3.192 \(\int \frac {\text {sech}^2(a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=18 \[ \frac {\tanh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3767, 8} \[ \frac {\tanh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*Log[c*x^n]]^2/x,x]

[Out]

Tanh[a + b*Log[c*x^n]]/(b*n)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 18, normalized size = 1.00 \[ \frac {\tanh \left (a+b \log \left (c x^n\right )\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*Log[c*x^n]]^2/x,x]

[Out]

Tanh[a + b*Log[c*x^n]]/(b*n)

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fricas [B]  time = 0.42, size = 70, normalized size = 3.89 \[ -\frac {2}{b n \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, b n \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + b n \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 28, normalized size = 1.56 \[ -\frac {2}{{\left (c^{2 \, b} x^{2 \, b n} e^{\left (2 \, a\right )} + 1\right )} b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 19, normalized size = 1.06 \[ \frac {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 28, normalized size = 1.56 \[ -\frac {2}{b c^{2 \, b} n e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )} + b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 24, normalized size = 1.33 \[ -\frac {2}{b\,n+b\,n\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*cosh(a + b*log(c*x^n))^2),x)

[Out]

-2/(b*n + b*n*exp(2*a)*(c*x^n)^(2*b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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