Integrand size = 22, antiderivative size = 139 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (e f-d g) (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {e^{-\frac {2 a}{b n}} g (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n} \]
(-d*g+e*f)*(e*x+d)*Ei((a+b*ln(c*(e*x+d)^n))/b/n)/b/e^2/exp(a/b/n)/n/((c*(e *x+d)^n)^(1/n))+g*(e*x+d)^2*Ei(2*(a+b*ln(c*(e*x+d)^n))/b/n)/b/e^2/exp(2*a/ b/n)/n/((c*(e*x+d)^n)^(2/n))
Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {2 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-2/n} \left (e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )}{b e^2 n} \]
((d + e*x)*(E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1)*ExpIntegralEi[( a + b*Log[c*(d + e*x)^n])/(b*n)] + g*(d + e*x)*ExpIntegralEi[(2*(a + b*Log [c*(d + e*x)^n]))/(b*n)]))/(b*e^2*E^((2*a)/(b*n))*n*(c*(d + e*x)^n)^(2/n))
Time = 0.40 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2846, 2009}
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.
\(\displaystyle \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx\) |
\(\Big \downarrow \) 2846 |
\(\displaystyle \int \left (\frac {e f-d g}{e \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac {g (d+e x)}{e \left (a+b \log \left (c (d+e x)^n\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^2 n}+\frac {g e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^2 n}\) |
((e*f - d*g)*(d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b *e^2*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1)) + (g*(d + e*x)^2*ExpIntegralEi[ (2*(a + b*Log[c*(d + e*x)^n]))/(b*n)])/(b*e^2*E^((2*a)/(b*n))*n*(c*(d + e* x)^n)^(2/n))
3.1.90.3.1 Defintions of rubi rules used
Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) ]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q/(a + b*Log[c*(d + e* x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] & & IGtQ[q, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.96 (sec) , antiderivative size = 937, normalized size of antiderivative = 6.74
-1/e^2*g/b/n*(e*x+d)^2*c^(-2/n)*((e*x+d)^n)^(-2/n)*exp(-(-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,-2*ln(e*x+d)-(-I*b*Pi*csgn(I*c*(e*x+d)^n)*csgn(I*c)*c sgn(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*b*ln(c)+2*b*(l n((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)-1/e*f/b/n*(e*x+d)*c^(-1/n)*((e*x+d)^n) ^(-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*(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*csg n(I*c*(e*x+d)^n)^3*b+2*b*ln(c)+2*b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)+1 /e^2*d*g/b/n*(e*x+d)*c^(-1/n)*((e*x+d)^n)^(-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*(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*b*ln(c)+2* b*(ln((e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\left ({\left (e f - d g\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right ) + g \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b e^{2} n} \]
((e*f - d*g)*e^((b*log(c) + a)/(b*n))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n))) + g*log_integral((e^2*x^2 + 2*d*e*x + d^2)*e^(2*(b*log(c) + a )/(b*n))))*e^(-2*(b*log(c) + a)/(b*n))/(b*e^2*n)
\[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {f + g x}{a + b \log {\left (c \left (d + e x\right )^{n} \right )}}\, dx \]
\[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int { \frac {g x + f}{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {f {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} - \frac {d g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{2} n} + \frac {g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{2} n} \]
f*Ei(log(c)/n + a/(b*n) + log(e*x + d))*e^(-a/(b*n))/(b*c^(1/n)*e*n) - d*g *Ei(log(c)/n + a/(b*n) + log(e*x + d))*e^(-a/(b*n))/(b*c^(1/n)*e^2*n) + g* Ei(2*log(c)/n + 2*a/(b*n) + 2*log(e*x + d))*e^(-2*a/(b*n))/(b*c^(2/n)*e^2* n)
Timed out. \[ \int \frac {f+g x}{a+b \log \left (c (d+e x)^n\right )} \, dx=\int \frac {f+g\,x}{a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )} \,d x \]