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

Optimal. Leaf size=75 \[ -\frac {\text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

[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)

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Rubi [A]  time = 0.13, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2280, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

Antiderivative was successfully verified.

[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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2280

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 2391

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*} \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 )-(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 {\operatorname {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*}

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Mathematica [A]  time = 0.00, size = 75, normalized size = 1.00 \[ -\frac {\text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+x \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-x \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.68, size = 106, normalized size = 1.41 \[ \frac {{\left (b c n x + a c n\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \relax (f) - {\left (b c n x + a c n\right )} \log \relax (f) \log \left (\frac {e f^{b c n x + a c n} + d}{d}\right ) - {\rm Li}_2\left (-\frac {e f^{b c n x + a c n} + d}{d} + 1\right )}{b c n \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + d\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.08, size = 82, normalized size = 1.09 \[ \frac {\ln \left (-\frac {e \left (f^{\left (b x +a \right ) c}\right )^{n}}{d}\right ) \ln \left (e \left (f^{\left (b x +a \right ) c}\right )^{n}+d \right )}{b c n \ln \relax (f )}+\frac {\dilog \left (-\frac {e \left (f^{\left (b x +a \right ) c}\right )^{n}}{d}\right )}{b c n \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b/c/ln(f)/n*ln(e*(f^((b*x+a)*c))^n+d)*ln(-e*(f^((b*x+a)*c))^n/d)+1/b/c/ln(f)/n*dilog(-e*(f^((b*x+a)*c))^n/d)

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maxima [A]  time = 0.83, size = 82, normalized size = 1.09 \[ x \log \left (e f^{{\left (b x + a\right )} c n} + d\right ) - \frac {b c n x \log \left (\frac {e f^{b c n x} f^{a c n}}{d} + 1\right ) \log \relax (f) + {\rm Li}_2\left (-\frac {e f^{b c n x} f^{a c n}}{d}\right )}{b c n \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (d+e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(d + e*(f^(c*(a + b*x)))^n),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - b c e n e^{a c n \log {\relax (f )}} \log {\relax (f )} \int \frac {x e^{b c n x \log {\relax (f )}}}{d + e e^{a c n \log {\relax (f )}} e^{b c n x \log {\relax (f )}}}\, dx + x \log {\left (d + e \left (f^{c \left (a + b x\right )}\right )^{n} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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