Integrand size = 24, antiderivative size = 119 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right )}{3 e^4 (d+e x)^3}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{2 e^4 (d+e x)^2}+\frac {3 c (2 c d-b e)}{e^4 (d+e x)}+\frac {2 c^2 \log (d+e x)}{e^4} \] Output:
1/3*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)/e^4/(e*x+d)^3-1/2*(6*c^2*d^2+b^2*e^2- 2*c*e*(-a*e+3*b*d))/e^4/(e*x+d)^2+3*c*(-b*e+2*c*d)/e^4/(e*x+d)+2*c^2*ln(e* x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {2 c^2 d \left (11 d^2+27 d e x+18 e^2 x^2\right )-b e^2 (2 a e+b (d+3 e x))-2 c e \left (a e (d+3 e x)+3 b \left (d^2+3 d e x+3 e^2 x^2\right )\right )+12 c^2 (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \] Input:
Integrate[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x)^4,x]
Output:
(2*c^2*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2) - b*e^2*(2*a*e + b*(d + 3*e*x)) - 2*c*e*(a*e*(d + 3*e*x) + 3*b*(d^2 + 3*d*e*x + 3*e^2*x^2)) + 12*c^2*(d + e*x)^3*Log[d + e*x])/(6*e^4*(d + e*x)^3)
Time = 0.51 (sec) , antiderivative size = 119, 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.
\(\displaystyle \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^3 (d+e x)^3}+\frac {(b e-2 c d) \left (a e^2-b d e+c d^2\right )}{e^3 (d+e x)^4}-\frac {3 c (2 c d-b e)}{e^3 (d+e x)^2}+\frac {2 c^2}{e^3 (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{2 e^4 (d+e x)^2}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{3 e^4 (d+e x)^3}+\frac {3 c (2 c d-b e)}{e^4 (d+e x)}+\frac {2 c^2 \log (d+e x)}{e^4}\) |
Input:
Int[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x)^4,x]
Output:
((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(3*e^4*(d + e*x)^3) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(2*e^4*(d + e*x)^2) + (3*c*(2*c*d - b*e))/ (e^4*(d + e*x)) + (2*c^2*Log[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 = 0.76 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {-\frac {3 c \left (b e -2 c d \right ) x^{2}}{e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x}{2 e^{3}}-\frac {2 a \,e^{3} b +2 a d \,e^{2} c +d \,e^{2} b^{2}+6 d^{2} e b c -22 c^{2} d^{3}}{6 e^{4}}}{\left (e x +d \right )^{3}}+\frac {2 c^{2} \ln \left (e x +d \right )}{e^{4}}\) | \(122\) |
norman | \(\frac {-\frac {2 a \,e^{3} b +2 a d \,e^{2} c +d \,e^{2} b^{2}+6 d^{2} e b c -22 c^{2} d^{3}}{6 e^{4}}-\frac {3 \left (b c e -2 c^{2} d \right ) x^{2}}{e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {2 c^{2} \ln \left (e x +d \right )}{e^{4}}\) | \(124\) |
default | \(-\frac {a \,e^{3} b -2 a d \,e^{2} c -d \,e^{2} b^{2}+3 d^{2} e b c -2 c^{2} d^{3}}{3 e^{4} \left (e x +d \right )^{3}}+\frac {2 c^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {3 c \left (b e -2 c d \right )}{e^{4} \left (e x +d \right )}-\frac {2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(130\) |
parallelrisch | \(\frac {12 \ln \left (e x +d \right ) x^{3} c^{2} e^{3}+36 \ln \left (e x +d \right ) x^{2} c^{2} d \,e^{2}+36 \ln \left (e x +d \right ) x \,c^{2} d^{2} e -18 e^{3} x^{2} c b +36 d \,e^{2} c^{2} x^{2}+12 \ln \left (e x +d \right ) c^{2} d^{3}-6 a c \,e^{3} x -3 e^{3} x \,b^{2}-18 d \,e^{2} c b x +54 d^{2} e \,c^{2} x -2 a \,e^{3} b -2 a d \,e^{2} c -d \,e^{2} b^{2}-6 d^{2} e b c +22 c^{2} d^{3}}{6 e^{4} \left (e x +d \right )^{3}}\) | \(177\) |
Input:
int((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
(-3*c*(b*e-2*c*d)/e^2*x^2-1/2*(2*a*c*e^2+b^2*e^2+6*b*c*d*e-18*c^2*d^2)/e^3 *x-1/6*(2*a*b*e^3+2*a*c*d*e^2+b^2*d*e^2+6*b*c*d^2*e-22*c^2*d^3)/e^4)/(e*x+ d)^3+2*c^2*ln(e*x+d)/e^4
Time = 0.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.50 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {22 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} - {\left (b^{2} + 2 \, a c\right )} d e^{2} + 18 \, {\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} x^{2} + 3 \, {\left (18 \, c^{2} d^{2} e - 6 \, b c d e^{2} - {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x + 12 \, {\left (c^{2} e^{3} x^{3} + 3 \, c^{2} d e^{2} x^{2} + 3 \, c^{2} d^{2} e x + c^{2} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^4,x, algorithm="fricas")
Output:
1/6*(22*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^3 - (b^2 + 2*a*c)*d*e^2 + 18*(2*c^ 2*d*e^2 - b*c*e^3)*x^2 + 3*(18*c^2*d^2*e - 6*b*c*d*e^2 - (b^2 + 2*a*c)*e^3 )*x + 12*(c^2*e^3*x^3 + 3*c^2*d*e^2*x^2 + 3*c^2*d^2*e*x + c^2*d^3)*log(e*x + d))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Time = 1.75 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.33 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {2 c^{2} \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 a b e^{3} - 2 a c d e^{2} - b^{2} d e^{2} - 6 b c d^{2} e + 22 c^{2} d^{3} + x^{2} \left (- 18 b c e^{3} + 36 c^{2} d e^{2}\right ) + x \left (- 6 a c e^{3} - 3 b^{2} e^{3} - 18 b c d e^{2} + 54 c^{2} d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \] Input:
integrate((2*c*x+b)*(c*x**2+b*x+a)/(e*x+d)**4,x)
Output:
2*c**2*log(d + e*x)/e**4 + (-2*a*b*e**3 - 2*a*c*d*e**2 - b**2*d*e**2 - 6*b *c*d**2*e + 22*c**2*d**3 + x**2*(-18*b*c*e**3 + 36*c**2*d*e**2) + x*(-6*a* c*e**3 - 3*b**2*e**3 - 18*b*c*d*e**2 + 54*c**2*d**2*e))/(6*d**3*e**4 + 18* d**2*e**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)
Time = 0.04 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.23 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {22 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} - {\left (b^{2} + 2 \, a c\right )} d e^{2} + 18 \, {\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} x^{2} + 3 \, {\left (18 \, c^{2} d^{2} e - 6 \, b c d e^{2} - {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {2 \, c^{2} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^4,x, algorithm="maxima")
Output:
1/6*(22*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^3 - (b^2 + 2*a*c)*d*e^2 + 18*(2*c^ 2*d*e^2 - b*c*e^3)*x^2 + 3*(18*c^2*d^2*e - 6*b*c*d*e^2 - (b^2 + 2*a*c)*e^3 )*x)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + 2*c^2*log(e*x + d)/ e^4
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {2 \, c^{2} \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {18 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} + 3 \, {\left (18 \, c^{2} d^{2} - 6 \, b c d e - b^{2} e^{2} - 2 \, a c e^{2}\right )} x + \frac {22 \, c^{2} d^{3} - 6 \, b c d^{2} e - b^{2} d e^{2} - 2 \, a c d e^{2} - 2 \, a b e^{3}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \] Input:
integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^4,x, algorithm="giac")
Output:
2*c^2*log(abs(e*x + d))/e^4 + 1/6*(18*(2*c^2*d*e - b*c*e^2)*x^2 + 3*(18*c^ 2*d^2 - 6*b*c*d*e - b^2*e^2 - 2*a*c*e^2)*x + (22*c^2*d^3 - 6*b*c*d^2*e - b ^2*d*e^2 - 2*a*c*d*e^2 - 2*a*b*e^3)/e)/((e*x + d)^3*e^3)
Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.21 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {2\,c^2\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {b^2\,d\,e^2+6\,b\,c\,d^2\,e+2\,a\,b\,e^3-22\,c^2\,d^3+2\,a\,c\,d\,e^2}{6\,e^4}+\frac {x\,\left (b^2\,e^2+6\,b\,c\,d\,e-18\,c^2\,d^2+2\,a\,c\,e^2\right )}{2\,e^3}+\frac {3\,c\,x^2\,\left (b\,e-2\,c\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \] Input:
int(((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x)^4,x)
Output:
(2*c^2*log(d + e*x))/e^4 - ((b^2*d*e^2 - 22*c^2*d^3 + 2*a*b*e^3 + 2*a*c*d* e^2 + 6*b*c*d^2*e)/(6*e^4) + (x*(b^2*e^2 - 18*c^2*d^2 + 2*a*c*e^2 + 6*b*c* d*e))/(2*e^3) + (3*c*x^2*(b*e - 2*c*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)
Time = 0.23 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.63 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {12 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{4}+36 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{3} e x +36 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{2} e^{2} x^{2}+12 \,\mathrm {log}\left (e x +d \right ) c^{2} d \,e^{3} x^{3}-2 a b d \,e^{3}-2 a c \,d^{2} e^{2}-6 a c d \,e^{3} x -b^{2} d^{2} e^{2}-3 b^{2} d \,e^{3} x +6 b c \,e^{4} x^{3}+10 c^{2} d^{4}+18 c^{2} d^{3} e x -12 c^{2} d \,e^{3} x^{3}}{6 d \,e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^4,x)
Output:
(12*log(d + e*x)*c**2*d**4 + 36*log(d + e*x)*c**2*d**3*e*x + 36*log(d + e* x)*c**2*d**2*e**2*x**2 + 12*log(d + e*x)*c**2*d*e**3*x**3 - 2*a*b*d*e**3 - 2*a*c*d**2*e**2 - 6*a*c*d*e**3*x - b**2*d**2*e**2 - 3*b**2*d*e**3*x + 6*b *c*e**4*x**3 + 10*c**2*d**4 + 18*c**2*d**3*e*x - 12*c**2*d*e**3*x**3)/(6*d *e**4*(d**3 + 3*d**2*e*x + 3*d*e**2*x**2 + e**3*x**3))