3.91 \(\int \frac{1}{a+b \log (c (d+e x)^n)} \, dx\)

Optimal. Leaf size=63 \[ \frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b*e*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1))

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Rubi [A]  time = 0.0463784, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2389, 2300, 2178} \[ \frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^(-1),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b*e*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rubi steps

\begin{align*} \int \frac{1}{a+b \log \left (c (d+e x)^n\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{e}\\ &=\frac{\left ((d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int \frac{e^{\frac{x}{n}}}{a+b x} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{e n}\\ &=\frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e n}\\ \end{align*}

Mathematica [F]  time = 0.0131231, size = 0, normalized size = 0. \[ \int \frac{1}{a+b \log \left (c (d+e x)^n\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^(-1),x]

[Out]

Integrate[(a + b*Log[c*(d + e*x)^n])^(-1), x]

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Maple [C]  time = 0.082, size = 312, normalized size = 5. \begin{align*} -{\frac{1}{enb}{\it Ei} \left ( 1,-\ln \left ( ex+d \right ) -{\frac{-ib\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ){\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) +ib\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}+ib\pi \,{\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}-ib\pi \, \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{3}+2\,b\ln \left ( c \right ) +2\,b \left ( \ln \left ( \left ( ex+d \right ) ^{n} \right ) -n\ln \left ( ex+d \right ) \right ) +2\,a}{2\,bn}} \right ){{\rm e}^{{\frac{ib\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ){\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) -ib\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}-ib\pi \,{\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}+ib\pi \, \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{3}+2\,bn\ln \left ( ex+d \right ) -2\,b\ln \left ( c \right ) -2\,b\ln \left ( \left ( ex+d \right ) ^{n} \right ) -2\,a}{2\,bn}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*ln(c*(e*x+d)^n)),x)

[Out]

-1/b/e/n*Ei(1,-ln(e*x+d)-1/2*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*
c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*b*(ln((
e*x+d)^n)-n*ln(e*x+d))+2*a)/b/n)*exp(1/2*(I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*b*Pi*csgn(I
*c)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*n*ln
(e*x+d)-2*b*ln(c)-2*b*ln((e*x+d)^n)-2*a)/b/n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")

[Out]

integrate(1/(b*log((e*x + d)^n*c) + a), x)

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Fricas [A]  time = 1.92413, size = 113, normalized size = 1.79 \begin{align*} \frac{e^{\left (-\frac{b \log \left (c\right ) + a}{b n}\right )} \logintegral \left ({\left (e x + d\right )} e^{\left (\frac{b \log \left (c\right ) + a}{b n}\right )}\right )}{b e n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

e^(-(b*log(c) + a)/(b*n))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n)))/(b*e*n)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \log{\left (c \left (d + e x\right )^{n} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Integral(1/(a + b*log(c*(d + e*x)**n)), x)

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Giac [A]  time = 1.27308, size = 66, normalized size = 1.05 \begin{align*} \frac{{\rm Ei}\left (\frac{\log \left (c\right )}{n} + \frac{a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac{a}{b n} - 1\right )}}{b c^{\left (\frac{1}{n}\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")

[Out]

Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n) - 1)/(b*c^(1/n)*n)