Integrand size = 18, antiderivative size = 107 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=-\frac {1}{2 a c x^2}+\frac {b c+a d}{a^2 c^2 x}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \log (x)}{a^3 c^3}-\frac {b^3 \log (a+b x)}{a^3 (b c-a d)}+\frac {d^3 \log (c+d x)}{c^3 (b c-a d)} \] Output:
-1/2/a/c/x^2+(a*d+b*c)/a^2/c^2/x+(a^2*d^2+a*b*c*d+b^2*c^2)*ln(x)/a^3/c^3-b ^3*ln(b*x+a)/a^3/(-a*d+b*c)+d^3*ln(d*x+c)/c^3/(-a*d+b*c)
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=-\frac {1}{2 a c x^2}+\frac {b c+a d}{a^2 c^2 x}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \log (x)}{a^3 c^3}+\frac {b^3 \log (a+b x)}{a^3 (-b c+a d)}+\frac {d^3 \log (c+d x)}{c^3 (b c-a d)} \] Input:
Integrate[1/(x^3*(a + b*x)*(c + d*x)),x]
Output:
-1/2*1/(a*c*x^2) + (b*c + a*d)/(a^2*c^2*x) + ((b^2*c^2 + a*b*c*d + a^2*d^2 )*Log[x])/(a^3*c^3) + (b^3*Log[a + b*x])/(a^3*(-(b*c) + a*d)) + (d^3*Log[c + d*x])/(c^3*(b*c - a*d))
Time = 0.26 (sec) , antiderivative size = 107, 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 = {93, 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.
\(\displaystyle \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \int \left (\frac {b^4}{a^3 (a+b x) (a d-b c)}+\frac {-a d-b c}{a^2 c^2 x^2}+\frac {a^2 d^2+a b c d+b^2 c^2}{a^3 c^3 x}+\frac {d^4}{c^3 (c+d x) (b c-a d)}+\frac {1}{a c x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^3 \log (a+b x)}{a^3 (b c-a d)}+\frac {a d+b c}{a^2 c^2 x}+\frac {\log (x) \left (a^2 d^2+a b c d+b^2 c^2\right )}{a^3 c^3}+\frac {d^3 \log (c+d x)}{c^3 (b c-a d)}-\frac {1}{2 a c x^2}\) |
Input:
Int[1/(x^3*(a + b*x)*(c + d*x)),x]
Output:
-1/2*1/(a*c*x^2) + (b*c + a*d)/(a^2*c^2*x) + ((b^2*c^2 + a*b*c*d + a^2*d^2 )*Log[x])/(a^3*c^3) - (b^3*Log[a + b*x])/(a^3*(b*c - a*d)) + (d^3*Log[c + d*x])/(c^3*(b*c - a*d))
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {\frac {\left (a d +b c \right ) x}{a^{2} c^{2}}-\frac {1}{2 a c}}{x^{2}}+\frac {b^{3} \ln \left (b x +a \right )}{\left (a d -b c \right ) a^{3}}+\frac {\left (a^{2} d^{2}+a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3} c^{3}}-\frac {d^{3} \ln \left (x d +c \right )}{c^{3} \left (a d -b c \right )}\) | \(106\) |
default | \(-\frac {d^{3} \ln \left (x d +c \right )}{c^{3} \left (a d -b c \right )}-\frac {1}{2 a c \,x^{2}}-\frac {-a d -b c}{a^{2} c^{2} x}+\frac {\left (a^{2} d^{2}+a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3} c^{3}}+\frac {b^{3} \ln \left (b x +a \right )}{\left (a d -b c \right ) a^{3}}\) | \(109\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) x^{2} a^{3} d^{3}-2 \ln \left (x \right ) x^{2} b^{3} c^{3}+2 \ln \left (b x +a \right ) x^{2} b^{3} c^{3}-2 d^{3} \ln \left (x d +c \right ) a^{3} x^{2}+2 x \,a^{3} c \,d^{2}-2 x a \,b^{2} c^{3}-a^{3} c^{2} d +a^{2} b \,c^{3}}{2 a^{3} c^{3} x^{2} \left (a d -b c \right )}\) | \(120\) |
risch | \(\frac {\frac {\left (a d +b c \right ) x}{a^{2} c^{2}}-\frac {1}{2 a c}}{x^{2}}+\frac {b^{3} \ln \left (b x +a \right )}{\left (a d -b c \right ) a^{3}}-\frac {d^{3} \ln \left (-x d -c \right )}{c^{3} \left (a d -b c \right )}+\frac {\ln \left (-x \right ) d^{2}}{c^{3} a}+\frac {\ln \left (-x \right ) b d}{c^{2} a^{2}}+\frac {\ln \left (-x \right ) b^{2}}{c \,a^{3}}\) | \(121\) |
Input:
int(1/x^3/(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)
Output:
((a*d+b*c)/a^2/c^2*x-1/2/a/c)/x^2+b^3/(a*d-b*c)/a^3*ln(b*x+a)+(a^2*d^2+a*b *c*d+b^2*c^2)*ln(x)/a^3/c^3-d^3/c^3/(a*d-b*c)*ln(d*x+c)
Time = 1.35 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=-\frac {2 \, b^{3} c^{3} x^{2} \log \left (b x + a\right ) - 2 \, a^{3} d^{3} x^{2} \log \left (d x + c\right ) + a^{2} b c^{3} - a^{3} c^{2} d - 2 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{2} \log \left (x\right ) - 2 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x}{2 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{2}} \] Input:
integrate(1/x^3/(b*x+a)/(d*x+c),x, algorithm="fricas")
Output:
-1/2*(2*b^3*c^3*x^2*log(b*x + a) - 2*a^3*d^3*x^2*log(d*x + c) + a^2*b*c^3 - a^3*c^2*d - 2*(b^3*c^3 - a^3*d^3)*x^2*log(x) - 2*(a*b^2*c^3 - a^3*c*d^2) *x)/((a^3*b*c^4 - a^4*c^3*d)*x^2)
Timed out. \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(1/x**3/(b*x+a)/(d*x+c),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=-\frac {b^{3} \log \left (b x + a\right )}{a^{3} b c - a^{4} d} + \frac {d^{3} \log \left (d x + c\right )}{b c^{4} - a c^{3} d} + \frac {{\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} \log \left (x\right )}{a^{3} c^{3}} - \frac {a c - 2 \, {\left (b c + a d\right )} x}{2 \, a^{2} c^{2} x^{2}} \] Input:
integrate(1/x^3/(b*x+a)/(d*x+c),x, algorithm="maxima")
Output:
-b^3*log(b*x + a)/(a^3*b*c - a^4*d) + d^3*log(d*x + c)/(b*c^4 - a*c^3*d) + (b^2*c^2 + a*b*c*d + a^2*d^2)*log(x)/(a^3*c^3) - 1/2*(a*c - 2*(b*c + a*d) *x)/(a^2*c^2*x^2)
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=-\frac {b^{4} \log \left ({\left | b x + a \right |}\right )}{a^{3} b^{2} c - a^{4} b d} + \frac {d^{4} \log \left ({\left | d x + c \right |}\right )}{b c^{4} d - a c^{3} d^{2}} + \frac {{\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{3} c^{3}} - \frac {a^{2} c^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} x}{2 \, a^{3} c^{3} x^{2}} \] Input:
integrate(1/x^3/(b*x+a)/(d*x+c),x, algorithm="giac")
Output:
-b^4*log(abs(b*x + a))/(a^3*b^2*c - a^4*b*d) + d^4*log(abs(d*x + c))/(b*c^ 4*d - a*c^3*d^2) + (b^2*c^2 + a*b*c*d + a^2*d^2)*log(abs(x))/(a^3*c^3) - 1 /2*(a^2*c^2 - 2*(a*b*c^2 + a^2*c*d)*x)/(a^3*c^3*x^2)
Time = 0.60 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=\frac {b^3\,\ln \left (a+b\,x\right )}{a^3\,\left (a\,d-b\,c\right )}-\frac {\frac {1}{2\,a\,c}-\frac {x\,\left (a\,d+b\,c\right )}{a^2\,c^2}}{x^2}-\frac {d^3\,\ln \left (c+d\,x\right )}{c^3\,\left (a\,d-b\,c\right )}+\frac {\ln \left (x\right )\,\left (a^2\,d^2+a\,b\,c\,d+b^2\,c^2\right )}{a^3\,c^3} \] Input:
int(1/(x^3*(a + b*x)*(c + d*x)),x)
Output:
(b^3*log(a + b*x))/(a^3*(a*d - b*c)) - (1/(2*a*c) - (x*(a*d + b*c))/(a^2*c ^2))/x^2 - (d^3*log(c + d*x))/(c^3*(a*d - b*c)) + (log(x)*(a^2*d^2 + b^2*c ^2 + a*b*c*d))/(a^3*c^3)
Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^3 (a+b x) (c+d x)} \, dx=\frac {2 \,\mathrm {log}\left (b x +a \right ) b^{3} c^{3} x^{2}-2 \,\mathrm {log}\left (d x +c \right ) a^{3} d^{3} x^{2}+2 \,\mathrm {log}\left (x \right ) a^{3} d^{3} x^{2}-2 \,\mathrm {log}\left (x \right ) b^{3} c^{3} x^{2}-a^{3} c^{2} d +2 a^{3} c \,d^{2} x +a^{2} b \,c^{3}-2 a \,b^{2} c^{3} x}{2 a^{3} c^{3} x^{2} \left (a d -b c \right )} \] Input:
int(1/x^3/(b*x+a)/(d*x+c),x)
Output:
(2*log(a + b*x)*b**3*c**3*x**2 - 2*log(c + d*x)*a**3*d**3*x**2 + 2*log(x)* a**3*d**3*x**2 - 2*log(x)*b**3*c**3*x**2 - a**3*c**2*d + 2*a**3*c*d**2*x + a**2*b*c**3 - 2*a*b**2*c**3*x)/(2*a**3*c**3*x**2*(a*d - b*c))