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

Optimal. Leaf size=70 \[ \frac {x e^{-\frac {a}{b n}} \left (c x^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b^2 n^2}-\frac {x}{b n \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2297, 2300, 2178} \[ \frac {x e^{-\frac {a}{b n}} \left (c x^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )}{b^2 n^2}-\frac {x}{b n \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])^(-2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 66, normalized size = 0.94 \[ \frac {x \left (e^{-\frac {a}{b n}} \left (c x^n\right )^{-1/n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{b n}\right )-\frac {b n}{a+b \log \left (c x^n\right )}\right )}{b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])^(-2),x]

[Out]

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

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fricas [A]  time = 0.40, size = 95, normalized size = 1.36 \[ -\frac {{\left (b n x e^{\left (\frac {b \log \relax (c) + a}{b n}\right )} - {\left (b n \log \relax (x) + b \log \relax (c) + a\right )} \operatorname {log\_integral}\left (x e^{\left (\frac {b \log \relax (c) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {b \log \relax (c) + a}{b n}\right )}}{b^{3} n^{3} \log \relax (x) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.31, size = 238, normalized size = 3.40 \[ \frac {b n {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \relax (x)\right ) e^{\left (-\frac {a}{b n}\right )} \log \relax (x)}{{\left (b^{3} n^{3} \log \relax (x) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )} c^{\left (\frac {1}{n}\right )}} - \frac {b n x}{b^{3} n^{3} \log \relax (x) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}} + \frac {b {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \relax (x)\right ) e^{\left (-\frac {a}{b n}\right )} \log \relax (c)}{{\left (b^{3} n^{3} \log \relax (x) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )} c^{\left (\frac {1}{n}\right )}} + \frac {a {\rm Ei}\left (\frac {\log \relax (c)}{n} + \frac {a}{b n} + \log \relax (x)\right ) e^{\left (-\frac {a}{b n}\right )}}{{\left (b^{3} n^{3} \log \relax (x) + b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )} c^{\left (\frac {1}{n}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.49, size = 350, normalized size = 5.00 \[ -\frac {x \,c^{-\frac {1}{n}} \left (x^{n}\right )^{-\frac {1}{n}} \Ei \left (1, -\ln \relax (x )-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \relax (c )+2 a +2 \left (-n \ln \relax (x )+\ln \left (x^{n}\right )\right ) b}{2 b n}\right ) {\mathrm e}^{-\frac {-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 a}{2 b n}}}{b^{2} n^{2}}-\frac {2 x}{\left (-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \relax (c )+2 b \ln \left (x^{n}\right )+2 a \right ) b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/b/n*x/(2*a+2*b*ln(c)+2*b*ln(x^n)+I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x
^n)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2)-1/b^2/n^2*x*(x^n)^(-1/n)*c^(-1/n)*exp(-1/2*(-I*Pi
*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2+I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*P
i*b*csgn(I*c*x^n)^3+2*a)/b/n)*Ei(1,-ln(x)-1/2*(-I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*Pi*b*csgn(I*c)*cs
gn(I*c*x^n)^2+I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*c*x^n)^3+2*b*ln(c)+2*a+2*(-n*ln(x)+ln(x^n))*b)/
b/n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*x^n))^2,x)

[Out]

int(1/(a + b*log(c*x^n))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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