Integrand size = 38, antiderivative size = 255 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac {(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac {4 (d+e x)^{3+m}}{5 e^3 (3+m)}-\frac {\left (6412 i-423 \sqrt {14}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-e+i \sqrt {14} e}\right )}{3500 \left (5 i d-\left (i+\sqrt {14}\right ) e\right ) (1+m)}-\frac {\left (6412 i+423 \sqrt {14}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 \left (5 i d-\left (i-\sqrt {14}\right ) e\right ) (1+m)} \]
1/125*(100*d^2+165*d*e+81*e^2)*(e*x+d)^(1+m)/e^3/(1+m)-1/25*(40*d+33*e)*(e *x+d)^(2+m)/e^3/(2+m)+4/5*(e*x+d)^(3+m)/e^3/(3+m)-1/3500*(e*x+d)^(1+m)*hyp ergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e*(1+I*14^(1/2))))*(6412*I+423*14^(1/ 2))/(1+m)/(5*I*d-e*(I-14^(1/2)))-1/3500*(e*x+d)^(1+m)*hypergeom([1, 1+m],[ 2+m],5*(e*x+d)/(5*d-e+I*14^(1/2)*e))*(6412*I-423*14^(1/2))/(1+m)/(5*I*d-e* (I+14^(1/2)))
Time = 0.59 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {28 \left (100 d^2+165 d e+81 e^2\right )}{e^3 (1+m)}-\frac {140 (40 d+33 e) (d+e x)}{e^3 (2+m)}+\frac {2800 (d+e x)^2}{e^3 (3+m)}-\frac {\left (6412 i+423 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+\left (-1-i \sqrt {14}\right ) e}\right )}{\left (5 i d+\left (-i+\sqrt {14}\right ) e\right ) (1+m)}-\frac {\left (-6412 i+423 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+i \left (i+\sqrt {14}\right ) e}\right )}{\left (-5 i d+\left (i+\sqrt {14}\right ) e\right ) (1+m)}\right )}{3500} \]
((d + e*x)^(1 + m)*((28*(100*d^2 + 165*d*e + 81*e^2))/(e^3*(1 + m)) - (140 *(40*d + 33*e)*(d + e*x))/(e^3*(2 + m)) + (2800*(d + e*x)^2)/(e^3*(3 + m)) - ((6412*I + 423*Sqrt[14])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x ))/(5*d + (-1 - I*Sqrt[14])*e)])/(((5*I)*d + (-I + Sqrt[14])*e)*(1 + m)) - ((-6412*I + 423*Sqrt[14])*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))))/3 500
Time = 0.62 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {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.
| \(\displaystyle \int \frac {\left (4 x^4-5 x^3+3 x^2+x+2\right ) (d+e x)^m}{5 x^2+2 x+3} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^m}{125 e^2}+\frac {(-40 d-33 e) (d+e x)^{m+1}}{25 e^2}+\frac {4 (d+e x)^{m+2}}{5 e^2}+\frac {\left (\frac {458}{125}+\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{10 x-2 i \sqrt {14}+2}+\frac {\left (\frac {458}{125}-\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{10 x+2 i \sqrt {14}+2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac {(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac {4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac {\left (-423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d+i \sqrt {14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt {14}+i\right ) e\right )}-\frac {\left (423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt {14}+i\right ) e\right )}\) |
((100*d^2 + 165*d*e + 81*e^2)*(d + e*x)^(1 + m))/(125*e^3*(1 + m)) - ((40* d + 33*e)*(d + e*x)^(2 + m))/(25*e^3*(2 + m)) + (4*(d + e*x)^(3 + m))/(5*e ^3*(3 + m)) - ((6412*I - 423*Sqrt[14])*(d + e*x)^(1 + m)*Hypergeometric2F1 [1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - e + I*Sqrt[14]*e)])/(3500*((5*I)*d - (I + Sqrt[14])*e)*(1 + m)) - ((6412*I + 423*Sqrt[14])*(d + e*x)^(1 + m)* Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - (1 + I*Sqrt[14])*e )])/(3500*((5*I)*d - (I - Sqrt[14])*e)*(1 + m))
3.4.70.3.1 Defintions of rubi rules used
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]
\[\int \frac {\left (e x +d \right )^{m} \left (4 x^{4}-5 x^{3}+3 x^{2}+x +2\right )}{5 x^{2}+2 x +3}d x\]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \]
\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,\left (4\,x^4-5\,x^3+3\,x^2+x+2\right )}{5\,x^2+2\,x+3} \,d x \]