\(\int \frac {(e+f x)^{-1+p}}{\log (d (e+f x)^p)} \, dx\) [33]

Optimal result

Integrand size = 22, antiderivative size = 20 \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {\operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2437, 2344, 2335} \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {\operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {\operatorname {LogIntegral}\left (d (e+f x)^p\right )}{d f p} \]

[In]

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

[Out]

LogIntegral[d*(e + f*x)^p]/(d*f*p)

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {\operatorname {Ei}_{1}\left (-\ln \left (d \left (f x +e \right )^{p}\right )\right )}{p f d}\)	\(26\)
risch	\(-\frac {{\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right ) \left (-\operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right )+\operatorname {csgn}\left (i d \right )\right ) \left (-\operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right )+\operatorname {csgn}\left (i \left (f x +e \right )^{p}\right )\right )}{2}} \operatorname {Ei}_{1}\left (-\ln \left (d \right )-\ln \left (\left (f x +e \right )^{p}\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (f x +e \right )^{p}\right ) \operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right )^{2}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (f x +e \right )^{p}\right ) \operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right ) \operatorname {csgn}\left (i d \right )}{2}+\frac {i \pi \operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right )^{3}}{2}-\frac {i \pi \operatorname {csgn}\left (i d \left (f x +e \right )^{p}\right )^{2} \operatorname {csgn}\left (i d \right )}{2}\right )}{p f d}\)	\(194\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \]

[In]

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

[Out]

Ei(p*log(f*x + e) + log(d))/(d*f*p)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \]

[In]

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

[Out]

Ei(p*log(f*x + e) + log(d))/(d*f*p)

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx=\frac {\mathrm {logint}\left (d\,{\left (e+f\,x\right )}^p\right )}{d\,f\,p} \]

[In]

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

[Out]

logint(d*(e + f*x)^p)/(d*f*p)