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

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

[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 = 0.08, 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 = {2318, 2221, 2317, 2438} \begin {gather*} -\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 ) \end {gather*}

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 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*} \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 {\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*}

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Mathematica [A]
time = 0.00, size = 75, normalized size = 1.00 \begin {gather*} 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 {gather*}

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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Maple [A]
time = 0.06, size = 69, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\dilog \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )+\ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{c b \ln \left (f \right ) n}\)	\(69\)
default	\(\frac {\dilog \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )+\ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (-\frac {e \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{c b \ln \left (f \right ) n}\)	\(69\)
risch	\(x \ln \left (d +e \left (f^{c \left (b x +a \right )}\right )^{n}\right )-\frac {\dilog \left (\frac {d +e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right )}{c b \ln \left (f \right ) n}-\frac {\ln \left (\frac {d +e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}}{d}\right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}-\ln \left (d +e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) x +\frac {\ln \left (d +e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (f^{c \left (b x +a \right )}\right )}{c b \ln \left (f \right )}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c/b/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))

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Maxima [A]
time = 0.29, size = 87, normalized size = 1.16 \begin {gather*} x \log \left (f^{{\left (b x + a\right )} c n} e + d\right ) - \frac {b c n x \log \left (f\right ) \log \left (\frac {f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )}}{d} + 1\right ) + {\rm Li}_2\left (-\frac {f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )}}{d}\right )}{b c n \log \left (f\right )} \end {gather*}

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

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Fricas [A]
time = 0.37, size = 109, normalized size = 1.45 \begin {gather*} \frac {{\left (b c n x + a c n\right )} \log \left (f^{b c n x + a c n} e + d\right ) \log \left (f\right ) - {\left (b c n x + a c n\right )} \log \left (f\right ) \log \left (\frac {f^{b c n x + a c n} e + d}{d}\right ) - {\rm Li}_2\left (-\frac {f^{b c n x + a c n} e + d}{d} + 1\right )}{b c n \log \left (f\right )} \end {gather*}

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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((f^((b*x + a)*c))^n*e + d), x)

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

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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