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

Optimal. Leaf size=20 \[ -\frac {1}{b n \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2302, 30} \[ -\frac {1}{b n \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*Log[c*x^n])^2),x]

[Out]

-(1/(b*n*(a + b*Log[c*x^n])))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ -\frac {1}{b n \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*Log[c*x^n])^2),x]

[Out]

-(1/(b*n*(a + b*Log[c*x^n])))

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fricas [A]  time = 0.40, size = 25, normalized size = 1.25 \[ -\frac {1}{b^{2} n^{2} \log \relax (x) + b^{2} n \log \relax (c) + a b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 21, normalized size = 1.05 \[ -\frac {1}{{\left (b n \log \relax (x) + b \log \relax (c) + a\right )} b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((b*n*log(x) + b*log(c) + a)*b*n)

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maple [A]  time = 0.02, size = 21, normalized size = 1.05 \[ -\frac {1}{\left (b \ln \left (c \,x^{n}\right )+a \right ) b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 20, normalized size = 1.00 \[ -\frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((b*log(c*x^n) + a)*b*n)

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mupad [B]  time = 3.27, size = 20, normalized size = 1.00 \[ -\frac {1}{n\,\ln \left (c\,x^n\right )\,b^2+a\,n\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*log(c*x^n))^2),x)

[Out]

-1/(b^2*n*log(c*x^n) + a*b*n)

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sympy [A]  time = 22.74, size = 70, normalized size = 3.50 \[ \begin {cases} \frac {\tilde {\infty } \log {\relax (x )}}{\log {\relax (c )}^{2}} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\\tilde {\infty } n \log {\relax (x )} & \text {for}\: a = - b \left (n \log {\relax (x )} + \log {\relax (c )}\right ) \\\frac {\log {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \\\frac {\log {\relax (x )}}{\left (a + b \log {\relax (c )}\right )^{2}} & \text {for}\: n = 0 \\- \frac {1}{a b n + b^{2} n^{2} \log {\relax (x )} + b^{2} n \log {\relax (c )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*log(x)/log(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(n, 0)), (zoo*n*log(x), Eq(a, -b*(n*log(x) + log(c)))
), (log(x)/a**2, Eq(b, 0)), (log(x)/(a + b*log(c))**2, Eq(n, 0)), (-1/(a*b*n + b**2*n**2*log(x) + b**2*n*log(c
)), True))

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