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

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {2437, 2344, 2335} \begin {gather*} \frac {\text {li}\left (d (e+f x)^p\right )}{d f p} \end {gather*}

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

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Mathematica [A]
time = 0.03, size = 21, normalized size = 1.05 \begin {gather*} \frac {\text {Ei}\left (\log \left (d (e+f x)^p\right )\right )}{d f p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ExpIntegralEi[Log[d*(e + f*x)^p]]/(d*f*p)

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Maple [A]
time = 1.59, size = 26, normalized size = 1.30

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.36, size = 23, normalized size = 1.15 \begin {gather*} \frac {{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (14) = 28\).
time = 3.42, size = 42, normalized size = 2.10 \begin {gather*} \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 {gather*}

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 = 4.51, size = 23, normalized size = 1.15 \begin {gather*} \frac {{\rm Ei}\left (p \log \left (f x + e\right ) + \log \left (d\right )\right )}{d f p} \end {gather*}

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)

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Mupad [B]
time = 0.20, size = 20, normalized size = 1.00 \begin {gather*} \frac {\mathrm {logint}\left (d\,{\left (e+f\,x\right )}^p\right )}{d\,f\,p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

logint(d*(e + f*x)^p)/(d*f*p)

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