Integrand size = 16, antiderivative size = 63 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \] Output:
(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 ))
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \] Input:
Integrate[(a + b*Log[c*(d + e*x)^n])^(-1),x]
Output:
((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))
Time = 0.43 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2836, 2737, 2609}
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.
\(\displaystyle \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {\int \frac {1}{a+b \log \left (c (d+e x)^n\right )}d(d+e x)}{e}\) |
\(\Big \downarrow \) 2737 |
\(\displaystyle \frac {(d+e x) \left (c (d+e x)^n\right )^{-1/n} \int \frac {\left (c (d+e x)^n\right )^{\frac {1}{n}}}{a+b \log \left (c (d+e x)^n\right )}d\log \left (c (d+e x)^n\right )}{e n}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n}\) |
Input:
Int[(a + b*Log[c*(d + e*x)^n])^(-1),x]
Output:
((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))
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.72 (sec) , antiderivative size = 311, normalized size of antiderivative = 4.94
method | result | size |
risch | \(-\frac {\left (e x +d \right ) \left (\left (e x +d \right )^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} {\mathrm e}^{-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 a}{2 n b}} \operatorname {expIntegral}_{1}\left (-\ln \left (e x +d \right )-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 b \ln \left (c \right )+2 b \left (\ln \left (\left (e x +d \right )^{n}\right )-n \ln \left (e x +d \right )\right )+2 a}{2 n b}\right )}{e n b}\) | \(311\) |
Input:
int(1/(a+b*ln(c*(e*x+d)^n)),x,method=_RETURNVERBOSE)
Output:
-1/e/n/b*(e*x+d)*((e*x+d)^n)^(-1/n)*c^(-1/n)*exp(-1/2*(I*b*Pi*csgn(I*(e*x+ d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*c sgn(I*c)-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I* c)+2*a)/n/b)*Ei(1,-ln(e*x+d)-1/2*(I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d )^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-I*b*Pi*csgn( I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+2*b*ln(c)+2*b*(ln( (e*x+d)^n)-n*ln(e*x+d))+2*a)/n/b)
Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{\left (-\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )}{b e n} \] Input:
integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")
Output:
e^(-(b*log(c) + a)/(b*n))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n))) /(b*e*n)
\[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {1}{a + b \log {\left (c \left (d + e x\right )^{n} \right )}}\, dx \] Input:
integrate(1/(a+b*ln(c*(e*x+d)**n)),x)
Output:
Integral(1/(a + b*log(c*(d + e*x)**n)), x)
\[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int { \frac {1}{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a} \,d x } \] Input:
integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")
Output:
integrate(1/(b*log((e*x + d)^n*c) + a), x)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} \] Input:
integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")
Output:
Ei(log(c)/n + a/(b*n) + log(e*x + d))*e^(-a/(b*n))/(b*c^(1/n)*e*n)
Timed out. \[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {1}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \] Input:
int(1/(a + b*log(c*(d + e*x)^n)),x)
Output:
int(1/(a + b*log(c*(d + e*x)^n)), x)
\[ \int \frac {1}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {\left (\int \frac {x}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b \,e^{2} n +\mathrm {log}\left (\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b +a \right ) d}{b e n} \] Input:
int(1/(a+b*log(c*(e*x+d)^n)),x)
Output:
(int(x/(log((d + e*x)**n*c)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e*x) ,x)*b*e**2*n + log(log((d + e*x)**n*c)*b + a)*d)/(b*e*n)