Integrand size = 24, antiderivative size = 104 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {\left (2 c^2 d^2+b^2 e^2-c e (3 b d-2 a e)\right ) x}{e^3}-\frac {c (2 c d-3 b e) x^2}{2 e^2}+\frac {2 c^2 x^3}{3 e}-\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right ) \log (d+e x)}{e^4} \] Output:
(2*c^2*d^2+b^2*e^2-c*e*(-2*a*e+3*b*d))*x/e^3-1/2*c*(-3*b*e+2*c*d)*x^2/e^2+ 2/3*c^2*x^3/e-(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)*ln(e*x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {e x \left (6 b^2 e^2+3 c e (-6 b d+4 a e+3 b e x)+2 c^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (2 c d-b e) \left (c d^2+e (-b d+a e)\right ) \log (d+e x)}{6 e^4} \] Input:
Integrate[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x),x]
Output:
(e*x*(6*b^2*e^2 + 3*c*e*(-6*b*d + 4*a*e + 3*b*e*x) + 2*c^2*(6*d^2 - 3*d*e* x + 2*e^2*x^2)) - 6*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*Log[d + e*x]) /(6*e^4)
Time = 0.49 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {-c e (3 b d-2 a e)+b^2 e^2+2 c^2 d^2}{e^3}+\frac {(b e-2 c d) \left (a e^2-b d e+c d^2\right )}{e^3 (d+e x)}-\frac {c x (2 c d-3 b e)}{e^2}+\frac {2 c^2 x^2}{e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (-c e (3 b d-2 a e)+b^2 e^2+2 c^2 d^2\right )}{e^3}-\frac {(2 c d-b e) \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^4}-\frac {c x^2 (2 c d-3 b e)}{2 e^2}+\frac {2 c^2 x^3}{3 e}\) |
Input:
Int[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x),x]
Output:
((2*c^2*d^2 + b^2*e^2 - c*e*(3*b*d - 2*a*e))*x)/e^3 - (c*(2*c*d - 3*b*e)*x ^2)/(2*e^2) + (2*c^2*x^3)/(3*e) - ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)*L og[d + e*x])/e^4
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 = 0.88 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11
method | result | size |
norman | \(\frac {\left (2 a c \,e^{2}+b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right ) x}{e^{3}}+\frac {2 c^{2} x^{3}}{3 e}+\frac {c \left (3 b e -2 c d \right ) x^{2}}{2 e^{2}}+\frac {\left (a \,e^{3} b -2 a d \,e^{2} c -d \,e^{2} b^{2}+3 d^{2} e b c -2 c^{2} d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(115\) |
default | \(\frac {\frac {2}{3} c^{2} e^{2} x^{3}+\frac {3}{2} b c \,e^{2} x^{2}-c^{2} d e \,x^{2}+2 a c \,e^{2} x +b^{2} e^{2} x -3 b c d e x +2 c^{2} d^{2} x}{e^{3}}+\frac {\left (a \,e^{3} b -2 a d \,e^{2} c -d \,e^{2} b^{2}+3 d^{2} e b c -2 c^{2} d^{3}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(120\) |
risch | \(\frac {2 c^{2} x^{3}}{3 e}+\frac {3 b c \,x^{2}}{2 e}-\frac {c^{2} d \,x^{2}}{e^{2}}+\frac {2 a c x}{e}+\frac {b^{2} x}{e}-\frac {3 b c d x}{e^{2}}+\frac {2 c^{2} d^{2} x}{e^{3}}+\frac {\ln \left (e x +d \right ) a b}{e}-\frac {2 \ln \left (e x +d \right ) a d c}{e^{2}}-\frac {\ln \left (e x +d \right ) d \,b^{2}}{e^{2}}+\frac {3 \ln \left (e x +d \right ) d^{2} b c}{e^{3}}-\frac {2 \ln \left (e x +d \right ) c^{2} d^{3}}{e^{4}}\) | \(146\) |
parallelrisch | \(\frac {4 e^{3} x^{3} c^{2}+9 e^{3} x^{2} c b -6 d \,e^{2} c^{2} x^{2}+6 \ln \left (e x +d \right ) a b \,e^{3}-12 \ln \left (e x +d \right ) a c d \,e^{2}-6 \ln \left (e x +d \right ) b^{2} d \,e^{2}+18 \ln \left (e x +d \right ) b c \,d^{2} e -12 \ln \left (e x +d \right ) c^{2} d^{3}+12 a c \,e^{3} x +6 e^{3} x \,b^{2}-18 d \,e^{2} c b x +12 d^{2} e \,c^{2} x}{6 e^{4}}\) | \(146\) |
Input:
int((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
(2*a*c*e^2+b^2*e^2-3*b*c*d*e+2*c^2*d^2)/e^3*x+2/3*c^2*x^3/e+1/2*c/e^2*(3*b *e-2*c*d)*x^2+(a*b*e^3-2*a*c*d*e^2-b^2*d*e^2+3*b*c*d^2*e-2*c^2*d^3)/e^4*ln (e*x+d)
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {4 \, c^{2} e^{3} x^{3} - 3 \, {\left (2 \, c^{2} d e^{2} - 3 \, b c e^{3}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{2} e - 3 \, b c d e^{2} + {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x - 6 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d),x, algorithm="fricas")
Output:
1/6*(4*c^2*e^3*x^3 - 3*(2*c^2*d*e^2 - 3*b*c*e^3)*x^2 + 6*(2*c^2*d^2*e - 3* b*c*d*e^2 + (b^2 + 2*a*c)*e^3)*x - 6*(2*c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*log(e*x + d))/e^4
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {2 c^{2} x^{3}}{3 e} + x^{2} \cdot \left (\frac {3 b c}{2 e} - \frac {c^{2} d}{e^{2}}\right ) + x \left (\frac {2 a c}{e} + \frac {b^{2}}{e} - \frac {3 b c d}{e^{2}} + \frac {2 c^{2} d^{2}}{e^{3}}\right ) + \frac {\left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \] Input:
integrate((2*c*x+b)*(c*x**2+b*x+a)/(e*x+d),x)
Output:
2*c**2*x**3/(3*e) + x**2*(3*b*c/(2*e) - c**2*d/e**2) + x*(2*a*c/e + b**2/e - 3*b*c*d/e**2 + 2*c**2*d**2/e**3) + (b*e - 2*c*d)*(a*e**2 - b*d*e + c*d* *2)*log(d + e*x)/e**4
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {4 \, c^{2} e^{2} x^{3} - 3 \, {\left (2 \, c^{2} d e - 3 \, b c e^{2}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{2} - 3 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d),x, algorithm="maxima")
Output:
1/6*(4*c^2*e^2*x^3 - 3*(2*c^2*d*e - 3*b*c*e^2)*x^2 + 6*(2*c^2*d^2 - 3*b*c* d*e + (b^2 + 2*a*c)*e^2)*x)/e^3 - (2*c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^ 2 + 2*a*c)*d*e^2)*log(e*x + d)/e^4
Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {4 \, c^{2} e^{2} x^{3} - 6 \, c^{2} d e x^{2} + 9 \, b c e^{2} x^{2} + 12 \, c^{2} d^{2} x - 18 \, b c d e x + 6 \, b^{2} e^{2} x + 12 \, a c e^{2} x}{6 \, e^{3}} - \frac {{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2} + 2 \, a c d e^{2} - a b e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d),x, algorithm="giac")
Output:
1/6*(4*c^2*e^2*x^3 - 6*c^2*d*e*x^2 + 9*b*c*e^2*x^2 + 12*c^2*d^2*x - 18*b*c *d*e*x + 6*b^2*e^2*x + 12*a*c*e^2*x)/e^3 - (2*c^2*d^3 - 3*b*c*d^2*e + b^2* d*e^2 + 2*a*c*d*e^2 - a*b*e^3)*log(abs(e*x + d))/e^4
Time = 11.87 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=x\,\left (\frac {b^2+2\,a\,c}{e}+\frac {d\,\left (\frac {2\,c^2\,d}{e^2}-\frac {3\,b\,c}{e}\right )}{e}\right )-x^2\,\left (\frac {c^2\,d}{e^2}-\frac {3\,b\,c}{2\,e}\right )-\frac {\ln \left (d+e\,x\right )\,\left (b^2\,d\,e^2-3\,b\,c\,d^2\,e-a\,b\,e^3+2\,c^2\,d^3+2\,a\,c\,d\,e^2\right )}{e^4}+\frac {2\,c^2\,x^3}{3\,e} \] Input:
int(((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x),x)
Output:
x*((2*a*c + b^2)/e + (d*((2*c^2*d)/e^2 - (3*b*c)/e))/e) - x^2*((c^2*d)/e^2 - (3*b*c)/(2*e)) - (log(d + e*x)*(2*c^2*d^3 + b^2*d*e^2 - a*b*e^3 + 2*a*c *d*e^2 - 3*b*c*d^2*e))/e^4 + (2*c^2*x^3)/(3*e)
Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.39 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{d+e x} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) a b \,e^{3}-12 \,\mathrm {log}\left (e x +d \right ) a c d \,e^{2}-6 \,\mathrm {log}\left (e x +d \right ) b^{2} d \,e^{2}+18 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2} e -12 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{3}+12 a c \,e^{3} x +6 b^{2} e^{3} x -18 b c d \,e^{2} x +9 b c \,e^{3} x^{2}+12 c^{2} d^{2} e x -6 c^{2} d \,e^{2} x^{2}+4 c^{2} e^{3} x^{3}}{6 e^{4}} \] Input:
int((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d),x)
Output:
(6*log(d + e*x)*a*b*e**3 - 12*log(d + e*x)*a*c*d*e**2 - 6*log(d + e*x)*b** 2*d*e**2 + 18*log(d + e*x)*b*c*d**2*e - 12*log(d + e*x)*c**2*d**3 + 12*a*c *e**3*x + 6*b**2*e**3*x - 18*b*c*d*e**2*x + 9*b*c*e**3*x**2 + 12*c**2*d**2 *e*x - 6*c**2*d*e**2*x**2 + 4*c**2*e**3*x**3)/(6*e**4)