3.22 \(\int \frac{1}{(a+b \log (c (d+e x)^n))^2} \, dx\)

Optimal. Leaf size=96 \[ \frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e n^2}-\frac{d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b^2*e*E^(a/(b*n))*n^2*(c*(d + e*x)^n)^n^(-1)) - (
d + e*x)/(b*e*n*(a + b*Log[c*(d + e*x)^n]))

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Rubi [A]  time = 0.0613845, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2389, 2297, 2300, 2178} \[ \frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b^2 e n^2}-\frac{d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^(-2),x]

[Out]

((d + e*x)*ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)])/(b^2*e*E^(a/(b*n))*n^2*(c*(d + e*x)^n)^n^(-1)) - (
d + e*x)/(b*e*n*(a + b*Log[c*(d + e*x)^n]))

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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}{\left (a+b \log \left (c (d+e x)^n\right )\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \log \left (c x^n\right )\right )^2} \, dx,x,d+e x\right )}{e}\\ &=-\frac{d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{b e n}\\ &=-\frac{d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}+\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 )}{b e n^2}\\ &=\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^2 e n^2}-\frac{d+e x}{b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.0658548, size = 123, normalized size = 1.28 \[ -\frac{e^{-\frac{a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \left (b n e^{\frac{a}{b n}} \left (c (d+e x)^n\right )^{\frac{1}{n}}-\left (a+b \log \left (c (d+e x)^n\right )\right ) \text{Ei}\left (\frac{a+b \log \left (c (d+e x)^n\right )}{b n}\right )\right )}{b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^(-2),x]

[Out]

-(((d + e*x)*(b*E^(a/(b*n))*n*(c*(d + e*x)^n)^n^(-1) - ExpIntegralEi[(a + b*Log[c*(d + e*x)^n])/(b*n)]*(a + b*
Log[c*(d + e*x)^n])))/(b^2*e*E^(a/(b*n))*n^2*(c*(d + e*x)^n)^n^(-1)*(a + b*Log[c*(d + e*x)^n])))

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Maple [C]  time = 0.541, size = 457, normalized size = 4.8 \begin{align*} -2\,{\frac{ex+d}{ \left ( 2\,a+2\,b\ln \left ( c \right ) +2\,b\ln \left ( \left ( ex+d \right ) ^{n} \right ) -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} \right ) bne}}-{\frac{1}{{b}^{2}e{n}^{2}}{\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))^2,x)

[Out]

-2/(2*a+2*b*ln(c)+2*b*ln((e*x+d)^n)-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)*cs
gn(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)/b/n/e*(e*x+d)
-1/b^2/n^2/e*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)*csg
n(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*cs
gn(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*} -\frac{e x + d}{b^{2} e n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} e n \log \left (c\right ) + a b e n} + \int \frac{1}{b^{2} n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} n \log \left (c\right ) + a b n}\,{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))^2,x, algorithm="maxima")

[Out]

-(e*x + d)/(b^2*e*n*log((e*x + d)^n) + b^2*e*n*log(c) + a*b*e*n) + integrate(1/(b^2*n*log((e*x + d)^n) + b^2*n
*log(c) + a*b*n), x)

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Fricas [A]  time = 1.96283, size = 292, normalized size = 3.04 \begin{align*} -\frac{{\left ({\left (b e n x + b d n\right )} e^{\left (\frac{b \log \left (c\right ) + a}{b n}\right )} -{\left (b n \log \left (e x + d\right ) + b \log \left (c\right ) + a\right )} \logintegral \left ({\left (e x + d\right )} e^{\left (\frac{b \log \left (c\right ) + a}{b n}\right )}\right )\right )} e^{\left (-\frac{b \log \left (c\right ) + a}{b n}\right )}}{b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \left (c\right ) + a b^{2} e n^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^2,x, algorithm="fricas")

[Out]

-((b*e*n*x + b*d*n)*e^((b*log(c) + a)/(b*n)) - (b*n*log(e*x + d) + b*log(c) + a)*log_integral((e*x + d)*e^((b*
log(c) + a)/(b*n))))*e^(-(b*log(c) + a)/(b*n))/(b^3*e*n^3*log(e*x + d) + b^3*e*n^2*log(c) + a*b^2*e*n^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \log{\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(e*x+d)**n))**2,x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**(-2), x)

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Giac [B]  time = 1.24836, size = 414, normalized size = 4.31 \begin{align*} \frac{b n{\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}\right )} \log \left (x e + d\right )}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \left (c\right ) + a b^{2} n^{2} e\right )} c^{\left (\frac{1}{n}\right )}} - \frac{{\left (x e + d\right )} b n}{b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \left (c\right ) + a b^{2} n^{2} e} + \frac{b{\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}\right )} \log \left (c\right )}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \left (c\right ) + a b^{2} n^{2} e\right )} c^{\left (\frac{1}{n}\right )}} + \frac{a{\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}\right )}}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \left (c\right ) + a b^{2} n^{2} e\right )} c^{\left (\frac{1}{n}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^2,x, algorithm="giac")

[Out]

b*n*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(x*e + d)/((b^3*n^3*e*log(x*e + d) + b^3*n^2*e*log(c
) + a*b^2*n^2*e)*c^(1/n)) - (x*e + d)*b*n/(b^3*n^3*e*log(x*e + d) + b^3*n^2*e*log(c) + a*b^2*n^2*e) + b*Ei(log
(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))*log(c)/((b^3*n^3*e*log(x*e + d) + b^3*n^2*e*log(c) + a*b^2*n^2*e)
*c^(1/n)) + a*Ei(log(c)/n + a/(b*n) + log(x*e + d))*e^(-a/(b*n))/((b^3*n^3*e*log(x*e + d) + b^3*n^2*e*log(c) +
a*b^2*n^2*e)*c^(1/n))