Integrand size = 21, antiderivative size = 137 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{e^5}-\frac {d \left (c d^2-b d e+a e^2\right ) x^2}{2 e^4}+\frac {\left (c d^2-b d e+a e^2\right ) x^3}{3 e^3}-\frac {(c d-b e) x^4}{4 e^2}+\frac {c x^5}{5 e}-\frac {d^3 \left (c d^2-b d e+a e^2\right ) \log (d+e x)}{e^6} \] Output:
d^2*(a*e^2-b*d*e+c*d^2)*x/e^5-1/2*d*(a*e^2-b*d*e+c*d^2)*x^2/e^4+1/3*(a*e^2 -b*d*e+c*d^2)*x^3/e^3-1/4*(-b*e+c*d)*x^4/e^2+1/5*c*x^5/e-d^3*(a*e^2-b*d*e+ c*d^2)*ln(e*x+d)/e^6
Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {60 d^2 e \left (c d^2+e (-b d+a e)\right ) x-30 d e^2 \left (c d^2+e (-b d+a e)\right ) x^2+20 e^3 \left (c d^2+e (-b d+a e)\right ) x^3+15 e^4 (-c d+b e) x^4+12 c e^5 x^5-60 \left (c d^5+d^3 e (-b d+a e)\right ) \log (d+e x)}{60 e^6} \] Input:
Integrate[(x^3*(a + b*x + c*x^2))/(d + e*x),x]
Output:
(60*d^2*e*(c*d^2 + e*(-(b*d) + a*e))*x - 30*d*e^2*(c*d^2 + e*(-(b*d) + a*e ))*x^2 + 20*e^3*(c*d^2 + e*(-(b*d) + a*e))*x^3 + 15*e^4*(-(c*d) + b*e)*x^4 + 12*c*e^5*x^5 - 60*(c*d^5 + d^3*e*(-(b*d) + a*e))*Log[d + e*x])/(60*e^6)
Time = 0.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1195, 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.
\(\displaystyle \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {d^2 \left (a e^2-b d e+c d^2\right )}{e^5}-\frac {d x \left (a e^2-b d e+c d^2\right )}{e^4}+\frac {x^2 \left (a e^2-b d e+c d^2\right )}{e^3}-\frac {d^3 \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}+\frac {x^3 (b e-c d)}{e^2}+\frac {c x^4}{e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 x \left (a e^2-b d e+c d^2\right )}{e^5}-\frac {d x^2 \left (a e^2-b d e+c d^2\right )}{2 e^4}+\frac {x^3 \left (a e^2-b d e+c d^2\right )}{3 e^3}-\frac {d^3 \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^6}-\frac {x^4 (c d-b e)}{4 e^2}+\frac {c x^5}{5 e}\) |
Input:
Int[(x^3*(a + b*x + c*x^2))/(d + e*x),x]
Output:
(d^2*(c*d^2 - b*d*e + a*e^2)*x)/e^5 - (d*(c*d^2 - b*d*e + a*e^2)*x^2)/(2*e ^4) + ((c*d^2 - b*d*e + a*e^2)*x^3)/(3*e^3) - ((c*d - b*e)*x^4)/(4*e^2) + (c*x^5)/(5*e) - (d^3*(c*d^2 - b*d*e + a*e^2)*Log[d + e*x])/e^6
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.69 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.95
method | result | size |
norman | \(\frac {d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) x}{e^{5}}+\frac {c \,x^{5}}{5 e}+\frac {\left (b e -c d \right ) x^{4}}{4 e^{2}}+\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) x^{3}}{3 e^{3}}-\frac {d \left (a \,e^{2}-b d e +c \,d^{2}\right ) x^{2}}{2 e^{4}}-\frac {d^{3} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}\) | \(130\) |
default | \(\frac {\frac {1}{5} c \,x^{5} e^{4}+\frac {1}{4} b \,e^{4} x^{4}-\frac {1}{4} c d \,e^{3} x^{4}+\frac {1}{3} a \,e^{4} x^{3}-\frac {1}{3} b d \,e^{3} x^{3}+\frac {1}{3} c \,d^{2} e^{2} x^{3}-\frac {1}{2} a d \,e^{3} x^{2}+\frac {1}{2} b \,d^{2} e^{2} x^{2}-\frac {1}{2} c \,d^{3} e \,x^{2}+a \,d^{2} e^{2} x -b \,d^{3} e x +c \,d^{4} x}{e^{5}}-\frac {d^{3} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}\) | \(151\) |
risch | \(\frac {c \,x^{5}}{5 e}+\frac {b \,x^{4}}{4 e}-\frac {c d \,x^{4}}{4 e^{2}}+\frac {a \,x^{3}}{3 e}-\frac {b d \,x^{3}}{3 e^{2}}+\frac {c \,d^{2} x^{3}}{3 e^{3}}-\frac {a d \,x^{2}}{2 e^{2}}+\frac {b \,d^{2} x^{2}}{2 e^{3}}-\frac {c \,d^{3} x^{2}}{2 e^{4}}+\frac {a \,d^{2} x}{e^{3}}-\frac {b \,d^{3} x}{e^{4}}+\frac {c \,d^{4} x}{e^{5}}-\frac {d^{3} \ln \left (e x +d \right ) a}{e^{4}}+\frac {d^{4} \ln \left (e x +d \right ) b}{e^{5}}-\frac {d^{5} \ln \left (e x +d \right ) c}{e^{6}}\) | \(167\) |
parallelrisch | \(-\frac {-12 c \,x^{5} e^{5}-15 b \,e^{5} x^{4}+15 x^{4} c d \,e^{4}-20 x^{3} a \,e^{5}+20 x^{3} b d \,e^{4}-20 x^{3} c \,d^{2} e^{3}+30 x^{2} a d \,e^{4}-30 x^{2} b \,d^{2} e^{3}+30 x^{2} c \,d^{3} e^{2}+60 \ln \left (e x +d \right ) a \,d^{3} e^{2}-60 \ln \left (e x +d \right ) b \,d^{4} e +60 \ln \left (e x +d \right ) c \,d^{5}-60 x a \,d^{2} e^{3}+60 x b \,d^{3} e^{2}-60 x c \,d^{4} e}{60 e^{6}}\) | \(168\) |
Input:
int(x^3*(c*x^2+b*x+a)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
d^2*(a*e^2-b*d*e+c*d^2)*x/e^5+1/5*c*x^5/e+1/4/e^2*(b*e-c*d)*x^4+1/3*(a*e^2 -b*d*e+c*d^2)*x^3/e^3-1/2*d*(a*e^2-b*d*e+c*d^2)*x^2/e^4-d^3*(a*e^2-b*d*e+c *d^2)*ln(e*x+d)/e^6
Time = 0.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {12 \, c e^{5} x^{5} - 15 \, {\left (c d e^{4} - b e^{5}\right )} x^{4} + 20 \, {\left (c d^{2} e^{3} - b d e^{4} + a e^{5}\right )} x^{3} - 30 \, {\left (c d^{3} e^{2} - b d^{2} e^{3} + a d e^{4}\right )} x^{2} + 60 \, {\left (c d^{4} e - b d^{3} e^{2} + a d^{2} e^{3}\right )} x - 60 \, {\left (c d^{5} - b d^{4} e + a d^{3} e^{2}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \] Input:
integrate(x^3*(c*x^2+b*x+a)/(e*x+d),x, algorithm="fricas")
Output:
1/60*(12*c*e^5*x^5 - 15*(c*d*e^4 - b*e^5)*x^4 + 20*(c*d^2*e^3 - b*d*e^4 + a*e^5)*x^3 - 30*(c*d^3*e^2 - b*d^2*e^3 + a*d*e^4)*x^2 + 60*(c*d^4*e - b*d^ 3*e^2 + a*d^2*e^3)*x - 60*(c*d^5 - b*d^4*e + a*d^3*e^2)*log(e*x + d))/e^6
Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {c x^{5}}{5 e} - \frac {d^{3} \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} + x^{4} \left (\frac {b}{4 e} - \frac {c d}{4 e^{2}}\right ) + x^{3} \left (\frac {a}{3 e} - \frac {b d}{3 e^{2}} + \frac {c d^{2}}{3 e^{3}}\right ) + x^{2} \left (- \frac {a d}{2 e^{2}} + \frac {b d^{2}}{2 e^{3}} - \frac {c d^{3}}{2 e^{4}}\right ) + x \left (\frac {a d^{2}}{e^{3}} - \frac {b d^{3}}{e^{4}} + \frac {c d^{4}}{e^{5}}\right ) \] Input:
integrate(x**3*(c*x**2+b*x+a)/(e*x+d),x)
Output:
c*x**5/(5*e) - d**3*(a*e**2 - b*d*e + c*d**2)*log(d + e*x)/e**6 + x**4*(b/ (4*e) - c*d/(4*e**2)) + x**3*(a/(3*e) - b*d/(3*e**2) + c*d**2/(3*e**3)) + x**2*(-a*d/(2*e**2) + b*d**2/(2*e**3) - c*d**3/(2*e**4)) + x*(a*d**2/e**3 - b*d**3/e**4 + c*d**4/e**5)
Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {12 \, c e^{4} x^{5} - 15 \, {\left (c d e^{3} - b e^{4}\right )} x^{4} + 20 \, {\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x^{3} - 30 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x^{2} + 60 \, {\left (c d^{4} - b d^{3} e + a d^{2} e^{2}\right )} x}{60 \, e^{5}} - \frac {{\left (c d^{5} - b d^{4} e + a d^{3} e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \] Input:
integrate(x^3*(c*x^2+b*x+a)/(e*x+d),x, algorithm="maxima")
Output:
1/60*(12*c*e^4*x^5 - 15*(c*d*e^3 - b*e^4)*x^4 + 20*(c*d^2*e^2 - b*d*e^3 + a*e^4)*x^3 - 30*(c*d^3*e - b*d^2*e^2 + a*d*e^3)*x^2 + 60*(c*d^4 - b*d^3*e + a*d^2*e^2)*x)/e^5 - (c*d^5 - b*d^4*e + a*d^3*e^2)*log(e*x + d)/e^6
Time = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.14 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {12 \, c e^{4} x^{5} - 15 \, c d e^{3} x^{4} + 15 \, b e^{4} x^{4} + 20 \, c d^{2} e^{2} x^{3} - 20 \, b d e^{3} x^{3} + 20 \, a e^{4} x^{3} - 30 \, c d^{3} e x^{2} + 30 \, b d^{2} e^{2} x^{2} - 30 \, a d e^{3} x^{2} + 60 \, c d^{4} x - 60 \, b d^{3} e x + 60 \, a d^{2} e^{2} x}{60 \, e^{5}} - \frac {{\left (c d^{5} - b d^{4} e + a d^{3} e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} \] Input:
integrate(x^3*(c*x^2+b*x+a)/(e*x+d),x, algorithm="giac")
Output:
1/60*(12*c*e^4*x^5 - 15*c*d*e^3*x^4 + 15*b*e^4*x^4 + 20*c*d^2*e^2*x^3 - 20 *b*d*e^3*x^3 + 20*a*e^4*x^3 - 30*c*d^3*e*x^2 + 30*b*d^2*e^2*x^2 - 30*a*d*e ^3*x^2 + 60*c*d^4*x - 60*b*d^3*e*x + 60*a*d^2*e^2*x)/e^5 - (c*d^5 - b*d^4* e + a*d^3*e^2)*log(abs(e*x + d))/e^6
Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=x^4\,\left (\frac {b}{4\,e}-\frac {c\,d}{4\,e^2}\right )+x^3\,\left (\frac {a}{3\,e}-\frac {d\,\left (\frac {b}{e}-\frac {c\,d}{e^2}\right )}{3\,e}\right )+\frac {c\,x^5}{5\,e}-\frac {\ln \left (d+e\,x\right )\,\left (c\,d^5-b\,d^4\,e+a\,d^3\,e^2\right )}{e^6}-\frac {d\,x^2\,\left (\frac {a}{e}-\frac {d\,\left (\frac {b}{e}-\frac {c\,d}{e^2}\right )}{e}\right )}{2\,e}+\frac {d^2\,x\,\left (\frac {a}{e}-\frac {d\,\left (\frac {b}{e}-\frac {c\,d}{e^2}\right )}{e}\right )}{e^2} \] Input:
int((x^3*(a + b*x + c*x^2))/(d + e*x),x)
Output:
x^4*(b/(4*e) - (c*d)/(4*e^2)) + x^3*(a/(3*e) - (d*(b/e - (c*d)/e^2))/(3*e) ) + (c*x^5)/(5*e) - (log(d + e*x)*(c*d^5 + a*d^3*e^2 - b*d^4*e))/e^6 - (d* x^2*(a/e - (d*(b/e - (c*d)/e^2))/e))/(2*e) + (d^2*x*(a/e - (d*(b/e - (c*d) /e^2))/e))/e^2
Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22 \[ \int \frac {x^3 \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {-60 \,\mathrm {log}\left (e x +d \right ) a \,d^{3} e^{2}+60 \,\mathrm {log}\left (e x +d \right ) b \,d^{4} e -60 \,\mathrm {log}\left (e x +d \right ) c \,d^{5}+60 a \,d^{2} e^{3} x -30 a d \,e^{4} x^{2}+20 a \,e^{5} x^{3}-60 b \,d^{3} e^{2} x +30 b \,d^{2} e^{3} x^{2}-20 b d \,e^{4} x^{3}+15 b \,e^{5} x^{4}+60 c \,d^{4} e x -30 c \,d^{3} e^{2} x^{2}+20 c \,d^{2} e^{3} x^{3}-15 c d \,e^{4} x^{4}+12 c \,e^{5} x^{5}}{60 e^{6}} \] Input:
int(x^3*(c*x^2+b*x+a)/(e*x+d),x)
Output:
( - 60*log(d + e*x)*a*d**3*e**2 + 60*log(d + e*x)*b*d**4*e - 60*log(d + e* x)*c*d**5 + 60*a*d**2*e**3*x - 30*a*d*e**4*x**2 + 20*a*e**5*x**3 - 60*b*d* *3*e**2*x + 30*b*d**2*e**3*x**2 - 20*b*d*e**4*x**3 + 15*b*e**5*x**4 + 60*c *d**4*e*x - 30*c*d**3*e**2*x**2 + 20*c*d**2*e**3*x**3 - 15*c*d*e**4*x**4 + 12*c*e**5*x**5)/(60*e**6)