Integrand size = 16, antiderivative size = 75 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-x \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)} \]
x*ln(d+e*(f^(c*(b*x+a)))^n)-x*ln(1+e*(f^(c*(b*x+a)))^n/d)-polylog(2,-e*(f^ (c*(b*x+a)))^n/d)/b/c/n/ln(f)
Time = 0.00 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-x \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)} \]
x*Log[d + e*(f^(c*(a + b*x)))^n] - x*Log[1 + (e*(f^(c*(a + b*x)))^n)/d] - PolyLog[2, -((e*(f^(c*(a + b*x)))^n)/d)]/(b*c*n*Log[f])
Time = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2620, 2715, 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.
\(\displaystyle \int \log \left (e \left (f^{c (a+b x)}\right )^n+d\right ) \, dx\) |
\(\Big \downarrow \) 2716 |
\(\displaystyle x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-b c e n \log (f) \int \frac {\left (f^{c (a+b x)}\right )^n x}{e \left (f^{c (a+b x)}\right )^n+d}dx\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-b c e n \log (f) \left (\frac {x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right )}{b c e n \log (f)}-\frac {\int \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right )dx}{b c e n \log (f)}\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-b c e n \log (f) \left (\frac {x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right )}{b c e n \log (f)}-\frac {\int \left (f^{c (a+b x)}\right )^{-n} \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right )d\left (f^{c (a+b x)}\right )^n}{b^2 c^2 e n^2 \log ^2(f)}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-b c e n \log (f) \left (\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 e n^2 \log ^2(f)}+\frac {x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right )}{b c e n \log (f)}\right )\) |
x*Log[d + e*(f^(c*(a + b*x)))^n] - b*c*e*n*Log[f]*((x*Log[1 + (e*(f^(c*(a + b*x)))^n)/d])/(b*c*e*n*Log[f]) + PolyLog[2, -((e*(f^(c*(a + b*x)))^n)/d) ]/(b^2*c^2*e*n^2*Log[f]^2))
3.2.26.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[x*Log[a + b*(F^(e*(c + d*x)))^n], x] - Simp[b*d*e*n*Log[F] Int[x* ((F^(e*(c + d*x)))^n/(a + b*(F^(e*(c + d*x)))^n)), x], x] /; FreeQ[{F, a, b , c, d, e, n}, x] && !GtQ[a, 0]
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]
Time = 1.67 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )+\ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{b c \ln \left (f \right ) n}\) | \(69\) |
default | \(\frac {\operatorname {dilog}\left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )+\ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{b c \ln \left (f \right ) n}\) | \(69\) |
risch | \(x \ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right )-\frac {\operatorname {dilog}\left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right )}{c b \ln \left (f \right ) n}-\frac {\ln \left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}-\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x +\frac {\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}\) | \(213\) |
1/b/c/ln(f)/n*(dilog(-e*(f^(c*(b*x+a)))^n/d)+ln(d+e*(f^(c*(b*x+a)))^n)*ln( -e*(f^(c*(b*x+a)))^n/d))
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.41 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {{\left (b c n x + a c n\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \left (f\right ) - {\left (b c n x + a c n\right )} \log \left (f\right ) \log \left (\frac {e f^{b c n x + a c n} + d}{d}\right ) - {\rm Li}_2\left (-\frac {e f^{b c n x + a c n} + d}{d} + 1\right )}{b c n \log \left (f\right )} \]
((b*c*n*x + a*c*n)*log(e*f^(b*c*n*x + a*c*n) + d)*log(f) - (b*c*n*x + a*c* n)*log(f)*log((e*f^(b*c*n*x + a*c*n) + d)/d) - dilog(-(e*f^(b*c*n*x + a*c* n) + d)/d + 1))/(b*c*n*log(f))
\[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int \log {\left (d + e \left (f^{c \left (a + b x\right )}\right )^{n} \right )}\, dx \]
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (e f^{{\left (b x + a\right )} c n} + d\right ) - \frac {b c n x \log \left (\frac {e f^{b c n x} f^{a c n}}{d} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-\frac {e f^{b c n x} f^{a c n}}{d}\right )}{b c n \log \left (f\right )} \]
x*log(e*f^((b*x + a)*c*n) + d) - (b*c*n*x*log(e*f^(b*c*n*x)*f^(a*c*n)/d + 1)*log(f) + dilog(-e*f^(b*c*n*x)*f^(a*c*n)/d))/(b*c*n*log(f))
\[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + d\right ) \,d x } \]
Timed out. \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int \ln \left (d+e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \]