Integrand size = 19, antiderivative size = 213 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=-\frac {c^2 (5 c d-3 b e) x}{e^6}+\frac {c^3 x^2}{2 e^5}-\frac {d^3 (c d-b e)^3}{4 e^7 (d+e x)^4}+\frac {d^2 (c d-b e)^2 (2 c d-b e)}{e^7 (d+e x)^3}-\frac {3 d (c d-b e) \left (5 c^2 d^2-5 b c d e+b^2 e^2\right )}{2 e^7 (d+e x)^2}+\frac {(2 c d-b e) \left (10 c^2 d^2-10 b c d e+b^2 e^2\right )}{e^7 (d+e x)}+\frac {3 c \left (5 c^2 d^2-5 b c d e+b^2 e^2\right ) \log (d+e x)}{e^7} \]
-c^2*(-3*b*e+5*c*d)*x/e^6+1/2*c^3*x^2/e^5-1/4*d^3*(-b*e+c*d)^3/e^7/(e*x+d) ^4+d^2*(-b*e+c*d)^2*(-b*e+2*c*d)/e^7/(e*x+d)^3-3/2*d*(-b*e+c*d)*(b^2*e^2-5 *b*c*d*e+5*c^2*d^2)/e^7/(e*x+d)^2+(-b*e+2*c*d)*(b^2*e^2-10*b*c*d*e+10*c^2* d^2)/e^7/(e*x+d)+3*c*(b^2*e^2-5*b*c*d*e+5*c^2*d^2)*ln(e*x+d)/e^7
Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.99 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {-4 c^2 e (5 c d-3 b e) x+2 c^3 e^2 x^2-\frac {d^3 (c d-b e)^3}{(d+e x)^4}+\frac {4 d^2 (c d-b e)^2 (2 c d-b e)}{(d+e x)^3}+\frac {6 d \left (-5 c^3 d^3+10 b c^2 d^2 e-6 b^2 c d e^2+b^3 e^3\right )}{(d+e x)^2}+\frac {80 c^3 d^3-120 b c^2 d^2 e+48 b^2 c d e^2-4 b^3 e^3}{d+e x}+12 c \left (5 c^2 d^2-5 b c d e+b^2 e^2\right ) \log (d+e x)}{4 e^7} \]
(-4*c^2*e*(5*c*d - 3*b*e)*x + 2*c^3*e^2*x^2 - (d^3*(c*d - b*e)^3)/(d + e*x )^4 + (4*d^2*(c*d - b*e)^2*(2*c*d - b*e))/(d + e*x)^3 + (6*d*(-5*c^3*d^3 + 10*b*c^2*d^2*e - 6*b^2*c*d*e^2 + b^3*e^3))/(d + e*x)^2 + (80*c^3*d^3 - 12 0*b*c^2*d^2*e + 48*b^2*c*d*e^2 - 4*b^3*e^3)/(d + e*x) + 12*c*(5*c^2*d^2 - 5*b*c*d*e + b^2*e^2)*Log[d + e*x])/(4*e^7)
Time = 0.44 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1140, 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.
| \(\displaystyle \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \int \left (\frac {3 c \left (b^2 e^2-5 b c d e+5 c^2 d^2\right )}{e^6 (d+e x)}+\frac {(2 c d-b e) \left (-b^2 e^2+10 b c d e-10 c^2 d^2\right )}{e^6 (d+e x)^2}+\frac {3 d (c d-b e) \left (b^2 e^2-5 b c d e+5 c^2 d^2\right )}{e^6 (d+e x)^3}-\frac {c^2 (5 c d-3 b e)}{e^6}+\frac {d^3 (c d-b e)^3}{e^6 (d+e x)^5}-\frac {3 d^2 (c d-b e)^2 (2 c d-b e)}{e^6 (d+e x)^4}+\frac {c^3 x}{e^5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(2 c d-b e) \left (b^2 e^2-10 b c d e+10 c^2 d^2\right )}{e^7 (d+e x)}-\frac {3 d (c d-b e) \left (b^2 e^2-5 b c d e+5 c^2 d^2\right )}{2 e^7 (d+e x)^2}+\frac {3 c \left (b^2 e^2-5 b c d e+5 c^2 d^2\right ) \log (d+e x)}{e^7}-\frac {c^2 x (5 c d-3 b e)}{e^6}-\frac {d^3 (c d-b e)^3}{4 e^7 (d+e x)^4}+\frac {d^2 (c d-b e)^2 (2 c d-b e)}{e^7 (d+e x)^3}+\frac {c^3 x^2}{2 e^5}\) |
-((c^2*(5*c*d - 3*b*e)*x)/e^6) + (c^3*x^2)/(2*e^5) - (d^3*(c*d - b*e)^3)/( 4*e^7*(d + e*x)^4) + (d^2*(c*d - b*e)^2*(2*c*d - b*e))/(e^7*(d + e*x)^3) - (3*d*(c*d - b*e)*(5*c^2*d^2 - 5*b*c*d*e + b^2*e^2))/(2*e^7*(d + e*x)^2) + ((2*c*d - b*e)*(10*c^2*d^2 - 10*b*c*d*e + b^2*e^2))/(e^7*(d + e*x)) + (3* c*(5*c^2*d^2 - 5*b*c*d*e + b^2*e^2)*Log[d + e*x])/e^7
3.3.53.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 2.35 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.19
| method | result | size |
| norman | \(\frac {\frac {c^{3} x^{6}}{2 e}-\frac {d^{3} \left (b^{3} e^{3}-25 b^{2} d \,e^{2} c +125 b \,c^{2} d^{2} e -125 c^{3} d^{3}\right )}{4 e^{7}}-\frac {\left (b^{3} e^{3}-12 b^{2} d \,e^{2} c +60 b \,c^{2} d^{2} e -60 c^{3} d^{3}\right ) x^{3}}{e^{4}}+\frac {3 c^{2} \left (b e -c d \right ) x^{5}}{e^{2}}-\frac {3 d \left (b^{3} e^{3}-18 b^{2} d \,e^{2} c +90 b \,c^{2} d^{2} e -90 c^{3} d^{3}\right ) x^{2}}{2 e^{5}}-\frac {d^{2} \left (b^{3} e^{3}-22 b^{2} d \,e^{2} c +110 b \,c^{2} d^{2} e -110 c^{3} d^{3}\right ) x}{e^{6}}}{\left (e x +d \right )^{4}}+\frac {3 c \left (b^{2} e^{2}-5 b c d e +5 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(253\) |
| default | \(\frac {c^{2} \left (\frac {1}{2} c e \,x^{2}+3 b e x -5 c d x \right )}{e^{6}}-\frac {b^{3} e^{3}-12 b^{2} d \,e^{2} c +30 b \,c^{2} d^{2} e -20 c^{3} d^{3}}{e^{7} \left (e x +d \right )}-\frac {d^{2} \left (b^{3} e^{3}-4 b^{2} d \,e^{2} c +5 b \,c^{2} d^{2} e -2 c^{3} d^{3}\right )}{e^{7} \left (e x +d \right )^{3}}+\frac {d^{3} \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}{4 e^{7} \left (e x +d \right )^{4}}+\frac {3 d \left (b^{3} e^{3}-6 b^{2} d \,e^{2} c +10 b \,c^{2} d^{2} e -5 c^{3} d^{3}\right )}{2 e^{7} \left (e x +d \right )^{2}}+\frac {3 c \left (b^{2} e^{2}-5 b c d e +5 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\) | \(260\) |
| risch | \(\frac {c^{3} x^{2}}{2 e^{5}}+\frac {3 c^{2} b x}{e^{5}}-\frac {5 c^{3} d x}{e^{6}}+\frac {\left (-b^{3} e^{5}+12 b^{2} c d \,e^{4}-30 b \,c^{2} d^{2} e^{3}+20 c^{3} d^{3} e^{2}\right ) x^{3}-\frac {3 d e \left (b^{3} e^{3}-18 b^{2} d \,e^{2} c +50 b \,c^{2} d^{2} e -35 c^{3} d^{3}\right ) x^{2}}{2}-d^{2} \left (b^{3} e^{3}-22 b^{2} d \,e^{2} c +65 b \,c^{2} d^{2} e -47 c^{3} d^{3}\right ) x -\frac {d^{3} \left (b^{3} e^{3}-25 b^{2} d \,e^{2} c +77 b \,c^{2} d^{2} e -57 c^{3} d^{3}\right )}{4 e}}{e^{6} \left (e x +d \right )^{4}}+\frac {3 c \ln \left (e x +d \right ) b^{2}}{e^{5}}-\frac {15 c^{2} \ln \left (e x +d \right ) b d}{e^{6}}+\frac {15 c^{3} \ln \left (e x +d \right ) d^{2}}{e^{7}}\) | \(268\) |
| parallelrisch | \(\frac {-240 \ln \left (e x +d \right ) x b \,c^{2} d^{4} e^{2}-b^{3} d^{3} e^{3}+48 \ln \left (e x +d \right ) x^{3} b^{2} c d \,e^{5}+48 \ln \left (e x +d \right ) x \,b^{2} c \,d^{3} e^{3}+12 x^{5} b \,c^{2} e^{6}+240 x^{3} c^{3} d^{3} e^{3}-6 x^{2} b^{3} d \,e^{5}+540 x^{2} c^{3} d^{4} e^{2}-4 x \,b^{3} d^{2} e^{4}+440 x \,c^{3} d^{5} e +125 c^{3} d^{6}-360 \ln \left (e x +d \right ) x^{2} b \,c^{2} d^{3} e^{3}-240 \ln \left (e x +d \right ) x^{3} b \,c^{2} d^{2} e^{4}+240 \ln \left (e x +d \right ) x \,c^{3} d^{5} e +2 x^{6} c^{3} e^{6}+72 \ln \left (e x +d \right ) x^{2} b^{2} c \,d^{2} e^{4}-60 \ln \left (e x +d \right ) x^{4} b \,c^{2} d \,e^{5}+48 x^{3} b^{2} c d \,e^{5}+360 \ln \left (e x +d \right ) x^{2} c^{3} d^{4} e^{2}+240 \ln \left (e x +d \right ) x^{3} c^{3} d^{3} e^{3}+25 b^{2} c \,d^{4} e^{2}-125 b \,c^{2} d^{5} e -12 x^{5} c^{3} d \,e^{5}+60 \ln \left (e x +d \right ) x^{4} c^{3} d^{2} e^{4}-240 x^{3} b \,c^{2} d^{2} e^{4}+108 x^{2} b^{2} c \,d^{2} e^{4}-540 x^{2} b \,c^{2} d^{3} e^{3}+12 \ln \left (e x +d \right ) b^{2} c \,d^{4} e^{2}-60 \ln \left (e x +d \right ) b \,c^{2} d^{5} e +88 x \,b^{2} c \,d^{3} e^{3}+12 \ln \left (e x +d \right ) x^{4} b^{2} c \,e^{6}-4 x^{3} b^{3} e^{6}+60 \ln \left (e x +d \right ) c^{3} d^{6}-440 x b \,c^{2} d^{4} e^{2}}{4 e^{7} \left (e x +d \right )^{4}}\) | \(528\) |
(1/2*c^3*x^6/e-1/4*d^3*(b^3*e^3-25*b^2*c*d*e^2+125*b*c^2*d^2*e-125*c^3*d^3 )/e^7-(b^3*e^3-12*b^2*c*d*e^2+60*b*c^2*d^2*e-60*c^3*d^3)/e^4*x^3+3*c^2*(b* e-c*d)/e^2*x^5-3/2*d*(b^3*e^3-18*b^2*c*d*e^2+90*b*c^2*d^2*e-90*c^3*d^3)/e^ 5*x^2-d^2*(b^3*e^3-22*b^2*c*d*e^2+110*b*c^2*d^2*e-110*c^3*d^3)/e^6*x)/(e*x +d)^4+3*c*(b^2*e^2-5*b*c*d*e+5*c^2*d^2)*ln(e*x+d)/e^7
Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (207) = 414\).
Time = 0.26 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.26 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {2 \, c^{3} e^{6} x^{6} + 57 \, c^{3} d^{6} - 77 \, b c^{2} d^{5} e + 25 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3} - 12 \, {\left (c^{3} d e^{5} - b c^{2} e^{6}\right )} x^{5} - 4 \, {\left (17 \, c^{3} d^{2} e^{4} - 12 \, b c^{2} d e^{5}\right )} x^{4} - 4 \, {\left (8 \, c^{3} d^{3} e^{3} + 12 \, b c^{2} d^{2} e^{4} - 12 \, b^{2} c d e^{5} + b^{3} e^{6}\right )} x^{3} + 6 \, {\left (22 \, c^{3} d^{4} e^{2} - 42 \, b c^{2} d^{3} e^{3} + 18 \, b^{2} c d^{2} e^{4} - b^{3} d e^{5}\right )} x^{2} + 4 \, {\left (42 \, c^{3} d^{5} e - 62 \, b c^{2} d^{4} e^{2} + 22 \, b^{2} c d^{3} e^{3} - b^{3} d^{2} e^{4}\right )} x + 12 \, {\left (5 \, c^{3} d^{6} - 5 \, b c^{2} d^{5} e + b^{2} c d^{4} e^{2} + {\left (5 \, c^{3} d^{2} e^{4} - 5 \, b c^{2} d e^{5} + b^{2} c e^{6}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{3} - 5 \, b c^{2} d^{2} e^{4} + b^{2} c d e^{5}\right )} x^{3} + 6 \, {\left (5 \, c^{3} d^{4} e^{2} - 5 \, b c^{2} d^{3} e^{3} + b^{2} c d^{2} e^{4}\right )} x^{2} + 4 \, {\left (5 \, c^{3} d^{5} e - 5 \, b c^{2} d^{4} e^{2} + b^{2} c d^{3} e^{3}\right )} x\right )} \log \left (e x + d\right )}{4 \, {\left (e^{11} x^{4} + 4 \, d e^{10} x^{3} + 6 \, d^{2} e^{9} x^{2} + 4 \, d^{3} e^{8} x + d^{4} e^{7}\right )}} \]
1/4*(2*c^3*e^6*x^6 + 57*c^3*d^6 - 77*b*c^2*d^5*e + 25*b^2*c*d^4*e^2 - b^3* d^3*e^3 - 12*(c^3*d*e^5 - b*c^2*e^6)*x^5 - 4*(17*c^3*d^2*e^4 - 12*b*c^2*d* e^5)*x^4 - 4*(8*c^3*d^3*e^3 + 12*b*c^2*d^2*e^4 - 12*b^2*c*d*e^5 + b^3*e^6) *x^3 + 6*(22*c^3*d^4*e^2 - 42*b*c^2*d^3*e^3 + 18*b^2*c*d^2*e^4 - b^3*d*e^5 )*x^2 + 4*(42*c^3*d^5*e - 62*b*c^2*d^4*e^2 + 22*b^2*c*d^3*e^3 - b^3*d^2*e^ 4)*x + 12*(5*c^3*d^6 - 5*b*c^2*d^5*e + b^2*c*d^4*e^2 + (5*c^3*d^2*e^4 - 5* b*c^2*d*e^5 + b^2*c*e^6)*x^4 + 4*(5*c^3*d^3*e^3 - 5*b*c^2*d^2*e^4 + b^2*c* d*e^5)*x^3 + 6*(5*c^3*d^4*e^2 - 5*b*c^2*d^3*e^3 + b^2*c*d^2*e^4)*x^2 + 4*( 5*c^3*d^5*e - 5*b*c^2*d^4*e^2 + b^2*c*d^3*e^3)*x)*log(e*x + d))/(e^11*x^4 + 4*d*e^10*x^3 + 6*d^2*e^9*x^2 + 4*d^3*e^8*x + d^4*e^7)
Time = 11.67 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.48 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {c^{3} x^{2}}{2 e^{5}} + \frac {3 c \left (b^{2} e^{2} - 5 b c d e + 5 c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{7}} + x \left (\frac {3 b c^{2}}{e^{5}} - \frac {5 c^{3} d}{e^{6}}\right ) + \frac {- b^{3} d^{3} e^{3} + 25 b^{2} c d^{4} e^{2} - 77 b c^{2} d^{5} e + 57 c^{3} d^{6} + x^{3} \left (- 4 b^{3} e^{6} + 48 b^{2} c d e^{5} - 120 b c^{2} d^{2} e^{4} + 80 c^{3} d^{3} e^{3}\right ) + x^{2} \left (- 6 b^{3} d e^{5} + 108 b^{2} c d^{2} e^{4} - 300 b c^{2} d^{3} e^{3} + 210 c^{3} d^{4} e^{2}\right ) + x \left (- 4 b^{3} d^{2} e^{4} + 88 b^{2} c d^{3} e^{3} - 260 b c^{2} d^{4} e^{2} + 188 c^{3} d^{5} e\right )}{4 d^{4} e^{7} + 16 d^{3} e^{8} x + 24 d^{2} e^{9} x^{2} + 16 d e^{10} x^{3} + 4 e^{11} x^{4}} \]
c**3*x**2/(2*e**5) + 3*c*(b**2*e**2 - 5*b*c*d*e + 5*c**2*d**2)*log(d + e*x )/e**7 + x*(3*b*c**2/e**5 - 5*c**3*d/e**6) + (-b**3*d**3*e**3 + 25*b**2*c* d**4*e**2 - 77*b*c**2*d**5*e + 57*c**3*d**6 + x**3*(-4*b**3*e**6 + 48*b**2 *c*d*e**5 - 120*b*c**2*d**2*e**4 + 80*c**3*d**3*e**3) + x**2*(-6*b**3*d*e* *5 + 108*b**2*c*d**2*e**4 - 300*b*c**2*d**3*e**3 + 210*c**3*d**4*e**2) + x *(-4*b**3*d**2*e**4 + 88*b**2*c*d**3*e**3 - 260*b*c**2*d**4*e**2 + 188*c** 3*d**5*e))/(4*d**4*e**7 + 16*d**3*e**8*x + 24*d**2*e**9*x**2 + 16*d*e**10* x**3 + 4*e**11*x**4)
Time = 0.23 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.42 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {57 \, c^{3} d^{6} - 77 \, b c^{2} d^{5} e + 25 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3} + 4 \, {\left (20 \, c^{3} d^{3} e^{3} - 30 \, b c^{2} d^{2} e^{4} + 12 \, b^{2} c d e^{5} - b^{3} e^{6}\right )} x^{3} + 6 \, {\left (35 \, c^{3} d^{4} e^{2} - 50 \, b c^{2} d^{3} e^{3} + 18 \, b^{2} c d^{2} e^{4} - b^{3} d e^{5}\right )} x^{2} + 4 \, {\left (47 \, c^{3} d^{5} e - 65 \, b c^{2} d^{4} e^{2} + 22 \, b^{2} c d^{3} e^{3} - b^{3} d^{2} e^{4}\right )} x}{4 \, {\left (e^{11} x^{4} + 4 \, d e^{10} x^{3} + 6 \, d^{2} e^{9} x^{2} + 4 \, d^{3} e^{8} x + d^{4} e^{7}\right )}} + \frac {c^{3} e x^{2} - 2 \, {\left (5 \, c^{3} d - 3 \, b c^{2} e\right )} x}{2 \, e^{6}} + \frac {3 \, {\left (5 \, c^{3} d^{2} - 5 \, b c^{2} d e + b^{2} c e^{2}\right )} \log \left (e x + d\right )}{e^{7}} \]
1/4*(57*c^3*d^6 - 77*b*c^2*d^5*e + 25*b^2*c*d^4*e^2 - b^3*d^3*e^3 + 4*(20* c^3*d^3*e^3 - 30*b*c^2*d^2*e^4 + 12*b^2*c*d*e^5 - b^3*e^6)*x^3 + 6*(35*c^3 *d^4*e^2 - 50*b*c^2*d^3*e^3 + 18*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2 + 4*(47*c^ 3*d^5*e - 65*b*c^2*d^4*e^2 + 22*b^2*c*d^3*e^3 - b^3*d^2*e^4)*x)/(e^11*x^4 + 4*d*e^10*x^3 + 6*d^2*e^9*x^2 + 4*d^3*e^8*x + d^4*e^7) + 1/2*(c^3*e*x^2 - 2*(5*c^3*d - 3*b*c^2*e)*x)/e^6 + 3*(5*c^3*d^2 - 5*b*c^2*d*e + b^2*c*e^2)* log(e*x + d)/e^7
Time = 0.27 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.83 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {{\left (c^{3} - \frac {6 \, {\left (2 \, c^{3} d e - b c^{2} e^{2}\right )}}{{\left (e x + d\right )} e}\right )} {\left (e x + d\right )}^{2}}{2 \, e^{7}} - \frac {3 \, {\left (5 \, c^{3} d^{2} - 5 \, b c^{2} d e + b^{2} c e^{2}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{7}} + \frac {\frac {80 \, c^{3} d^{3} e^{29}}{e x + d} - \frac {30 \, c^{3} d^{4} e^{29}}{{\left (e x + d\right )}^{2}} + \frac {8 \, c^{3} d^{5} e^{29}}{{\left (e x + d\right )}^{3}} - \frac {c^{3} d^{6} e^{29}}{{\left (e x + d\right )}^{4}} - \frac {120 \, b c^{2} d^{2} e^{30}}{e x + d} + \frac {60 \, b c^{2} d^{3} e^{30}}{{\left (e x + d\right )}^{2}} - \frac {20 \, b c^{2} d^{4} e^{30}}{{\left (e x + d\right )}^{3}} + \frac {3 \, b c^{2} d^{5} e^{30}}{{\left (e x + d\right )}^{4}} + \frac {48 \, b^{2} c d e^{31}}{e x + d} - \frac {36 \, b^{2} c d^{2} e^{31}}{{\left (e x + d\right )}^{2}} + \frac {16 \, b^{2} c d^{3} e^{31}}{{\left (e x + d\right )}^{3}} - \frac {3 \, b^{2} c d^{4} e^{31}}{{\left (e x + d\right )}^{4}} - \frac {4 \, b^{3} e^{32}}{e x + d} + \frac {6 \, b^{3} d e^{32}}{{\left (e x + d\right )}^{2}} - \frac {4 \, b^{3} d^{2} e^{32}}{{\left (e x + d\right )}^{3}} + \frac {b^{3} d^{3} e^{32}}{{\left (e x + d\right )}^{4}}}{4 \, e^{36}} \]
1/2*(c^3 - 6*(2*c^3*d*e - b*c^2*e^2)/((e*x + d)*e))*(e*x + d)^2/e^7 - 3*(5 *c^3*d^2 - 5*b*c^2*d*e + b^2*c*e^2)*log(abs(e*x + d)/((e*x + d)^2*abs(e))) /e^7 + 1/4*(80*c^3*d^3*e^29/(e*x + d) - 30*c^3*d^4*e^29/(e*x + d)^2 + 8*c^ 3*d^5*e^29/(e*x + d)^3 - c^3*d^6*e^29/(e*x + d)^4 - 120*b*c^2*d^2*e^30/(e* x + d) + 60*b*c^2*d^3*e^30/(e*x + d)^2 - 20*b*c^2*d^4*e^30/(e*x + d)^3 + 3 *b*c^2*d^5*e^30/(e*x + d)^4 + 48*b^2*c*d*e^31/(e*x + d) - 36*b^2*c*d^2*e^3 1/(e*x + d)^2 + 16*b^2*c*d^3*e^31/(e*x + d)^3 - 3*b^2*c*d^4*e^31/(e*x + d) ^4 - 4*b^3*e^32/(e*x + d) + 6*b^3*d*e^32/(e*x + d)^2 - 4*b^3*d^2*e^32/(e*x + d)^3 + b^3*d^3*e^32/(e*x + d)^4)/e^36
Time = 0.12 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.42 \[ \int \frac {\left (b x+c x^2\right )^3}{(d+e x)^5} \, dx=x\,\left (\frac {3\,b\,c^2}{e^5}-\frac {5\,c^3\,d}{e^6}\right )-\frac {x^2\,\left (\frac {3\,b^3\,d\,e^4}{2}-27\,b^2\,c\,d^2\,e^3+75\,b\,c^2\,d^3\,e^2-\frac {105\,c^3\,d^4\,e}{2}\right )-x\,\left (-b^3\,d^2\,e^3+22\,b^2\,c\,d^3\,e^2-65\,b\,c^2\,d^4\,e+47\,c^3\,d^5\right )-\frac {-b^3\,d^3\,e^3+25\,b^2\,c\,d^4\,e^2-77\,b\,c^2\,d^5\,e+57\,c^3\,d^6}{4\,e}+x^3\,\left (b^3\,e^5-12\,b^2\,c\,d\,e^4+30\,b\,c^2\,d^2\,e^3-20\,c^3\,d^3\,e^2\right )}{d^4\,e^6+4\,d^3\,e^7\,x+6\,d^2\,e^8\,x^2+4\,d\,e^9\,x^3+e^{10}\,x^4}+\frac {\ln \left (d+e\,x\right )\,\left (3\,b^2\,c\,e^2-15\,b\,c^2\,d\,e+15\,c^3\,d^2\right )}{e^7}+\frac {c^3\,x^2}{2\,e^5} \]
x*((3*b*c^2)/e^5 - (5*c^3*d)/e^6) - (x^2*((3*b^3*d*e^4)/2 - (105*c^3*d^4*e )/2 + 75*b*c^2*d^3*e^2 - 27*b^2*c*d^2*e^3) - x*(47*c^3*d^5 - b^3*d^2*e^3 + 22*b^2*c*d^3*e^2 - 65*b*c^2*d^4*e) - (57*c^3*d^6 - b^3*d^3*e^3 + 25*b^2*c *d^4*e^2 - 77*b*c^2*d^5*e)/(4*e) + x^3*(b^3*e^5 - 20*c^3*d^3*e^2 + 30*b*c^ 2*d^2*e^3 - 12*b^2*c*d*e^4))/(d^4*e^6 + e^10*x^4 + 4*d^3*e^7*x + 4*d*e^9*x ^3 + 6*d^2*e^8*x^2) + (log(d + e*x)*(15*c^3*d^2 + 3*b^2*c*e^2 - 15*b*c^2*d *e))/e^7 + (c^3*x^2)/(2*e^5)