\(\int \frac {(e g+f g x)^{-1+p}}{\log (d (e+f x)^p)} \, dx\) [34]

Optimal result

Integrand size = 25, antiderivative size = 42 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {(e+f x)^{1-p} (g (e+f x))^{-1+p} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[Out]

(f*x+e)^(1-p)*(g*(f*x+e))^(-1+p)*Li(d*(f*x+e)^p)/d/f/p

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2437, 2345, 2344, 2335} \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {(e+f x)^{1-p} (g (e+f x))^{p-1} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[In]

Int[(e*g + f*g*x)^(-1 + p)/Log[d*(e + f*x)^p],x]

[Out]

((e + f*x)^(1 - p)*(g*(e + f*x))^(-1 + p)*LogIntegral[d*(e + f*x)^p])/(d*f*p)

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2344

Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[1/n, Subst[Int[1/Log[c*x], x], x, x^n], x] /; FreeQ[{c,
m, n}, x] && EqQ[m, n - 1]

Rule 2345

Int[((d_)*(x_))^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[(d*x)^m/x^m, Int[x^m/Log[c*x^n], x], x] /; FreeQ
[{c, d, m, n}, x] && EqQ[m, n - 1]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(g x)^{-1+p}}{\log \left (d x^p\right )} \, dx,x,e+f x\right )}{f} \\ & = \frac {\left ((e+f x)^{1-p} (g (e+f x))^{-1+p}\right ) \text {Subst}\left (\int \frac {x^{-1+p}}{\log \left (d x^p\right )} \, dx,x,e+f x\right )}{f} \\ & = \frac {\left ((e+f x)^{1-p} (g (e+f x))^{-1+p}\right ) \text {Subst}\left (\int \frac {1}{\log (d x)} \, dx,x,(e+f x)^p\right )}{f p} \\ & = \frac {(e+f x)^{1-p} (g (e+f x))^{-1+p} \text {li}\left (d (e+f x)^p\right )}{d f p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {(e+f x)^{1-p} (g (e+f x))^{-1+p} \operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[In]

Integrate[(e*g + f*g*x)^(-1 + p)/Log[d*(e + f*x)^p],x]

[Out]

((e + f*x)^(1 - p)*(g*(e + f*x))^(-1 + p)*LogIntegral[d*(e + f*x)^p])/(d*f*p)

Maple [F]

\[\int \frac {\left (f g x +e g \right )^{-1+p}}{\ln \left (d \left (f x +e \right )^{p}\right )}d x\]

[In]

int((f*g*x+e*g)^(-1+p)/ln(d*(f*x+e)^p),x)

[Out]

int((f*g*x+e*g)^(-1+p)/ln(d*(f*x+e)^p),x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64 \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {g^{p - 1} {\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \]

[In]

integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="fricas")

[Out]

g^(p - 1)*Ei(p*log(f*x + e) + log(d))/(d*f*p)

Sympy [F]

\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int \frac {\left (g \left (e + f x\right )\right )^{p - 1}}{\log {\left (d \left (e + f x\right )^{p} \right )}}\, dx \]

[In]

integrate((f*g*x+e*g)**(-1+p)/ln(d*(f*x+e)**p),x)

[Out]

Integral((g*(e + f*x))**(p - 1)/log(d*(e + f*x)**p), x)

Maxima [F]

\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int { \frac {{\left (f g x + e g\right )}^{p - 1}}{\log \left ({\left (f x + e\right )}^{p} d\right )} \,d x } \]

[In]

integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="maxima")

[Out]

integrate((f*g*x + e*g)^(p - 1)/log((f*x + e)^p*d), x)

Giac [F]

\[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int { \frac {{\left (f g x + e g\right )}^{p - 1}}{\log \left ({\left (f x + e\right )}^{p} d\right )} \,d x } \]

[In]

integrate((f*g*x+e*g)^(-1+p)/log(d*(f*x+e)^p),x, algorithm="giac")

[Out]

integrate((f*g*x + e*g)^(p - 1)/log((f*x + e)^p*d), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e g+f g x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\int \frac {{\left (e\,g+f\,g\,x\right )}^{p-1}}{\ln \left (d\,{\left (e+f\,x\right )}^p\right )} \,d x \]

[In]

int((e*g + f*g*x)^(p - 1)/log(d*(e + f*x)^p),x)

[Out]

int((e*g + f*g*x)^(p - 1)/log(d*(e + f*x)^p), x)