Integrand size = 18, antiderivative size = 114 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=-\frac {c^2}{4 a x^4}+\frac {c (b c-2 a d)}{3 a^2 x^3}-\frac {(b c-a d)^2}{2 a^3 x^2}+\frac {b (b c-a d)^2}{a^4 x}+\frac {b^2 (b c-a d)^2 \log (x)}{a^5}-\frac {b^2 (b c-a d)^2 \log (a+b x)}{a^5} \]
-1/4*c^2/a/x^4+1/3*c*(-2*a*d+b*c)/a^2/x^3-1/2*(-a*d+b*c)^2/a^3/x^2+b*(-a*d +b*c)^2/a^4/x+b^2*(-a*d+b*c)^2*ln(x)/a^5-b^2*(-a*d+b*c)^2*ln(b*x+a)/a^5
Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=\frac {\frac {a \left (12 b^3 c^2 x^3-6 a b^2 c x^2 (c+4 d x)+4 a^2 b x \left (c^2+3 c d x+3 d^2 x^2\right )-a^3 \left (3 c^2+8 c d x+6 d^2 x^2\right )\right )}{x^4}+12 b^2 (b c-a d)^2 \log (x)-12 b^2 (b c-a d)^2 \log (a+b x)}{12 a^5} \]
((a*(12*b^3*c^2*x^3 - 6*a*b^2*c*x^2*(c + 4*d*x) + 4*a^2*b*x*(c^2 + 3*c*d*x + 3*d^2*x^2) - a^3*(3*c^2 + 8*c*d*x + 6*d^2*x^2)))/x^4 + 12*b^2*(b*c - a* d)^2*Log[x] - 12*b^2*(b*c - a*d)^2*Log[a + b*x])/(12*a^5)
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {99, 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.
\(\displaystyle \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {b^3 (a d-b c)^2}{a^5 (a+b x)}+\frac {b^2 (a d-b c)^2}{a^5 x}-\frac {b (a d-b c)^2}{a^4 x^2}+\frac {(a d-b c)^2}{a^3 x^3}+\frac {c (2 a d-b c)}{a^2 x^4}+\frac {c^2}{a x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 \log (x) (b c-a d)^2}{a^5}-\frac {b^2 (b c-a d)^2 \log (a+b x)}{a^5}+\frac {b (b c-a d)^2}{a^4 x}-\frac {(b c-a d)^2}{2 a^3 x^2}+\frac {c (b c-2 a d)}{3 a^2 x^3}-\frac {c^2}{4 a x^4}\) |
-1/4*c^2/(a*x^4) + (c*(b*c - 2*a*d))/(3*a^2*x^3) - (b*c - a*d)^2/(2*a^3*x^ 2) + (b*(b*c - a*d)^2)/(a^4*x) + (b^2*(b*c - a*d)^2*Log[x])/a^5 - (b^2*(b* c - a*d)^2*Log[a + b*x])/a^5
3.3.21.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 0.45 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.35
method | result | size |
default | \(-\frac {c^{2}}{4 a \,x^{4}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 a^{3} x^{2}}-\frac {c \left (2 a d -b c \right )}{3 a^{2} x^{3}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \ln \left (x \right )}{a^{5}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b}{a^{4} x}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \ln \left (b x +a \right )}{a^{5}}\) | \(154\) |
norman | \(\frac {\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b \,x^{3}}{a^{4}}-\frac {c^{2}}{4 a}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 a^{3}}-\frac {c \left (2 a d -b c \right ) x}{3 a^{2}}}{x^{4}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \ln \left (x \right )}{a^{5}}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \ln \left (b x +a \right )}{a^{5}}\) | \(154\) |
risch | \(\frac {\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b \,x^{3}}{a^{4}}-\frac {c^{2}}{4 a}-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 a^{3}}-\frac {c \left (2 a d -b c \right ) x}{3 a^{2}}}{x^{4}}+\frac {b^{2} \ln \left (-x \right ) d^{2}}{a^{3}}-\frac {2 b^{3} \ln \left (-x \right ) c d}{a^{4}}+\frac {b^{4} \ln \left (-x \right ) c^{2}}{a^{5}}-\frac {b^{2} \ln \left (b x +a \right ) d^{2}}{a^{3}}+\frac {2 b^{3} \ln \left (b x +a \right ) c d}{a^{4}}-\frac {b^{4} \ln \left (b x +a \right ) c^{2}}{a^{5}}\) | \(181\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) x^{4} a^{2} b^{2} d^{2}-24 \ln \left (x \right ) x^{4} a \,b^{3} c d +12 \ln \left (x \right ) x^{4} b^{4} c^{2}-12 \ln \left (b x +a \right ) x^{4} a^{2} b^{2} d^{2}+24 \ln \left (b x +a \right ) x^{4} a \,b^{3} c d -12 \ln \left (b x +a \right ) x^{4} b^{4} c^{2}+12 a^{3} b \,d^{2} x^{3}-24 a^{2} b^{2} c d \,x^{3}+12 a \,b^{3} c^{2} x^{3}-6 a^{4} d^{2} x^{2}+12 a^{3} b c d \,x^{2}-6 a^{2} b^{2} c^{2} x^{2}-8 a^{4} c d x +4 a^{3} b \,c^{2} x -3 c^{2} a^{4}}{12 a^{5} x^{4}}\) | \(205\) |
-1/4*c^2/a/x^4-1/2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/a^3/x^2-1/3*c*(2*a*d-b*c)/a ^2/x^3+(a^2*d^2-2*a*b*c*d+b^2*c^2)/a^5*b^2*ln(x)+(a^2*d^2-2*a*b*c*d+b^2*c^ 2)/a^4*b/x-(a^2*d^2-2*a*b*c*d+b^2*c^2)/a^5*b^2*ln(b*x+a)
Time = 0.22 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.48 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=-\frac {3 \, a^{4} c^{2} + 12 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} \log \left (b x + a\right ) - 12 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} \log \left (x\right ) - 12 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2} - 4 \, {\left (a^{3} b c^{2} - 2 \, a^{4} c d\right )} x}{12 \, a^{5} x^{4}} \]
-1/12*(3*a^4*c^2 + 12*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^4*log(b*x + a) - 12*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^4*log(x) - 12*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^3 + 6*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)* x^2 - 4*(a^3*b*c^2 - 2*a^4*c*d)*x)/(a^5*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (100) = 200\).
Time = 0.46 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.52 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=\frac {- 3 a^{3} c^{2} + x^{3} \cdot \left (12 a^{2} b d^{2} - 24 a b^{2} c d + 12 b^{3} c^{2}\right ) + x^{2} \left (- 6 a^{3} d^{2} + 12 a^{2} b c d - 6 a b^{2} c^{2}\right ) + x \left (- 8 a^{3} c d + 4 a^{2} b c^{2}\right )}{12 a^{4} x^{4}} + \frac {b^{2} \left (a d - b c\right )^{2} \log {\left (x + \frac {a^{3} b^{2} d^{2} - 2 a^{2} b^{3} c d + a b^{4} c^{2} - a b^{2} \left (a d - b c\right )^{2}}{2 a^{2} b^{3} d^{2} - 4 a b^{4} c d + 2 b^{5} c^{2}} \right )}}{a^{5}} - \frac {b^{2} \left (a d - b c\right )^{2} \log {\left (x + \frac {a^{3} b^{2} d^{2} - 2 a^{2} b^{3} c d + a b^{4} c^{2} + a b^{2} \left (a d - b c\right )^{2}}{2 a^{2} b^{3} d^{2} - 4 a b^{4} c d + 2 b^{5} c^{2}} \right )}}{a^{5}} \]
(-3*a**3*c**2 + x**3*(12*a**2*b*d**2 - 24*a*b**2*c*d + 12*b**3*c**2) + x** 2*(-6*a**3*d**2 + 12*a**2*b*c*d - 6*a*b**2*c**2) + x*(-8*a**3*c*d + 4*a**2 *b*c**2))/(12*a**4*x**4) + b**2*(a*d - b*c)**2*log(x + (a**3*b**2*d**2 - 2 *a**2*b**3*c*d + a*b**4*c**2 - a*b**2*(a*d - b*c)**2)/(2*a**2*b**3*d**2 - 4*a*b**4*c*d + 2*b**5*c**2))/a**5 - b**2*(a*d - b*c)**2*log(x + (a**3*b**2 *d**2 - 2*a**2*b**3*c*d + a*b**4*c**2 + a*b**2*(a*d - b*c)**2)/(2*a**2*b** 3*d**2 - 4*a*b**4*c*d + 2*b**5*c**2))/a**5
Time = 0.19 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=-\frac {{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left (b x + a\right )}{a^{5}} + \frac {{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left (x\right )}{a^{5}} - \frac {3 \, a^{3} c^{2} - 12 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + 6 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2} - 4 \, {\left (a^{2} b c^{2} - 2 \, a^{3} c d\right )} x}{12 \, a^{4} x^{4}} \]
-(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*log(b*x + a)/a^5 + (b^4*c^2 - 2*a*b ^3*c*d + a^2*b^2*d^2)*log(x)/a^5 - 1/12*(3*a^3*c^2 - 12*(b^3*c^2 - 2*a*b^2 *c*d + a^2*b*d^2)*x^3 + 6*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2 - 4*(a^2 *b*c^2 - 2*a^3*c*d)*x)/(a^4*x^4)
Time = 0.27 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.53 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=\frac {{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{5}} - \frac {{\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b} - \frac {3 \, a^{4} c^{2} - 12 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2} - 4 \, {\left (a^{3} b c^{2} - 2 \, a^{4} c d\right )} x}{12 \, a^{5} x^{4}} \]
(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*log(abs(x))/a^5 - (b^5*c^2 - 2*a*b^4 *c*d + a^2*b^3*d^2)*log(abs(b*x + a))/(a^5*b) - 1/12*(3*a^4*c^2 - 12*(a*b^ 3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^3 + 6*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^ 4*d^2)*x^2 - 4*(a^3*b*c^2 - 2*a^4*c*d)*x)/(a^5*x^4)
Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^2}{x^5 (a+b x)} \, dx=-\frac {\frac {c^2}{4\,a}+\frac {x^2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,a^3}-\frac {b\,x^3\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{a^4}+\frac {c\,x\,\left (2\,a\,d-b\,c\right )}{3\,a^2}}{x^4}-\frac {2\,b^2\,\mathrm {atanh}\left (\frac {b^2\,{\left (a\,d-b\,c\right )}^2\,\left (a+2\,b\,x\right )}{a\,\left (a^2\,b^2\,d^2-2\,a\,b^3\,c\,d+b^4\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{a^5} \]