\(\int \log (d+e (f^{c (a+b x)})^n) \, dx\) [126]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-x \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2318, 2221, 2317, 2438} \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

[In]

Int[Log[d + e*(f^(c*(a + b*x)))^n],x]

[Out]

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])

Rule 2221

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]
- Dist[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]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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]

Rule 2318

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[x*Log[a + b*(F^(e*(c + d*x)
))^n], x] - Dist[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]

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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-x \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {\operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)} \]

[In]

Integrate[Log[d + e*(f^(c*(a + b*x)))^n],x]

[Out]

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])

Maple [A] (verified)

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	\(x \ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right )-\frac {\operatorname {dilog}\left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right )}{c b \ln \left (f \right ) n}-\frac {\ln \left (\frac {d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e}{d}\right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}-\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x +\frac {\ln \left (d +f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}\)	\(213\)

[In]

int(ln(d+e*(f^(c*(b*x+a)))^n),x,method=_RETURNVERBOSE)

[Out]

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))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.41 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {{\left (b c n x + a c n\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \left (f\right ) - {\left (b c n x + a c n\right )} \log \left (f\right ) \log \left (\frac {e f^{b c n x + a c n} + d}{d}\right ) - {\rm Li}_2\left (-\frac {e f^{b c n x + a c n} + d}{d} + 1\right )}{b c n \log \left (f\right )} \]

[In]

integrate(log(d+e*(f^(c*(b*x+a)))^n),x, algorithm="fricas")

[Out]

((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))

Sympy [F]

\[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int \log {\left (d + e \left (f^{c \left (a + b x\right )}\right )^{n} \right )}\, dx \]

[In]

integrate(ln(d+e*(f**(c*(b*x+a)))**n),x)

[Out]

Integral(log(d + e*(f**(c*(a + b*x)))**n), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=x \log \left (e f^{{\left (b x + a\right )} c n} + d\right ) - \frac {b c n x \log \left (\frac {e f^{b c n x} f^{a c n}}{d} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-\frac {e f^{b c n x} f^{a c n}}{d}\right )}{b c n \log \left (f\right )} \]

[In]

integrate(log(d+e*(f^(c*(b*x+a)))^n),x, algorithm="maxima")

[Out]

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))

Giac [F]

\[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + d\right ) \,d x } \]

[In]

integrate(log(d+e*(f^(c*(b*x+a)))^n),x, algorithm="giac")

[Out]

integrate(log(e*(f^((b*x + a)*c))^n + d), x)

Mupad [F(-1)]

Timed out. \[ \int \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int \ln \left (d+e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \]

[In]

int(log(d + e*(f^(c*(a + b*x)))^n),x)

[Out]

int(log(d + e*(f^(c*(a + b*x)))^n), x)