\(\int \frac {1}{a+b \log (c (d+e x)^n)} \, dx\) [21]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 63 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \]

[Out]

(e*x+d)*Ei((a+b*ln(c*(e*x+d)^n))/b/n)/b/e/exp(a/b/n)/n/((c*(e*x+d)^n)^(1/n))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2436, 2337, 2209} \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \]

[In]

Int[(a + b*Log[c*(d + e*x)^n])^(-1),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b*e*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \]

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^(-1),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b*e*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.87 (sec) , antiderivative size = 309, normalized size of antiderivative = 4.90

method result size
risch \(-\frac {\left (e x +d \right ) \left (\left (e x +d \right )^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} {\mathrm e}^{-\frac {-i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )+i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b +i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b -i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b +2 a}{2 b n}} \operatorname {Ei}_{1}\left (-\ln \left (e x +d \right )+\frac {i \left (b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )-b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 i b \ln \left (c \right )+2 i b \left (\ln \left (\left (e x +d \right )^{n}\right )-n \ln \left (e x +d \right )\right )+2 i a \right )}{2 b n}\right )}{e b n}\) \(309\)

[In]

int(1/(a+b*ln(c*(e*x+d)^n)),x,method=_RETURNVERBOSE)

[Out]

-1/e/b/n*(e*x+d)*((e*x+d)^n)^(-1/n)*c^(-1/n)*exp(-1/2*(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*csgn(I*(e*x+d)^n)
+I*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*b+I*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*b-I*Pi*csgn(I*c*(e*x+d)^n
)^3*b+2*a)/b/n)*Ei(1,-ln(e*x+d)+1/2*I*(b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-b*Pi*csgn(I*c)*csg
n(I*c*(e*x+d)^n)^2-b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+b*Pi*csgn(I*c*(e*x+d)^n)^3+2*I*b*ln(c)+2*I*b*(
ln((e*x+d)^n)-n*ln(e*x+d))+2*I*a)/b/n)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{\left (-\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )}{b e n} \]

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

e^(-(b*log(c) + a)/(b*n))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n)))/(b*e*n)

Sympy [F]

\[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {1}{a + b \log {\left (c \left (d + e x\right )^{n} \right )}}\, dx \]

[In]

integrate(1/(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Integral(1/(a + b*log(c*(d + e*x)**n)), x)

Maxima [F]

\[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int { \frac {1}{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a} \,d x } \]

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")

[Out]

integrate(1/(b*log((e*x + d)^n*c) + a), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} \]

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {1}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \]

[In]

int(1/(a + b*log(c*(d + e*x)^n)),x)

[Out]

int(1/(a + b*log(c*(d + e*x)^n)), x)