d-1367 e)+1367 d-293 e) (d+e x)^{m+1}}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}+\frac {\int \left (112 \left (5 d^2-2 e d+3 e^2\right ) (d+e x)^m+\frac {\left (-5740 d^2+2296 e d+423 e m d-3444 e^2-1367 e^2 m-\frac {i \left (6565 d^2-2626 e d+6412 e m d+3939 e^2-98 e^2 m\right )}{\sqrt {14}}\right ) (d+e x)^m}{10 x-2 i \sqrt {14}+2}+\frac {\left (-5740 d^2+2296 e d+423 e m d-3444 e^2-1367 e^2 m+\frac {i \left (6565 d^2-2626 e d+6412 e m d+3939 e^2-98 e^2 m\right )}{\sqrt {14}}\right ) (d+e x)^m}{10 x+2 i \sqrt {14}+2}\right )dx}{700 \left (5 d^2-2 d e+3 e^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {(x (423 d-1367 e)+1367 d-293 e) (d+e x)^{m+1}}{700 \left (5 x^2+2 x+3\right ) \left (5 d^2-2 d e+3 e^2\right )}+\frac {\frac {\left (i \sqrt {14} \left (6565 d^2-2 d e (1313-3206 m)+e^2 (3939-98 m)\right )+80360 d^2-5922 d e m-32144 d e+19138 e^2 m+48216 e^2\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d+i \sqrt {14} e-e}\right )}{28 (m+1) \left (5 d+i \left (\sqrt {14}+i\right ) e\right )}+\frac {\left (-i \sqrt {14} \left (6565 d^2-2 d e (1313-3206 m)+e^2 (3939-98 m)\right )+80360 d^2-5922 d e m-32144 d e+19138 e^2 m+48216 e^2\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{28 (m+1) \left (5 d-\left (1+i \sqrt {14}\right ) e\right )}+\frac {112 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^{m+1}}{e (m+1)}}{700 \left (5 d^2-2 d e+3 e^2\right )}\) |
-1/700*((1367*d - 293*e + (423*d - 1367*e)*x)*(d + e*x)^(1 + m))/((5*d^2 - 2*d*e + 3*e^2)*(3 + 2*x + 5*x^2)) + ((112*(5*d^2 - 2*d*e + 3*e^2)*(d + e* x)^(1 + m))/(e*(1 + m)) + ((80360*d^2 - 32144*d*e + 48216*e^2 + I*Sqrt[14] *(6565*d^2 - 2*d*e*(1313 - 3206*m) + e^2*(3939 - 98*m)) - 5922*d*e*m + 191 38*e^2*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x ))/(5*d - e + I*Sqrt[14]*e)])/(28*(5*d + I*(I + Sqrt[14])*e)*(1 + m)) + (( 80360*d^2 - 32144*d*e + 48216*e^2 - I*Sqrt[14]*(6565*d^2 - 2*d*e*(1313 - 3 206*m) + e^2*(3939 - 98*m)) - 5922*d*e*m + 19138*e^2*m)*(d + e*x)^(1 + m)* Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - (1 + I*Sqrt[14])*e )])/(28*(5*d - (1 + I*Sqrt[14])*e)*(1 + m)))/(700*(5*d^2 - 2*d*e + 3*e^2))
3.4.71.3.1 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[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, a + b*x + c*x^2, x], R = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[Polyno mialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1)*((R*(b*c*d - b^2*e + 2*a*c*e) - a*S*(2*c*d - b*e) + c* (R*(2*c*d - b*e) - S*(b*d - 2*a*e))*x)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d* e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c )*(c*d^2 - b*d*e + a*e^2)*Qx + R*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - S*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(S*(b*d - 2*a*e) - R *(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x ] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] & & LtQ[p, -1] && !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] | | ILtQ[p + 1/2, 0]))
\[\int \frac {\left (e x +d \right )^{m} \left (4 x^{4}-5 x^{3}+3 x^{2}+x +2\right )}{\left (5 x^{2}+2 x +3\right )^{2}}d x\]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{{\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \,d x } \]
integral((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(25*x^4 + 20*x^3 + 34 *x^2 + 12*x + 9), x)
Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{{\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \,d x } \]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{{\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{\left (3+2 x+5 x^2\right )^2} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,\left (4\,x^4-5\,x^3+3\,x^2+x+2\right )}{{\left (5\,x^2+2\,x+3\right )}^2} \,d x \]