Integrand size = 32, antiderivative size = 27 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (b+a x^{-m}\right )}{b}\right )}{a m} \] Output:
polylog(2,(1-c)*(b+a/(x^m))/b)/a/m
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {(-1+c) x^{-m} \left (a+b x^m\right )}{b}\right )}{a m} \] Input:
Integrate[Log[c - (a*(1 - c))/(b*x^m)]/(x*(a + b*x^m)),x]
Output:
PolyLog[2, -(((-1 + c)*(a + b*x^m))/(b*x^m))]/(a*m)
Time = 0.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2925, 2005, 2840, 2838}
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.
\(\displaystyle \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle -\frac {\int \frac {x^m \log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{b x^m+a}dx^{-m}}{m}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle -\frac {\int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{a x^{-m}+b}dx^{-m}}{m}\) |
\(\Big \downarrow \) 2840 |
\(\displaystyle -\frac {\int x^m \log \left (1-\frac {(1-c) \left (a x^{-m}+b\right )}{b}\right )d\left (a x^{-m}+b\right )}{a m}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (a x^{-m}+b\right )}{b}\right )}{a m}\) |
Input:
Int[Log[c - (a*(1 - c))/(b*x^m)]/(x*(a + b*x^m)),x]
Output:
PolyLog[2, ((1 - c)*(b + a/x^m))/b]/(a*m)
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_ Symbol] :> Simp[1/g Subst[Int[(a + b*Log[1 + c*e*(x/g)])/x, x], x, f + g* x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g + c *(e*f - d*g), 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 0])
Time = 6.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (\frac {\left (a c -a \right ) x^{-m}}{b}+c \right )}{m a}\) | \(27\) |
default | \(\frac {\operatorname {dilog}\left (\frac {\left (a c -a \right ) x^{-m}}{b}+c \right )}{m a}\) | \(27\) |
risch | \(\text {Expression too large to display}\) | \(1267\) |
Input:
int(ln(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x,method=_RETURNVERBOSE)
Output:
1/m*dilog(1/b*(a*c-a)/(x^m)+c)/a
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {{\rm Li}_2\left (-\frac {b c x^{m} + a c - a}{b x^{m}} + 1\right )}{a m} \] Input:
integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="fricas")
Output:
dilog(-(b*c*x^m + a*c - a)/(b*x^m) + 1)/(a*m)
Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\text {Timed out} \] Input:
integrate(ln(c-a*(1-c)/b/(x**m))/x/(a+b*x**m),x)
Output:
Timed out
\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \] Input:
integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="maxima")
Output:
(c*m - m)*integrate(log(x)/(b*c*x*x^m + a*(c - 1)*x), x) + (log(b*c*x^m + a*c - a)*log(x) - log(b)*log(x) - log(x)*log(x^m))/a + log(b)*log((b*x^m + a)/b)/(a*m) + (log(x^m)*log(b*x^m/a + 1) + dilog(-b*x^m/a))/(a*m) - (log( b*c*x^m + a*c - a)*log((b*c*x^m + a*(c - 1))/a + 1) + dilog(-(b*c*x^m + a* (c - 1))/a))/(a*m)
\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \] Input:
integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="giac")
Output:
integrate(log(c + a*(c - 1)/(b*x^m))/((b*x^m + a)*x), x)
Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int \frac {\ln \left (c+\frac {a\,\left (c-1\right )}{b\,x^m}\right )}{x\,\left (a+b\,x^m\right )} \,d x \] Input:
int(log(c + (a*(c - 1))/(b*x^m))/(x*(a + b*x^m)),x)
Output:
int(log(c + (a*(c - 1))/(b*x^m))/(x*(a + b*x^m)), x)
\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {-2 \left (\int \frac {\mathrm {log}\left (\frac {x^{m} b c +a c -a}{x^{m} b}\right )}{x^{2 m} b^{2} c x +2 x^{m} a b c x -x^{m} a b x +a^{2} c x -a^{2} x}d x \right ) a^{2} c m +2 \left (\int \frac {\mathrm {log}\left (\frac {x^{m} b c +a c -a}{x^{m} b}\right )}{x^{2 m} b^{2} c x +2 x^{m} a b c x -x^{m} a b x +a^{2} c x -a^{2} x}d x \right ) a^{2} m -\mathrm {log}\left (\frac {x^{m} b c +a c -a}{x^{m} b}\right )^{2} c}{2 a m \left (c -1\right )} \] Input:
int(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x)
Output:
( - 2*int(log((x**m*b*c + a*c - a)/(x**m*b))/(x**(2*m)*b**2*c*x + 2*x**m*a *b*c*x - x**m*a*b*x + a**2*c*x - a**2*x),x)*a**2*c*m + 2*int(log((x**m*b*c + a*c - a)/(x**m*b))/(x**(2*m)*b**2*c*x + 2*x**m*a*b*c*x - x**m*a*b*x + a **2*c*x - a**2*x),x)*a**2*m - log((x**m*b*c + a*c - a)/(x**m*b))**2*c)/(2* a*m*(c - 1))