\(\int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx\) [12]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 68 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {a \log (a-x)}{(a-b) (a-c)}-\frac {b \log (b-x)}{(a-b) (b-c)}+\frac {c \log (c-x)}{(a-c) (b-c)} \]

[Out]

a*ln(a-x)/(a-b)/(a-c)-b*ln(b-x)/(a-b)/(b-c)+c*ln(c-x)/(a-c)/(b-c)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {153} \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {a \log (a-x)}{(a-b) (a-c)}-\frac {b \log (b-x)}{(a-b) (b-c)}+\frac {c \log (c-x)}{(a-c) (b-c)} \]

[In]

Int[x/((-a + x)*(-b + x)*(-c + x)),x]

[Out]

(a*Log[a - x])/((a - b)*(a - c)) - (b*Log[b - x])/((a - b)*(b - c)) + (c*Log[c - x])/((a - c)*(b - c))

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a}{(a-b) (a-c) (a-x)}+\frac {b}{(a-b) (b-c) (b-x)}+\frac {c}{(a-c) (-b+c) (c-x)}\right ) \, dx \\ & = \frac {a \log (a-x)}{(a-b) (a-c)}-\frac {b \log (b-x)}{(a-b) (b-c)}+\frac {c \log (c-x)}{(a-c) (b-c)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {a (b-c) \log (-a+x)+b (-a+c) \log (-b+x)+(a-b) c \log (-c+x)}{(a-b) (a-c) (b-c)} \]

[In]

Integrate[x/((-a + x)*(-b + x)*(-c + x)),x]

[Out]

(a*(b - c)*Log[-a + x] + b*(-a + c)*Log[-b + x] + (a - b)*c*Log[-c + x])/((a - b)*(a - c)*(b - c))

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01

method result size
default \(\frac {a \ln \left (a -x \right )}{\left (a -b \right ) \left (a -c \right )}-\frac {b \ln \left (b -x \right )}{\left (a -b \right ) \left (b -c \right )}+\frac {c \ln \left (c -x \right )}{\left (a -c \right ) \left (b -c \right )}\) \(69\)
norman \(\frac {c \ln \left (c -x \right )}{a b -a c -b c +c^{2}}+\frac {a \ln \left (a -x \right )}{\left (a -b \right ) \left (a -c \right )}-\frac {b \ln \left (b -x \right )}{\left (a -b \right ) \left (b -c \right )}\) \(72\)
risch \(-\frac {b \ln \left (-b +x \right )}{a b -a c -b^{2}+b c}+\frac {a \ln \left (-a +x \right )}{a^{2}-a b -a c +b c}+\frac {c \ln \left (c -x \right )}{a b -a c -b c +c^{2}}\) \(79\)
parallelrisch \(\frac {\ln \left (-a +x \right ) a b -\ln \left (-a +x \right ) a c -\ln \left (-b +x \right ) a b +\ln \left (-b +x \right ) b c +\ln \left (-c +x \right ) a c -\ln \left (-c +x \right ) b c}{\left (a b -a c -b c +c^{2}\right ) \left (a -b \right )}\) \(84\)

[In]

int(x/(-a+x)/(-b+x)/(-c+x),x,method=_RETURNVERBOSE)

[Out]

a*ln(a-x)/(a-b)/(a-c)-b*ln(b-x)/(a-b)/(b-c)+c*ln(c-x)/(a-c)/(b-c)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {{\left (a - b\right )} c \log \left (-c + x\right ) + {\left (a b - a c\right )} \log \left (-a + x\right ) - {\left (a b - b c\right )} \log \left (-b + x\right )}{a^{2} b - a b^{2} + {\left (a - b\right )} c^{2} - {\left (a^{2} - b^{2}\right )} c} \]

[In]

integrate(x/(-a+x)/(-b+x)/(-c+x),x, algorithm="fricas")

[Out]

((a - b)*c*log(-c + x) + (a*b - a*c)*log(-a + x) - (a*b - b*c)*log(-b + x))/(a^2*b - a*b^2 + (a - b)*c^2 - (a^
2 - b^2)*c)

Sympy [F(-1)]

Timed out. \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\text {Timed out} \]

[In]

integrate(x/(-a+x)/(-b+x)/(-c+x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {a \log \left (-a + x\right )}{a^{2} - a b - {\left (a - b\right )} c} - \frac {b \log \left (-b + x\right )}{a b - b^{2} - {\left (a - b\right )} c} + \frac {c \log \left (-c + x\right )}{a b - {\left (a + b\right )} c + c^{2}} \]

[In]

integrate(x/(-a+x)/(-b+x)/(-c+x),x, algorithm="maxima")

[Out]

a*log(-a + x)/(a^2 - a*b - (a - b)*c) - b*log(-b + x)/(a*b - b^2 - (a - b)*c) + c*log(-c + x)/(a*b - (a + b)*c
 + c^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\frac {a \log \left ({\left | -a + x \right |}\right )}{a^{2} - a b - a c + b c} - \frac {b \log \left ({\left | -b + x \right |}\right )}{a b - b^{2} - a c + b c} + \frac {c \log \left ({\left | -c + x \right |}\right )}{a b - a c - b c + c^{2}} \]

[In]

integrate(x/(-a+x)/(-b+x)/(-c+x),x, algorithm="giac")

[Out]

a*log(abs(-a + x))/(a^2 - a*b - a*c + b*c) - b*log(abs(-b + x))/(a*b - b^2 - a*c + b*c) + c*log(abs(-c + x))/(
a*b - a*c - b*c + c^2)

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.28 \[ \int \frac {x}{(-a+x) (-b+x) (-c+x)} \, dx=\ln \left (x-a\right )\,\left (\frac {b}{\left (a-b\right )\,\left (b-c\right )}-\frac {c}{\left (a-c\right )\,\left (b-c\right )}\right )-\frac {b\,\ln \left (x-b\right )}{\left (a-b\right )\,\left (b-c\right )}+\frac {c\,\ln \left (x-c\right )}{\left (a-c\right )\,\left (b-c\right )} \]

[In]

int(-x/((a - x)*(b - x)*(c - x)),x)

[Out]

log(x - a)*(b/((a - b)*(b - c)) - c/((a - c)*(b - c))) - (b*log(x - b))/((a - b)*(b - c)) + (c*log(x - c))/((a
 - c)*(b - c))