\(\int \frac {\log (c-\frac {a (1-c) x^{-m}}{b})}{x (a+b x^m)} \, dx\) [333]

Optimal result

Integrand size = 32, antiderivative size = 27 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (b+a x^{-m}\right )}{b}\right )}{a m} \]

[Out]

polylog(2,(1-c)*(b+a/(x^m))/b)/a/m

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2525, 2459, 2440, 2438} \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (a x^{-m}+b\right )}{b}\right )}{a m} \]

[In]

Int[Log[c - (a*(1 - c))/(b*x^m)]/(x*(a + b*x^m)),x]

[Out]

PolyLog[2, ((1 - c)*(b + a/x^m))/b]/(a*m)

Rule 2438

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]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[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]

Rule 2459

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol]
:> Int[(g + f*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q}, x] && EqQ[m,
q] && IntegerQ[q]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(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] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\log \left (c-\frac {a (1-c) x}{b}\right )}{\left (a+\frac {b}{x}\right ) x} \, dx,x,x^{-m}\right )}{m} \\ & = -\frac {\text {Subst}\left (\int \frac {\log \left (c-\frac {a (1-c) x}{b}\right )}{b+a x} \, dx,x,x^{-m}\right )}{m} \\ & = -\frac {\text {Subst}\left (\int \frac {\log \left (1-\frac {(1-c) x}{b}\right )}{x} \, dx,x,b+a x^{-m}\right )}{a m} \\ & = \frac {\text {Li}_2\left (\frac {(1-c) \left (b+a x^{-m}\right )}{b}\right )}{a m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {(-1+c) x^{-m} \left (a+b x^m\right )}{b}\right )}{a m} \]

[In]

Integrate[Log[c - (a*(1 - c))/(b*x^m)]/(x*(a + b*x^m)),x]

[Out]

PolyLog[2, -(((-1 + c)*(a + b*x^m))/(b*x^m))]/(a*m)

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\operatorname {dilog}\left (\frac {\left (c a -a \right ) x^{-m}}{b}+c \right )}{m a}\)	\(27\)
default	\(\frac {\operatorname {dilog}\left (\frac {\left (c a -a \right ) x^{-m}}{b}+c \right )}{m a}\)	\(27\)
risch	\(\text {Expression too large to display}\)	\(1267\)

[In]

int(ln(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x,method=_RETURNVERBOSE)

[Out]

1/m/a*dilog(1/b*(a*c-a)/(x^m)+c)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {{\rm Li}_2\left (-\frac {b c x^{m} + a c - a}{b x^{m}} + 1\right )}{a m} \]

[In]

integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="fricas")

[Out]

dilog(-(b*c*x^m + a*c - a)/(b*x^m) + 1)/(a*m)

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\text {Timed out} \]

[In]

integrate(ln(c-a*(1-c)/b/(x**m))/x/(a+b*x**m),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \]

[In]

integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="maxima")

[Out]

(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) - lo
g(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) - (lo
g(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)

Giac [F]

\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \]

[In]

integrate(log(c-a*(1-c)/b/(x^m))/x/(a+b*x^m),x, algorithm="giac")

[Out]

integrate(log(c + a*(c - 1)/(b*x^m))/((b*x^m + a)*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int \frac {\ln \left (c+\frac {a\,\left (c-1\right )}{b\,x^m}\right )}{x\,\left (a+b\,x^m\right )} \,d x \]

[In]

int(log(c + (a*(c - 1))/(b*x^m))/(x*(a + b*x^m)),x)

[Out]

int(log(c + (a*(c - 1))/(b*x^m))/(x*(a + b*x^m)), x)