Integrand size = 25, antiderivative size = 42 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {(e+f x)^{1-p} (g (e+f x))^{-1+p} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \] Output:
(f*x+e)^(1-p)*(g*(f*x+e))^(-1+p)*Li(d*(f*x+e)^p)/d/f/p
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {(e+f x)^{1-p} (g (e+f x))^{-1+p} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \] Input:
Integrate[(e*g + f*g*x)^(-1 + p)/Log[d*(e + f*x)^p],x]
Output:
((e + f*x)^(1 - p)*(g*(e + f*x))^(-1 + p)*LogIntegral[d*(e + f*x)^p])/(d*f *p)
Time = 0.48 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2837, 2745, 2744, 2735}
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.
\(\displaystyle \int \frac {(e g+f g x)^{p-1}}{\log \left (d (e+f x)^p\right )} \, dx\) |
\(\Big \downarrow \) 2837 |
\(\displaystyle \frac {\int \frac {(g (e+f x))^{p-1}}{\log \left (d (e+f x)^p\right )}d(e+f x)}{f}\) |
\(\Big \downarrow \) 2745 |
\(\displaystyle \frac {(e+f x)^{1-p} (g (e+f x))^{p-1} \int \frac {(e+f x)^{p-1}}{\log \left (d (e+f x)^p\right )}d(e+f x)}{f}\) |
\(\Big \downarrow \) 2744 |
\(\displaystyle \frac {(e+f x)^{1-p} (g (e+f x))^{p-1} \int \frac {1}{\log \left (d (e+f x)^p\right )}d(e+f x)^p}{f p}\) |
\(\Big \downarrow \) 2735 |
\(\displaystyle \frac {(e+f x)^{1-p} (g (e+f x))^{p-1} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p}\) |
Input:
Int[(e*g + f*g*x)^(-1 + p)/Log[d*(e + f*x)^p],x]
Output:
((e + f*x)^(1 - p)*(g*(e + f*x))^(-1 + p)*LogIntegral[d*(e + f*x)^p])/(d*f *p)
Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[1/n Subst[Int[1/Lo g[c*x], x], x, x^n], x] /; FreeQ[{c, m, n}, x] && EqQ[m, n - 1]
Int[((d_)*(x_))^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[(d*x)^m/x^m Int[x^m/Log[c*x^n], x], x] /; FreeQ[{c, d, m, n}, x] && EqQ[m, n - 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[1/e Subst[Int[(f*(x/d))^q*(a + b*Log[c*x ^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && EqQ[e*f - d*g, 0]
\[\int \frac {\left (f g x +e g \right )^{p -1}}{\ln \left (d \left (f x +e \right )^{p}\right )}d x\]
Input:
int((f*g*x+e*g)^(p-1)/ln(d*(f*x+e)^p),x)
Output:
int((f*g*x+e*g)^(p-1)/ln(d*(f*x+e)^p),x)
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {g^{p - 1} {\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \] Input:
integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="fricas")
Output:
g^(p - 1)*Ei(p*log(f*x + e) + log(d))/(d*f*p)
\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int \frac {\left (g \left (e + f x\right )\right )^{p - 1}}{\log {\left (d \left (e + f x\right )^{p} \right )}}\, dx \] Input:
integrate((f*g*x+e*g)**(-1+p)/ln(d*(f*x+e)**p),x)
Output:
Integral((g*(e + f*x))**(p - 1)/log(d*(e + f*x)**p), x)
\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int { \frac {{\left (f g x + e g\right )}^{p - 1}}{\log \left ({\left (f x + e\right )}^{p} d\right )} \,d x } \] Input:
integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="maxima")
Output:
integrate((f*g*x + e*g)^(p - 1)/log((f*x + e)^p*d), x)
\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int { \frac {{\left (f g x + e g\right )}^{p - 1}}{\log \left ({\left (f x + e\right )}^{p} d\right )} \,d x } \] Input:
integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="giac")
Output:
integrate((f*g*x + e*g)^(p - 1)/log((f*x + e)^p*d), x)
Timed out. \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int \frac {{\left (e\,g+f\,g\,x\right )}^{p-1}}{\ln \left (d\,{\left (e+f\,x\right )}^p\right )} \,d x \] Input:
int((e*g + f*g*x)^(p - 1)/log(d*(e + f*x)^p),x)
Output:
int((e*g + f*g*x)^(p - 1)/log(d*(e + f*x)^p), x)
\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {\int \frac {\left (f g x +e g \right )^{p}}{\mathrm {log}\left (\left (f x +e \right )^{p} d \right ) e +\mathrm {log}\left (\left (f x +e \right )^{p} d \right ) f x}d x}{g} \] Input:
int((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x)
Output:
int((e*g + f*g*x)**p/(log((e + f*x)**p*d)*e + log((e + f*x)**p*d)*f*x),x)/ g