Integrand size = 33, antiderivative size = 25 \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=-\frac {\operatorname {PolyLog}\left (2,1-c \left (d+e x^n\right )\right )}{c e n} \] Output:
-polylog(2,1-c*(d+e*x^n))/c/e/n
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=-\frac {\operatorname {PolyLog}\left (2,1-c d-c e x^n\right )}{c e n} \] Input:
Integrate[Log[c*(d + e*x^n)]/(x*(c*e - (1 - c*d)/x^n)),x]
Output:
-(PolyLog[2, 1 - c*d - c*e*x^n]/(c*e*n))
Time = 0.54 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2005, 2925, 25, 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.
\(\displaystyle \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \int \frac {x^{n-1} \log \left (c \left (d+e x^n\right )\right )}{c d+c e x^n-1}dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {\int -\frac {\log \left (c \left (e x^n+d\right )\right )}{-c e x^n-c d+1}dx^n}{n}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\log \left (c \left (e x^n+d\right )\right )}{-c e x^n-c d+1}dx^n}{n}\) |
\(\Big \downarrow \) 2840 |
\(\displaystyle \frac {\int x^{-n} \log \left (c e x^n+c d\right )d\left (-c e x^n-c d+1\right )}{c e n}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\frac {\operatorname {PolyLog}\left (2,-c e x^n-c d+1\right )}{c e n}\) |
Input:
Int[Log[c*(d + e*x^n)]/(x*(c*e - (1 - c*d)/x^n)),x]
Output:
-(PolyLog[2, 1 - c*d - c*e*x^n]/(c*e*n))
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 = 4.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(-\frac {\operatorname {dilog}\left (c e \,x^{n}+c d \right )}{n c e}\) | \(23\) |
default | \(-\frac {\operatorname {dilog}\left (c e \,x^{n}+c d \right )}{n c e}\) | \(23\) |
risch | \(\frac {\ln \left (1-c \left (d +e \,x^{n}\right )\right ) \ln \left (d +e \,x^{n}\right )}{n e c}-\frac {\ln \left (1-c \left (d +e \,x^{n}\right )\right ) \ln \left (c \left (d +e \,x^{n}\right )\right )}{n e c}-\frac {\operatorname {dilog}\left (c \left (d +e \,x^{n}\right )\right )}{n e c}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (-1+c d +c e \,x^{n}\right )}{n c e}\) | \(215\) |
Input:
int(ln(c*(d+e*x^n))/x/(c*e-(-c*d+1)/(x^n)),x,method=_RETURNVERBOSE)
Output:
-1/n*dilog(c*e*x^n+c*d)/c/e
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=-\frac {{\rm Li}_2\left (-c e x^{n} - c d + 1\right )}{c e n} \] Input:
integrate(log(c*(d+e*x^n))/x/(c*e-(-c*d+1)/(x^n)),x, algorithm="fricas")
Output:
-dilog(-c*e*x^n - c*d + 1)/(c*e*n)
Exception generated. \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(ln(c*(d+e*x**n))/x/(c*e-(-c*d+1)/(x**n)),x)
Output:
Exception raised: TypeError >> Invalid comparison of non-real zoo
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.24 \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx={\left (\frac {\log \left (c e + \frac {c d - 1}{x^{n}}\right )}{c e n} - \frac {\log \left (\frac {1}{x^{n}}\right )}{c e n}\right )} \log \left ({\left (e x^{n} + d\right )} c\right ) - \frac {\log \left (c e x^{n} + c d\right ) \log \left (c e x^{n} + c d - 1\right ) + {\rm Li}_2\left (-c e x^{n} - c d + 1\right )}{c e n} \] Input:
integrate(log(c*(d+e*x^n))/x/(c*e-(-c*d+1)/(x^n)),x, algorithm="maxima")
Output:
(log(c*e + (c*d - 1)/x^n)/(c*e*n) - log(1/(x^n))/(c*e*n))*log((e*x^n + d)* c) - (log(c*e*x^n + c*d)*log(c*e*x^n + c*d - 1) + dilog(-c*e*x^n - c*d + 1 ))/(c*e*n)
\[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )} c\right )}{{\left (c e + \frac {c d - 1}{x^{n}}\right )} x} \,d x } \] Input:
integrate(log(c*(d+e*x^n))/x/(c*e-(-c*d+1)/(x^n)),x, algorithm="giac")
Output:
integrate(log((e*x^n + d)*c)/((c*e + (c*d - 1)/x^n)*x), x)
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=\int \frac {\ln \left (c\,\left (d+e\,x^n\right )\right )}{x\,\left (c\,e+\frac {c\,d-1}{x^n}\right )} \,d x \] Input:
int(log(c*(d + e*x^n))/(x*(c*e + (c*d - 1)/x^n)),x)
Output:
int(log(c*(d + e*x^n))/(x*(c*e + (c*d - 1)/x^n)), x)
\[ \int \frac {\log \left (c \left (d+e x^n\right )\right )}{x \left (c e-(1-c d) x^{-n}\right )} \, dx=\frac {-2 \left (\int \frac {x^{2 n} \mathrm {log}\left (x^{n} c e +c d \right )}{x^{2 n} c^{2} d \,e^{2} x -x^{2 n} c \,e^{2} x +2 x^{n} c^{2} d^{2} e x -3 x^{n} c d e x +x^{n} e x +c^{2} d^{3} x -2 c \,d^{2} x +d x}d x \right ) c d \,e^{2} n +2 \left (\int \frac {x^{2 n} \mathrm {log}\left (x^{n} c e +c d \right )}{x^{2 n} c^{2} d \,e^{2} x -x^{2 n} c \,e^{2} x +2 x^{n} c^{2} d^{2} e x -3 x^{n} c d e x +x^{n} e x +c^{2} d^{3} x -2 c \,d^{2} x +d x}d x \right ) e^{2} n +\mathrm {log}\left (x^{n} c e +c d \right )^{2} d}{2 e n \left (c d -1\right )} \] Input:
int(log(c*(d+e*x^n))/x/(c*e-(-c*d+1)/(x^n)),x)
Output:
( - 2*int((x**(2*n)*log(x**n*c*e + c*d))/(x**(2*n)*c**2*d*e**2*x - x**(2*n )*c*e**2*x + 2*x**n*c**2*d**2*e*x - 3*x**n*c*d*e*x + x**n*e*x + c**2*d**3* x - 2*c*d**2*x + d*x),x)*c*d*e**2*n + 2*int((x**(2*n)*log(x**n*c*e + c*d)) /(x**(2*n)*c**2*d*e**2*x - x**(2*n)*c*e**2*x + 2*x**n*c**2*d**2*e*x - 3*x* *n*c*d*e*x + x**n*e*x + c**2*d**3*x - 2*c*d**2*x + d*x),x)*e**2*n + log(x* *n*c*e + c*d)**2*d)/(2*e*n*(c*d - 1))