Integrand size = 36, antiderivative size = 228 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=-\frac {\left (20 d^5+17 d^4 e+17 d^3 e^2+4 d^2 e^3+21 d e^4-7 e^5\right ) x}{e^6}+\frac {\left (20 d^4+17 d^3 e+17 d^2 e^2+4 d e^3+21 e^4\right ) x^2}{2 e^5}-\frac {\left (20 d^3+17 d^2 e+17 d e^2+4 e^3\right ) x^3}{3 e^4}+\frac {\left (20 d^2+17 d e+17 e^2\right ) x^4}{4 e^3}-\frac {(20 d+17 e) x^5}{5 e^2}+\frac {10 x^6}{3 e}+\frac {\left (5 d^2-2 d e+3 e^2\right ) \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e^7} \] Output:
-(20*d^5+17*d^4*e+17*d^3*e^2+4*d^2*e^3+21*d*e^4-7*e^5)*x/e^6+1/2*(20*d^4+1 7*d^3*e+17*d^2*e^2+4*d*e^3+21*e^4)*x^2/e^5-1/3*(20*d^3+17*d^2*e+17*d*e^2+4 *e^3)*x^3/e^4+1/4*(20*d^2+17*d*e+17*e^2)*x^4/e^3-1/5*(20*d+17*e)*x^5/e^2+1 0/3*x^6/e+(5*d^2-2*d*e+3*e^2)*(4*d^4+5*d^3*e+3*d^2*e^2-d*e^3+2*e^4)*ln(e*x +d)/e^7
Time = 0.06 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.79 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=\frac {e x \left (-1200 d^5+60 d^4 e (-17+10 x)-10 d^3 e^2 \left (102-51 x+40 x^2\right )+10 d^2 e^3 \left (-24+51 x-34 x^2+30 x^3\right )-5 d e^4 \left (252-24 x+68 x^2-51 x^3+48 x^4\right )+e^5 \left (420+630 x-80 x^2+255 x^3-204 x^4+200 x^5\right )\right )+60 \left (20 d^6+17 d^5 e+17 d^4 e^2+4 d^3 e^3+21 d^2 e^4-7 d e^5+6 e^6\right ) \log (d+e x)}{60 e^7} \] Input:
Integrate[((3 + 2*x + 5*x^2)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(d + e*x),x]
Output:
(e*x*(-1200*d^5 + 60*d^4*e*(-17 + 10*x) - 10*d^3*e^2*(102 - 51*x + 40*x^2) + 10*d^2*e^3*(-24 + 51*x - 34*x^2 + 30*x^3) - 5*d*e^4*(252 - 24*x + 68*x^ 2 - 51*x^3 + 48*x^4) + e^5*(420 + 630*x - 80*x^2 + 255*x^3 - 204*x^4 + 200 *x^5)) + 60*(20*d^6 + 17*d^5*e + 17*d^4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d *e^5 + 6*e^6)*Log[d + e*x])/(60*e^7)
Time = 0.72 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, 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.
| \(\displaystyle \int \frac {\left (5 x^2+2 x+3\right ) \left (4 x^4-5 x^3+3 x^2+x+2\right )}{d+e x} \, dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \int \left (\frac {x^3 \left (20 d^2+17 d e+17 e^2\right )}{e^3}-\frac {x^2 \left (20 d^3+17 d^2 e+17 d e^2+4 e^3\right )}{e^4}+\frac {x \left (20 d^4+17 d^3 e+17 d^2 e^2+4 d e^3+21 e^4\right )}{e^5}+\frac {-20 d^5-17 d^4 e-17 d^3 e^2-4 d^2 e^3-21 d e^4+7 e^5}{e^6}+\frac {20 d^6+17 d^5 e+17 d^4 e^2+4 d^3 e^3+21 d^2 e^4-7 d e^5+6 e^6}{e^6 (d+e x)}-\frac {x^4 (20 d+17 e)}{e^2}+\frac {20 x^5}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^4 \left (20 d^2+17 d e+17 e^2\right )}{4 e^3}-\frac {x^3 \left (20 d^3+17 d^2 e+17 d e^2+4 e^3\right )}{3 e^4}+\frac {\left (5 d^2-2 d e+3 e^2\right ) \left (4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4\right ) \log (d+e x)}{e^7}+\frac {x^2 \left (20 d^4+17 d^3 e+17 d^2 e^2+4 d e^3+21 e^4\right )}{2 e^5}-\frac {x \left (20 d^5+17 d^4 e+17 d^3 e^2+4 d^2 e^3+21 d e^4-7 e^5\right )}{e^6}-\frac {x^5 (20 d+17 e)}{5 e^2}+\frac {10 x^6}{3 e}\) |
Input:
Int[((3 + 2*x + 5*x^2)*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(d + e*x),x]
Output:
-(((20*d^5 + 17*d^4*e + 17*d^3*e^2 + 4*d^2*e^3 + 21*d*e^4 - 7*e^5)*x)/e^6) + ((20*d^4 + 17*d^3*e + 17*d^2*e^2 + 4*d*e^3 + 21*e^4)*x^2)/(2*e^5) - ((2 0*d^3 + 17*d^2*e + 17*d*e^2 + 4*e^3)*x^3)/(3*e^4) + ((20*d^2 + 17*d*e + 17 *e^2)*x^4)/(4*e^3) - ((20*d + 17*e)*x^5)/(5*e^2) + (10*x^6)/(3*e) + ((5*d^ 2 - 2*d*e + 3*e^2)*(4*d^4 + 5*d^3*e + 3*d^2*e^2 - d*e^3 + 2*e^4)*Log[d + e *x])/e^7
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]
Time = 0.10 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.96
| method | result | size |
| norman | \(\frac {10 x^{6}}{3 e}-\frac {\left (20 d +17 e \right ) x^{5}}{5 e^{2}}+\frac {\left (20 d^{2}+17 d e +17 e^{2}\right ) x^{4}}{4 e^{3}}-\frac {\left (20 d^{3}+17 d^{2} e +17 d \,e^{2}+4 e^{3}\right ) x^{3}}{3 e^{4}}+\frac {\left (20 d^{4}+17 d^{3} e +17 d^{2} e^{2}+4 d \,e^{3}+21 e^{4}\right ) x^{2}}{2 e^{5}}-\frac {\left (20 d^{5}+17 d^{4} e +17 d^{3} e^{2}+4 d^{2} e^{3}+21 d \,e^{4}-7 e^{5}\right ) x}{e^{6}}+\frac {\left (20 d^{6}+17 d^{5} e +17 d^{4} e^{2}+4 d^{3} e^{3}+21 d^{2} e^{4}-7 d \,e^{5}+6 e^{6}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(220\) |
| default | \(-\frac {-\frac {10}{3} x^{6} e^{5}+4 x^{5} e^{4} d +\frac {17}{5} x^{5} e^{5}-5 d^{2} e^{3} x^{4}-\frac {17}{4} x^{4} e^{4} d -\frac {17}{4} x^{4} e^{5}+\frac {20}{3} d^{3} e^{2} x^{3}+\frac {17}{3} d^{2} e^{3} x^{3}+\frac {17}{3} d \,e^{4} x^{3}+\frac {4}{3} e^{5} x^{3}-10 d^{4} e \,x^{2}-\frac {17}{2} d^{3} e^{2} x^{2}-\frac {17}{2} d^{2} e^{3} x^{2}-2 d \,e^{4} x^{2}-\frac {21}{2} e^{5} x^{2}+20 x \,d^{5}+17 d^{4} e x +17 d^{3} e^{2} x +4 d^{2} e^{3} x +21 d \,e^{4} x -7 e^{5} x}{e^{6}}+\frac {\left (20 d^{6}+17 d^{5} e +17 d^{4} e^{2}+4 d^{3} e^{3}+21 d^{2} e^{4}-7 d \,e^{5}+6 e^{6}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(249\) |
| parallelrisch | \(\frac {300 x^{4} d^{2} e^{4}+255 x^{4} d \,e^{5}-400 x^{3} d^{3} e^{3}-340 x^{3} d^{2} e^{4}+1020 \ln \left (e x +d \right ) d^{4} e^{2}+240 \ln \left (e x +d \right ) d^{3} e^{3}+1260 \ln \left (e x +d \right ) d^{2} e^{4}-420 \ln \left (e x +d \right ) d \,e^{5}-340 x^{3} d \,e^{5}+600 x^{2} d^{4} e^{2}+510 x^{2} d^{3} e^{3}+510 x^{2} d^{2} e^{4}+120 x^{2} d \,e^{5}-1200 x \,d^{5} e -1020 x \,d^{4} e^{2}-1020 x \,d^{3} e^{3}-240 x \,d^{2} e^{4}-1260 x d \,e^{5}-240 x^{5} e^{5} d +1020 \ln \left (e x +d \right ) d^{5} e +420 x \,e^{6}+1200 \ln \left (e x +d \right ) d^{6}+360 \ln \left (e x +d \right ) e^{6}+200 x^{6} e^{6}-204 x^{5} e^{6}+255 x^{4} e^{6}-80 x^{3} e^{6}+630 x^{2} e^{6}}{60 e^{7}}\) | \(284\) |
| risch | \(-\frac {17 x^{5}}{5 e}+\frac {21 x^{2}}{2 e}+\frac {7 x}{e}+\frac {6 \ln \left (e x +d \right )}{e}+\frac {10 x^{6}}{3 e}+\frac {17 x^{4}}{4 e}-\frac {4 x^{3}}{3 e}-\frac {4 x^{5} d}{e^{2}}+\frac {5 d^{2} x^{4}}{e^{3}}+\frac {17 x^{4} d}{4 e^{2}}-\frac {20 d^{3} x^{3}}{3 e^{4}}-\frac {17 d^{2} x^{3}}{3 e^{3}}-\frac {17 d \,x^{3}}{3 e^{2}}+\frac {10 d^{4} x^{2}}{e^{5}}+\frac {17 d^{3} x^{2}}{2 e^{4}}+\frac {17 d^{2} x^{2}}{2 e^{3}}+\frac {2 d \,x^{2}}{e^{2}}-\frac {20 x \,d^{5}}{e^{6}}-\frac {17 d^{4} x}{e^{5}}-\frac {17 d^{3} x}{e^{4}}-\frac {4 d^{2} x}{e^{3}}-\frac {21 d x}{e^{2}}+\frac {20 \ln \left (e x +d \right ) d^{6}}{e^{7}}+\frac {17 \ln \left (e x +d \right ) d^{5}}{e^{6}}+\frac {17 \ln \left (e x +d \right ) d^{4}}{e^{5}}+\frac {4 \ln \left (e x +d \right ) d^{3}}{e^{4}}+\frac {21 \ln \left (e x +d \right ) d^{2}}{e^{3}}-\frac {7 \ln \left (e x +d \right ) d}{e^{2}}\) | \(286\) |
Input:
int((5*x^2+2*x+3)*(4*x^4-5*x^3+3*x^2+x+2)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
10/3*x^6/e-1/5*(20*d+17*e)*x^5/e^2+1/4*(20*d^2+17*d*e+17*e^2)*x^4/e^3-1/3* (20*d^3+17*d^2*e+17*d*e^2+4*e^3)*x^3/e^4+1/2*(20*d^4+17*d^3*e+17*d^2*e^2+4 *d*e^3+21*e^4)*x^2/e^5-(20*d^5+17*d^4*e+17*d^3*e^2+4*d^2*e^3+21*d*e^4-7*e^ 5)*x/e^6+(20*d^6+17*d^5*e+17*d^4*e^2+4*d^3*e^3+21*d^2*e^4-7*d*e^5+6*e^6)/e ^7*ln(e*x+d)
Time = 0.06 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.01 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=\frac {200 \, e^{6} x^{6} - 12 \, {\left (20 \, d e^{5} + 17 \, e^{6}\right )} x^{5} + 15 \, {\left (20 \, d^{2} e^{4} + 17 \, d e^{5} + 17 \, e^{6}\right )} x^{4} - 20 \, {\left (20 \, d^{3} e^{3} + 17 \, d^{2} e^{4} + 17 \, d e^{5} + 4 \, e^{6}\right )} x^{3} + 30 \, {\left (20 \, d^{4} e^{2} + 17 \, d^{3} e^{3} + 17 \, d^{2} e^{4} + 4 \, d e^{5} + 21 \, e^{6}\right )} x^{2} - 60 \, {\left (20 \, d^{5} e + 17 \, d^{4} e^{2} + 17 \, d^{3} e^{3} + 4 \, d^{2} e^{4} + 21 \, d e^{5} - 7 \, e^{6}\right )} x + 60 \, {\left (20 \, d^{6} + 17 \, d^{5} e + 17 \, d^{4} e^{2} + 4 \, d^{3} e^{3} + 21 \, d^{2} e^{4} - 7 \, d e^{5} + 6 \, e^{6}\right )} \log \left (e x + d\right )}{60 \, e^{7}} \] Input:
integrate((5*x^2+2*x+3)*(4*x^4-5*x^3+3*x^2+x+2)/(e*x+d),x, algorithm="fric as")
Output:
1/60*(200*e^6*x^6 - 12*(20*d*e^5 + 17*e^6)*x^5 + 15*(20*d^2*e^4 + 17*d*e^5 + 17*e^6)*x^4 - 20*(20*d^3*e^3 + 17*d^2*e^4 + 17*d*e^5 + 4*e^6)*x^3 + 30* (20*d^4*e^2 + 17*d^3*e^3 + 17*d^2*e^4 + 4*d*e^5 + 21*e^6)*x^2 - 60*(20*d^5 *e + 17*d^4*e^2 + 17*d^3*e^3 + 4*d^2*e^4 + 21*d*e^5 - 7*e^6)*x + 60*(20*d^ 6 + 17*d^5*e + 17*d^4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d*e^5 + 6*e^6)*log( e*x + d))/e^7
Time = 0.25 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.03 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=x^{5} \left (- \frac {4 d}{e^{2}} - \frac {17}{5 e}\right ) + x^{4} \cdot \left (\frac {5 d^{2}}{e^{3}} + \frac {17 d}{4 e^{2}} + \frac {17}{4 e}\right ) + x^{3} \left (- \frac {20 d^{3}}{3 e^{4}} - \frac {17 d^{2}}{3 e^{3}} - \frac {17 d}{3 e^{2}} - \frac {4}{3 e}\right ) + x^{2} \cdot \left (\frac {10 d^{4}}{e^{5}} + \frac {17 d^{3}}{2 e^{4}} + \frac {17 d^{2}}{2 e^{3}} + \frac {2 d}{e^{2}} + \frac {21}{2 e}\right ) + x \left (- \frac {20 d^{5}}{e^{6}} - \frac {17 d^{4}}{e^{5}} - \frac {17 d^{3}}{e^{4}} - \frac {4 d^{2}}{e^{3}} - \frac {21 d}{e^{2}} + \frac {7}{e}\right ) + \frac {10 x^{6}}{3 e} + \frac {\left (5 d^{2} - 2 d e + 3 e^{2}\right ) \left (4 d^{4} + 5 d^{3} e + 3 d^{2} e^{2} - d e^{3} + 2 e^{4}\right ) \log {\left (d + e x \right )}}{e^{7}} \] Input:
integrate((5*x**2+2*x+3)*(4*x**4-5*x**3+3*x**2+x+2)/(e*x+d),x)
Output:
x**5*(-4*d/e**2 - 17/(5*e)) + x**4*(5*d**2/e**3 + 17*d/(4*e**2) + 17/(4*e) ) + x**3*(-20*d**3/(3*e**4) - 17*d**2/(3*e**3) - 17*d/(3*e**2) - 4/(3*e)) + x**2*(10*d**4/e**5 + 17*d**3/(2*e**4) + 17*d**2/(2*e**3) + 2*d/e**2 + 21 /(2*e)) + x*(-20*d**5/e**6 - 17*d**4/e**5 - 17*d**3/e**4 - 4*d**2/e**3 - 2 1*d/e**2 + 7/e) + 10*x**6/(3*e) + (5*d**2 - 2*d*e + 3*e**2)*(4*d**4 + 5*d* *3*e + 3*d**2*e**2 - d*e**3 + 2*e**4)*log(d + e*x)/e**7
Time = 0.03 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=\frac {200 \, e^{5} x^{6} - 12 \, {\left (20 \, d e^{4} + 17 \, e^{5}\right )} x^{5} + 15 \, {\left (20 \, d^{2} e^{3} + 17 \, d e^{4} + 17 \, e^{5}\right )} x^{4} - 20 \, {\left (20 \, d^{3} e^{2} + 17 \, d^{2} e^{3} + 17 \, d e^{4} + 4 \, e^{5}\right )} x^{3} + 30 \, {\left (20 \, d^{4} e + 17 \, d^{3} e^{2} + 17 \, d^{2} e^{3} + 4 \, d e^{4} + 21 \, e^{5}\right )} x^{2} - 60 \, {\left (20 \, d^{5} + 17 \, d^{4} e + 17 \, d^{3} e^{2} + 4 \, d^{2} e^{3} + 21 \, d e^{4} - 7 \, e^{5}\right )} x}{60 \, e^{6}} + \frac {{\left (20 \, d^{6} + 17 \, d^{5} e + 17 \, d^{4} e^{2} + 4 \, d^{3} e^{3} + 21 \, d^{2} e^{4} - 7 \, d e^{5} + 6 \, e^{6}\right )} \log \left (e x + d\right )}{e^{7}} \] Input:
integrate((5*x^2+2*x+3)*(4*x^4-5*x^3+3*x^2+x+2)/(e*x+d),x, algorithm="maxi ma")
Output:
1/60*(200*e^5*x^6 - 12*(20*d*e^4 + 17*e^5)*x^5 + 15*(20*d^2*e^3 + 17*d*e^4 + 17*e^5)*x^4 - 20*(20*d^3*e^2 + 17*d^2*e^3 + 17*d*e^4 + 4*e^5)*x^3 + 30* (20*d^4*e + 17*d^3*e^2 + 17*d^2*e^3 + 4*d*e^4 + 21*e^5)*x^2 - 60*(20*d^5 + 17*d^4*e + 17*d^3*e^2 + 4*d^2*e^3 + 21*d*e^4 - 7*e^5)*x)/e^6 + (20*d^6 + 17*d^5*e + 17*d^4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d*e^5 + 6*e^6)*log(e*x + d)/e^7
Time = 0.18 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.09 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=\frac {200 \, e^{5} x^{6} - 240 \, d e^{4} x^{5} - 204 \, e^{5} x^{5} + 300 \, d^{2} e^{3} x^{4} + 255 \, d e^{4} x^{4} + 255 \, e^{5} x^{4} - 400 \, d^{3} e^{2} x^{3} - 340 \, d^{2} e^{3} x^{3} - 340 \, d e^{4} x^{3} - 80 \, e^{5} x^{3} + 600 \, d^{4} e x^{2} + 510 \, d^{3} e^{2} x^{2} + 510 \, d^{2} e^{3} x^{2} + 120 \, d e^{4} x^{2} + 630 \, e^{5} x^{2} - 1200 \, d^{5} x - 1020 \, d^{4} e x - 1020 \, d^{3} e^{2} x - 240 \, d^{2} e^{3} x - 1260 \, d e^{4} x + 420 \, e^{5} x}{60 \, e^{6}} + \frac {{\left (20 \, d^{6} + 17 \, d^{5} e + 17 \, d^{4} e^{2} + 4 \, d^{3} e^{3} + 21 \, d^{2} e^{4} - 7 \, d e^{5} + 6 \, e^{6}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{7}} \] Input:
integrate((5*x^2+2*x+3)*(4*x^4-5*x^3+3*x^2+x+2)/(e*x+d),x, algorithm="giac ")
Output:
1/60*(200*e^5*x^6 - 240*d*e^4*x^5 - 204*e^5*x^5 + 300*d^2*e^3*x^4 + 255*d* e^4*x^4 + 255*e^5*x^4 - 400*d^3*e^2*x^3 - 340*d^2*e^3*x^3 - 340*d*e^4*x^3 - 80*e^5*x^3 + 600*d^4*e*x^2 + 510*d^3*e^2*x^2 + 510*d^2*e^3*x^2 + 120*d*e ^4*x^2 + 630*e^5*x^2 - 1200*d^5*x - 1020*d^4*e*x - 1020*d^3*e^2*x - 240*d^ 2*e^3*x - 1260*d*e^4*x + 420*e^5*x)/e^6 + (20*d^6 + 17*d^5*e + 17*d^4*e^2 + 4*d^3*e^3 + 21*d^2*e^4 - 7*d*e^5 + 6*e^6)*log(abs(e*x + d))/e^7
Time = 0.07 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.14 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=x\,\left (\frac {7}{e}-\frac {d\,\left (\frac {21}{e}+\frac {d\,\left (\frac {4}{e}+\frac {d\,\left (\frac {17}{e}+\frac {d\,\left (\frac {20\,d}{e^2}+\frac {17}{e}\right )}{e}\right )}{e}\right )}{e}\right )}{e}\right )-x^5\,\left (\frac {4\,d}{e^2}+\frac {17}{5\,e}\right )+x^4\,\left (\frac {17}{4\,e}+\frac {d\,\left (\frac {20\,d}{e^2}+\frac {17}{e}\right )}{4\,e}\right )-x^3\,\left (\frac {4}{3\,e}+\frac {d\,\left (\frac {17}{e}+\frac {d\,\left (\frac {20\,d}{e^2}+\frac {17}{e}\right )}{e}\right )}{3\,e}\right )+x^2\,\left (\frac {21}{2\,e}+\frac {d\,\left (\frac {4}{e}+\frac {d\,\left (\frac {17}{e}+\frac {d\,\left (\frac {20\,d}{e^2}+\frac {17}{e}\right )}{e}\right )}{e}\right )}{2\,e}\right )+\frac {10\,x^6}{3\,e}+\frac {\ln \left (d+e\,x\right )\,\left (20\,d^6+17\,d^5\,e+17\,d^4\,e^2+4\,d^3\,e^3+21\,d^2\,e^4-7\,d\,e^5+6\,e^6\right )}{e^7} \] Input:
int(((2*x + 5*x^2 + 3)*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(d + e*x),x)
Output:
x*(7/e - (d*(21/e + (d*(4/e + (d*(17/e + (d*((20*d)/e^2 + 17/e))/e))/e))/e ))/e) - x^5*((4*d)/e^2 + 17/(5*e)) + x^4*(17/(4*e) + (d*((20*d)/e^2 + 17/e ))/(4*e)) - x^3*(4/(3*e) + (d*(17/e + (d*((20*d)/e^2 + 17/e))/e))/(3*e)) + x^2*(21/(2*e) + (d*(4/e + (d*(17/e + (d*((20*d)/e^2 + 17/e))/e))/e))/(2*e )) + (10*x^6)/(3*e) + (log(d + e*x)*(17*d^5*e - 7*d*e^5 + 20*d^6 + 6*e^6 + 21*d^2*e^4 + 4*d^3*e^3 + 17*d^4*e^2))/e^7
Time = 0.16 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.24 \[ \int \frac {\left (3+2 x+5 x^2\right ) \left (2+x+3 x^2-5 x^3+4 x^4\right )}{d+e x} \, dx=\frac {1200 \,\mathrm {log}\left (e x +d \right ) d^{6}+360 \,\mathrm {log}\left (e x +d \right ) e^{6}+200 e^{6} x^{6}-204 e^{6} x^{5}+255 e^{6} x^{4}-80 e^{6} x^{3}+630 e^{6} x^{2}+420 e^{6} x +1020 \,\mathrm {log}\left (e x +d \right ) d^{5} e +1020 \,\mathrm {log}\left (e x +d \right ) d^{4} e^{2}+240 \,\mathrm {log}\left (e x +d \right ) d^{3} e^{3}+1260 \,\mathrm {log}\left (e x +d \right ) d^{2} e^{4}-420 \,\mathrm {log}\left (e x +d \right ) d \,e^{5}-1200 d^{5} e x +600 d^{4} e^{2} x^{2}-1020 d^{4} e^{2} x -400 d^{3} e^{3} x^{3}+510 d^{3} e^{3} x^{2}-1020 d^{3} e^{3} x +300 d^{2} e^{4} x^{4}-340 d^{2} e^{4} x^{3}+510 d^{2} e^{4} x^{2}-240 d^{2} e^{4} x -240 d \,e^{5} x^{5}+255 d \,e^{5} x^{4}-340 d \,e^{5} x^{3}+120 d \,e^{5} x^{2}-1260 d \,e^{5} x}{60 e^{7}} \] Input:
int((5*x^2+2*x+3)*(4*x^4-5*x^3+3*x^2+x+2)/(e*x+d),x)
Output:
(1200*log(d + e*x)*d**6 + 1020*log(d + e*x)*d**5*e + 1020*log(d + e*x)*d** 4*e**2 + 240*log(d + e*x)*d**3*e**3 + 1260*log(d + e*x)*d**2*e**4 - 420*lo g(d + e*x)*d*e**5 + 360*log(d + e*x)*e**6 - 1200*d**5*e*x + 600*d**4*e**2* x**2 - 1020*d**4*e**2*x - 400*d**3*e**3*x**3 + 510*d**3*e**3*x**2 - 1020*d **3*e**3*x + 300*d**2*e**4*x**4 - 340*d**2*e**4*x**3 + 510*d**2*e**4*x**2 - 240*d**2*e**4*x - 240*d*e**5*x**5 + 255*d*e**5*x**4 - 340*d*e**5*x**3 + 120*d*e**5*x**2 - 1260*d*e**5*x + 200*e**6*x**6 - 204*e**6*x**5 + 255*e**6 *x**4 - 80*e**6*x**3 + 630*e**6*x**2 + 420*e**6*x)/(60*e**7)