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

Optimal. Leaf size=20 \[ \frac{\text{li}\left (d (e+f x)^p\right )}{d f p} \]

[Out]

LogIntegral[d*(e + f*x)^p]/(d*f*p)

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Rubi [A]  time = 0.0467459, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2390, 2307, 2298} \[ \frac{\text{li}\left (d (e+f x)^p\right )}{d f p} \]

Antiderivative was successfully verified.

[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 2390

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]

Rule 2307

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 2298

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

Rubi steps

\begin{align*} \int \frac{(e+f x)^{-1+p}}{\log \left (d (e+f x)^p\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{-1+p}}{\log \left (d x^p\right )} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{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]  time = 0.0268894, size = 20, normalized size = 1. \[ \frac{\text{li}\left (d (e+f x)^p\right )}{d f p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

LogIntegral[d*(e + f*x)^p]/(d*f*p)

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Maple [A]  time = 0.11, size = 26, normalized size = 1.3 \begin{align*} -{\frac{{\it Ei} \left ( 1,-\ln \left ( d \left ( fx+e \right ) ^{p} \right ) \right ) }{pfd}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{p - 1}}{\log \left ({\left (f x + e\right )}^{p} d\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.96998, size = 50, normalized size = 2.5 \begin{align*} \frac{{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 45.0141, size = 42, normalized size = 2.1 \begin{align*} \begin{cases} - \frac{\begin{cases} - \frac{\log{\left (e + f x \right )}}{\log{\left (d \right )}} & \text{for}\: p = 0 \\- \frac{\operatorname{li}{\left (d \left (e + f x\right )^{p} \right )}}{d p} & \text{otherwise} \end{cases}}{f} & \text{for}\: f \neq 0 \\\frac{e^{p - 1} x}{\log{\left (d e^{p} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-Piecewise((-log(e + f*x)/log(d), Eq(p, 0)), (-li(d*(e + f*x)**p)/(d*p), True))/f, Ne(f, 0)), (e**(
p - 1)*x/log(d*e**p), True))

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Giac [A]  time = 1.29802, size = 31, normalized size = 1.55 \begin{align*} \frac{{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)