Integrand size = 21, antiderivative size = 135 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=-\frac {a}{4 d x^4}-\frac {b d-a e}{3 d^2 x^3}-\frac {c d^2-b d e+a e^2}{2 d^3 x^2}+\frac {e \left (c d^2-b d e+a e^2\right )}{d^4 x}+\frac {e^2 \left (c d^2-b d e+a e^2\right ) \log (x)}{d^5}-\frac {e^2 \left (c d^2-b d e+a e^2\right ) \log (d+e x)}{d^5} \] Output:
-1/4*a/d/x^4-1/3*(-a*e+b*d)/d^2/x^3-1/2*(a*e^2-b*d*e+c*d^2)/d^3/x^2+e*(a*e ^2-b*d*e+c*d^2)/d^4/x+e^2*(a*e^2-b*d*e+c*d^2)*ln(x)/d^5-e^2*(a*e^2-b*d*e+c *d^2)*ln(e*x+d)/d^5
Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=\frac {\frac {d \left (a \left (-3 d^3+4 d^2 e x-6 d e^2 x^2+12 e^3 x^3\right )-2 d x \left (3 c d x (d-2 e x)+b \left (2 d^2-3 d e x+6 e^2 x^2\right )\right )\right )}{x^4}+12 e^2 \left (c d^2+e (-b d+a e)\right ) \log (x)-12 e^2 \left (c d^2+e (-b d+a e)\right ) \log (d+e x)}{12 d^5} \] Input:
Integrate[(a + b*x + c*x^2)/(x^5*(d + e*x)),x]
Output:
((d*(a*(-3*d^3 + 4*d^2*e*x - 6*d*e^2*x^2 + 12*e^3*x^3) - 2*d*x*(3*c*d*x*(d - 2*e*x) + b*(2*d^2 - 3*d*e*x + 6*e^2*x^2))))/x^4 + 12*e^2*(c*d^2 + e*(-( b*d) + a*e))*Log[x] - 12*e^2*(c*d^2 + e*(-(b*d) + a*e))*Log[d + e*x])/(12* d^5)
Time = 0.31 (sec) , antiderivative size = 135, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {e^2 \left (a e^2-b d e+c d^2\right )}{d^5 x}-\frac {e^3 \left (a e^2-b d e+c d^2\right )}{d^5 (d+e x)}-\frac {e \left (a e^2-b d e+c d^2\right )}{d^4 x^2}+\frac {a e^2-b d e+c d^2}{d^3 x^3}+\frac {b d-a e}{d^2 x^4}+\frac {a}{d x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 \log (x) \left (a e^2-b d e+c d^2\right )}{d^5}-\frac {e^2 \log (d+e x) \left (a e^2-b d e+c d^2\right )}{d^5}+\frac {e \left (a e^2-b d e+c d^2\right )}{d^4 x}-\frac {a e^2-b d e+c d^2}{2 d^3 x^2}-\frac {b d-a e}{3 d^2 x^3}-\frac {a}{4 d x^4}\) |
Input:
Int[(a + b*x + c*x^2)/(x^5*(d + e*x)),x]
Output:
-1/4*a/(d*x^4) - (b*d - a*e)/(3*d^2*x^3) - (c*d^2 - b*d*e + a*e^2)/(2*d^3* x^2) + (e*(c*d^2 - b*d*e + a*e^2))/(d^4*x) + (e^2*(c*d^2 - b*d*e + a*e^2)* Log[x])/d^5 - (e^2*(c*d^2 - b*d*e + a*e^2)*Log[d + e*x])/d^5
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 = 0.77 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {a}{4 d \,x^{4}}-\frac {-a e +b d}{3 d^{2} x^{3}}-\frac {a \,e^{2}-b d e +c \,d^{2}}{2 d^{3} x^{2}}+\frac {e \left (a \,e^{2}-b d e +c \,d^{2}\right )}{d^{4} x}+\frac {e^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (x \right )}{d^{5}}-\frac {e^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (e x +d \right )}{d^{5}}\) | \(130\) |
norman | \(\frac {\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) e \,x^{3}}{d^{4}}-\frac {a}{4 d}+\frac {\left (a e -b d \right ) x}{3 d^{2}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) x^{2}}{2 d^{3}}}{x^{4}}+\frac {e^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (x \right )}{d^{5}}-\frac {e^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (e x +d \right )}{d^{5}}\) | \(130\) |
risch | \(\frac {\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) e \,x^{3}}{d^{4}}-\frac {a}{4 d}+\frac {\left (a e -b d \right ) x}{3 d^{2}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) x^{2}}{2 d^{3}}}{x^{4}}-\frac {e^{4} \ln \left (e x +d \right ) a}{d^{5}}+\frac {e^{3} \ln \left (e x +d \right ) b}{d^{4}}-\frac {e^{2} \ln \left (e x +d \right ) c}{d^{3}}+\frac {e^{4} \ln \left (-x \right ) a}{d^{5}}-\frac {e^{3} \ln \left (-x \right ) b}{d^{4}}+\frac {e^{2} \ln \left (-x \right ) c}{d^{3}}\) | \(156\) |
parallelrisch | \(-\frac {12 \ln \left (e x +d \right ) x^{4} a \,e^{4}-12 \ln \left (e x +d \right ) x^{4} b d \,e^{3}+12 \ln \left (e x +d \right ) x^{4} c \,d^{2} e^{2}-12 \ln \left (x \right ) x^{4} a \,e^{4}+12 \ln \left (x \right ) x^{4} b d \,e^{3}-12 \ln \left (x \right ) x^{4} c \,d^{2} e^{2}-12 x^{3} a d \,e^{3}+12 x^{3} b \,d^{2} e^{2}-12 c \,d^{3} e \,x^{3}+6 x^{2} a \,d^{2} e^{2}-6 x^{2} b \,d^{3} e +6 c \,d^{4} x^{2}-4 x \,d^{3} e a +4 b \,d^{4} x +3 a \,d^{4}}{12 d^{5} x^{4}}\) | \(180\) |
Input:
int((c*x^2+b*x+a)/x^5/(e*x+d),x,method=_RETURNVERBOSE)
Output:
-1/4*a/d/x^4-1/3*(-a*e+b*d)/d^2/x^3-1/2*(a*e^2-b*d*e+c*d^2)/d^3/x^2+e*(a*e ^2-b*d*e+c*d^2)/d^4/x+e^2*(a*e^2-b*d*e+c*d^2)*ln(x)/d^5-e^2*(a*e^2-b*d*e+c *d^2)*ln(e*x+d)/d^5
Time = 0.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=-\frac {12 \, {\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x^{4} \log \left (e x + d\right ) - 12 \, {\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x^{4} \log \left (x\right ) + 3 \, a d^{4} - 12 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x^{3} + 6 \, {\left (c d^{4} - b d^{3} e + a d^{2} e^{2}\right )} x^{2} + 4 \, {\left (b d^{4} - a d^{3} e\right )} x}{12 \, d^{5} x^{4}} \] Input:
integrate((c*x^2+b*x+a)/x^5/(e*x+d),x, algorithm="fricas")
Output:
-1/12*(12*(c*d^2*e^2 - b*d*e^3 + a*e^4)*x^4*log(e*x + d) - 12*(c*d^2*e^2 - b*d*e^3 + a*e^4)*x^4*log(x) + 3*a*d^4 - 12*(c*d^3*e - b*d^2*e^2 + a*d*e^3 )*x^3 + 6*(c*d^4 - b*d^3*e + a*d^2*e^2)*x^2 + 4*(b*d^4 - a*d^3*e)*x)/(d^5* x^4)
Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (128) = 256\).
Time = 0.47 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.04 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=\frac {- 3 a d^{3} + x^{3} \cdot \left (12 a e^{3} - 12 b d e^{2} + 12 c d^{2} e\right ) + x^{2} \left (- 6 a d e^{2} + 6 b d^{2} e - 6 c d^{3}\right ) + x \left (4 a d^{2} e - 4 b d^{3}\right )}{12 d^{4} x^{4}} + \frac {e^{2} \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (x + \frac {a d e^{4} - b d^{2} e^{3} + c d^{3} e^{2} - d e^{2} \left (a e^{2} - b d e + c d^{2}\right )}{2 a e^{5} - 2 b d e^{4} + 2 c d^{2} e^{3}} \right )}}{d^{5}} - \frac {e^{2} \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (x + \frac {a d e^{4} - b d^{2} e^{3} + c d^{3} e^{2} + d e^{2} \left (a e^{2} - b d e + c d^{2}\right )}{2 a e^{5} - 2 b d e^{4} + 2 c d^{2} e^{3}} \right )}}{d^{5}} \] Input:
integrate((c*x**2+b*x+a)/x**5/(e*x+d),x)
Output:
(-3*a*d**3 + x**3*(12*a*e**3 - 12*b*d*e**2 + 12*c*d**2*e) + x**2*(-6*a*d*e **2 + 6*b*d**2*e - 6*c*d**3) + x*(4*a*d**2*e - 4*b*d**3))/(12*d**4*x**4) + e**2*(a*e**2 - b*d*e + c*d**2)*log(x + (a*d*e**4 - b*d**2*e**3 + c*d**3*e **2 - d*e**2*(a*e**2 - b*d*e + c*d**2))/(2*a*e**5 - 2*b*d*e**4 + 2*c*d**2* e**3))/d**5 - e**2*(a*e**2 - b*d*e + c*d**2)*log(x + (a*d*e**4 - b*d**2*e* *3 + c*d**3*e**2 + d*e**2*(a*e**2 - b*d*e + c*d**2))/(2*a*e**5 - 2*b*d*e** 4 + 2*c*d**2*e**3))/d**5
Time = 0.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=-\frac {{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} \log \left (e x + d\right )}{d^{5}} + \frac {{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} \log \left (x\right )}{d^{5}} - \frac {3 \, a d^{3} - 12 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )} x^{3} + 6 \, {\left (c d^{3} - b d^{2} e + a d e^{2}\right )} x^{2} + 4 \, {\left (b d^{3} - a d^{2} e\right )} x}{12 \, d^{4} x^{4}} \] Input:
integrate((c*x^2+b*x+a)/x^5/(e*x+d),x, algorithm="maxima")
Output:
-(c*d^2*e^2 - b*d*e^3 + a*e^4)*log(e*x + d)/d^5 + (c*d^2*e^2 - b*d*e^3 + a *e^4)*log(x)/d^5 - 1/12*(3*a*d^3 - 12*(c*d^2*e - b*d*e^2 + a*e^3)*x^3 + 6* (c*d^3 - b*d^2*e + a*d*e^2)*x^2 + 4*(b*d^3 - a*d^2*e)*x)/(d^4*x^4)
Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=\frac {{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} \log \left ({\left | x \right |}\right )}{d^{5}} - \frac {{\left (c d^{2} e^{3} - b d e^{4} + a e^{5}\right )} \log \left ({\left | e x + d \right |}\right )}{d^{5} e} - \frac {3 \, a d^{4} - 12 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x^{3} + 6 \, {\left (c d^{4} - b d^{3} e + a d^{2} e^{2}\right )} x^{2} + 4 \, {\left (b d^{4} - a d^{3} e\right )} x}{12 \, d^{5} x^{4}} \] Input:
integrate((c*x^2+b*x+a)/x^5/(e*x+d),x, algorithm="giac")
Output:
(c*d^2*e^2 - b*d*e^3 + a*e^4)*log(abs(x))/d^5 - (c*d^2*e^3 - b*d*e^4 + a*e ^5)*log(abs(e*x + d))/(d^5*e) - 1/12*(3*a*d^4 - 12*(c*d^3*e - b*d^2*e^2 + a*d*e^3)*x^3 + 6*(c*d^4 - b*d^3*e + a*d^2*e^2)*x^2 + 4*(b*d^4 - a*d^3*e)*x )/(d^5*x^4)
Time = 10.88 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.13 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=-\frac {\frac {a}{4\,d}-\frac {x\,\left (a\,e-b\,d\right )}{3\,d^2}+\frac {x^2\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{2\,d^3}-\frac {e\,x^3\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{d^4}}{x^4}-\frac {2\,e^2\,\mathrm {atanh}\left (\frac {e^2\,\left (d+2\,e\,x\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{d\,\left (c\,d^2\,e^2-b\,d\,e^3+a\,e^4\right )}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{d^5} \] Input:
int((a + b*x + c*x^2)/(x^5*(d + e*x)),x)
Output:
- (a/(4*d) - (x*(a*e - b*d))/(3*d^2) + (x^2*(a*e^2 + c*d^2 - b*d*e))/(2*d^ 3) - (e*x^3*(a*e^2 + c*d^2 - b*d*e))/d^4)/x^4 - (2*e^2*atanh((e^2*(d + 2*e *x)*(a*e^2 + c*d^2 - b*d*e))/(d*(a*e^4 + c*d^2*e^2 - b*d*e^3)))*(a*e^2 + c *d^2 - b*d*e))/d^5
Time = 0.22 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{x^5 (d+e x)} \, dx=\frac {-12 \,\mathrm {log}\left (e x +d \right ) a \,e^{4} x^{4}+12 \,\mathrm {log}\left (e x +d \right ) b d \,e^{3} x^{4}-12 \,\mathrm {log}\left (e x +d \right ) c \,d^{2} e^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) a \,e^{4} x^{4}-12 \,\mathrm {log}\left (x \right ) b d \,e^{3} x^{4}+12 \,\mathrm {log}\left (x \right ) c \,d^{2} e^{2} x^{4}-3 a \,d^{4}+4 a \,d^{3} e x -6 a \,d^{2} e^{2} x^{2}+12 a d \,e^{3} x^{3}-4 b \,d^{4} x +6 b \,d^{3} e \,x^{2}-12 b \,d^{2} e^{2} x^{3}-6 c \,d^{4} x^{2}+12 c \,d^{3} e \,x^{3}}{12 d^{5} x^{4}} \] Input:
int((c*x^2+b*x+a)/x^5/(e*x+d),x)
Output:
( - 12*log(d + e*x)*a*e**4*x**4 + 12*log(d + e*x)*b*d*e**3*x**4 - 12*log(d + e*x)*c*d**2*e**2*x**4 + 12*log(x)*a*e**4*x**4 - 12*log(x)*b*d*e**3*x**4 + 12*log(x)*c*d**2*e**2*x**4 - 3*a*d**4 + 4*a*d**3*e*x - 6*a*d**2*e**2*x* *2 + 12*a*d*e**3*x**3 - 4*b*d**4*x + 6*b*d**3*e*x**2 - 12*b*d**2*e**2*x**3 - 6*c*d**4*x**2 + 12*c*d**3*e*x**3)/(12*d**5*x**4)