3.157 \(\int \frac {\tanh ^2(a+2 \log (x))}{x} \, dx\)

Optimal. Leaf size=14 \[ \log (x)-\frac {1}{2} \tanh (a+2 \log (x)) \]

[Out]

ln(x)-1/2*tanh(a+2*ln(x))

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3473, 8} \[ \log (x)-\frac {1}{2} \tanh (a+2 \log (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[a + 2*Log[x]]^2/x,x]

[Out]

Log[x] - Tanh[a + 2*Log[x]]/2

Rule 8

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 24, normalized size = 1.71 \[ \frac {1}{2} \tanh ^{-1}(\tanh (a+2 \log (x)))-\frac {1}{2} \tanh (a+2 \log (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]]^2/x,x]

[Out]

ArcTanh[Tanh[a + 2*Log[x]]]/2 - Tanh[a + 2*Log[x]]/2

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fricas [B]  time = 0.47, size = 28, normalized size = 2.00 \[ \frac {{\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \log \relax (x) + 1}{x^{4} e^{\left (2 \, a\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x,x, algorithm="fricas")

[Out]

((x^4*e^(2*a) + 1)*log(x) + 1)/(x^4*e^(2*a) + 1)

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giac [A]  time = 0.14, size = 19, normalized size = 1.36 \[ \frac {1}{x^{4} e^{\left (2 \, a\right )} + 1} + \frac {1}{4} \, \log \left (x^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x,x, algorithm="giac")

[Out]

1/(x^4*e^(2*a) + 1) + 1/4*log(x^4)

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maple [B]  time = 0.02, size = 35, normalized size = 2.50 \[ -\frac {\tanh \left (a +2 \ln \relax (x )\right )}{2}-\frac {\ln \left (\tanh \left (a +2 \ln \relax (x )\right )-1\right )}{4}+\frac {\ln \left (\tanh \left (a +2 \ln \relax (x )\right )+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+2*ln(x))^2/x,x)

[Out]

-1/2*tanh(a+2*ln(x))-1/4*ln(tanh(a+2*ln(x))-1)+1/4*ln(tanh(a+2*ln(x))+1)

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maxima [A]  time = 0.33, size = 21, normalized size = 1.50 \[ \frac {1}{2} \, a - \frac {1}{e^{\left (-2 \, a - 4 \, \log \relax (x)\right )} + 1} + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x,x, algorithm="maxima")

[Out]

1/2*a - 1/(e^(-2*a - 4*log(x)) + 1) + log(x)

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mupad [B]  time = 1.06, size = 28, normalized size = 2.00 \[ \ln \relax (x)-\frac {x^4\,{\mathrm {e}}^{2\,a}-1}{2\,\left ({\mathrm {e}}^{2\,a}\,x^4+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x))^2/x,x)

[Out]

log(x) - (x^4*exp(2*a) - 1)/(2*(x^4*exp(2*a) + 1))

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sympy [A]  time = 0.36, size = 12, normalized size = 0.86 \[ \log {\relax (x )} - \frac {\tanh {\left (a + 2 \log {\relax (x )} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*ln(x))**2/x,x)

[Out]

log(x) - tanh(a + 2*log(x))/2

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