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]

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

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Rubi [A]  time = 0.06, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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

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 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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.68, size = 117, normalized size = 1.22 \[ -\frac {{\left ({\left (b e n x + b d n\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )} - {\left (b n \log \left (e x + d\right ) + b \log \relax (c) + a\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \relax (c) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {b \log \relax (c) + a}{b n}\right )}}{b^{3} e n^{3} \log \left (e x + d\right ) + b^{3} e n^{2} \log \relax (c) + a b^{2} e n^{2}} \]

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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giac [B]  time = 0.20, size = 307, normalized size = 3.20 \[ \frac {b n {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac {a}{b n}\right )} \log \left (x e + d\right )}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \relax (c) + a b^{2} n^{2} e\right )} c^{\left (\frac {1}{n}\right )}} - \frac {{\left (x e + d\right )} b n}{b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \relax (c) + a b^{2} n^{2} e} + \frac {b {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac {a}{b n}\right )} \log \relax (c)}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \relax (c) + a b^{2} n^{2} e\right )} c^{\left (\frac {1}{n}\right )}} + \frac {a {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \left (x e + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{{\left (b^{3} n^{3} e \log \left (x e + d\right ) + b^{3} n^{2} e \log \relax (c) + a b^{2} n^{2} e\right )} c^{\left (\frac {1}{n}\right )}} \]

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))

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maple [C]  time = 0.66, size = 456, normalized size = 4.75 \[ -\frac {\left (e x +d \right ) c^{-\frac {1}{n}} \left (\left (e x +d \right )^{n}\right )^{-\frac {1}{n}} \Ei \left (1, -\ln \left (e x +d \right )-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 b \ln \relax (c )+2 a +2 \left (-n \ln \left (e x +d \right )+\ln \left (\left (e x +d \right )^{n}\right )\right ) b}{2 b n}\right ) {\mathrm e}^{-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 a}{2 b n}}}{b^{2} e \,n^{2}}-\frac {2 \left (e x +d \right )}{\left (-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+2 b \ln \relax (c )+2 b \ln \left (\left (e x +d \right )^{n}\right )+2 a \right ) b e n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {e x + d}{b^{2} e n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} e n \log \relax (c) + a b e n} + \int \frac {1}{b^{2} n \log \left ({\left (e x + d\right )}^{n}\right ) + b^{2} n \log \relax (c) + a b n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*(d + e*x)^n))^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}\, dx \]

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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