3.206 \(\int \log (a \cosh ^n(x)) \, dx\)

Optimal. Leaf size=44 \[ x \log \left (a \cosh ^n(x)\right )-\frac {1}{2} n \text {Li}_2\left (-e^{2 x}\right )+\frac {n x^2}{2}-n x \log \left (e^{2 x}+1\right ) \]

[Out]

1/2*n*x^2-n*x*ln(1+exp(2*x))+x*ln(a*cosh(x)^n)-1/2*n*polylog(2,-exp(2*x))

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Rubi [A]  time = 0.06, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2548, 12, 3718, 2190, 2279, 2391} \[ -\frac {1}{2} n \text {PolyLog}\left (2,-e^{2 x}\right )+x \log \left (a \cosh ^n(x)\right )+\frac {n x^2}{2}-n x \log \left (e^{2 x}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[a*Cosh[x]^n],x]

[Out]

(n*x^2)/2 - n*x*Log[1 + E^(2*x)] + x*Log[a*Cosh[x]^n] - (n*PolyLog[2, -E^(2*x)])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \log \left (a \cosh ^n(x)\right ) \, dx &=x \log \left (a \cosh ^n(x)\right )-\int n x \tanh (x) \, dx\\ &=x \log \left (a \cosh ^n(x)\right )-n \int x \tanh (x) \, dx\\ &=\frac {n x^2}{2}+x \log \left (a \cosh ^n(x)\right )-(2 n) \int \frac {e^{2 x} x}{1+e^{2 x}} \, dx\\ &=\frac {n x^2}{2}-n x \log \left (1+e^{2 x}\right )+x \log \left (a \cosh ^n(x)\right )+n \int \log \left (1+e^{2 x}\right ) \, dx\\ &=\frac {n x^2}{2}-n x \log \left (1+e^{2 x}\right )+x \log \left (a \cosh ^n(x)\right )+\frac {1}{2} n \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 x}\right )\\ &=\frac {n x^2}{2}-n x \log \left (1+e^{2 x}\right )+x \log \left (a \cosh ^n(x)\right )-\frac {1}{2} n \text {Li}_2\left (-e^{2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.98 \[ \frac {1}{2} \left (n \text {Li}_2\left (-e^{-2 x}\right )-x \left (-2 \log \left (a \cosh ^n(x)\right )+n x+2 n \log \left (e^{-2 x}+1\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a*Cosh[x]^n],x]

[Out]

(-(x*(n*x + 2*n*Log[1 + E^(-2*x)] - 2*Log[a*Cosh[x]^n])) + n*PolyLog[2, -E^(-2*x)])/2

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fricas [C]  time = 0.51, size = 73, normalized size = 1.66 \[ \frac {1}{2} \, n x^{2} - n x \log \left (i \, \cosh \relax (x) + i \, \sinh \relax (x) + 1\right ) - n x \log \left (-i \, \cosh \relax (x) - i \, \sinh \relax (x) + 1\right ) + n x \log \left (\cosh \relax (x)\right ) - n {\rm Li}_2\left (i \, \cosh \relax (x) + i \, \sinh \relax (x)\right ) - n {\rm Li}_2\left (-i \, \cosh \relax (x) - i \, \sinh \relax (x)\right ) + x \log \relax (a) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)^n),x, algorithm="fricas")

[Out]

1/2*n*x^2 - n*x*log(I*cosh(x) + I*sinh(x) + 1) - n*x*log(-I*cosh(x) - I*sinh(x) + 1) + n*x*log(cosh(x)) - n*di
log(I*cosh(x) + I*sinh(x)) - n*dilog(-I*cosh(x) - I*sinh(x)) + x*log(a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \cosh \relax (x)^{n}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)^n),x, algorithm="giac")

[Out]

integrate(log(a*cosh(x)^n), x)

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maple [F]  time = 1.98, size = 0, normalized size = 0.00 \[ \int \ln \left (a \left (\cosh ^{n}\relax (x )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*cosh(x)^n),x)

[Out]

int(ln(a*cosh(x)^n),x)

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maxima [A]  time = 1.15, size = 36, normalized size = 0.82 \[ \frac {1}{2} \, {\left (x^{2} - 2 \, x \log \left (e^{\left (2 \, x\right )} + 1\right ) - {\rm Li}_2\left (-e^{\left (2 \, x\right )}\right )\right )} n + x \log \left (a \cosh \relax (x)^{n}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*cosh(x)^n),x, algorithm="maxima")

[Out]

1/2*(x^2 - 2*x*log(e^(2*x) + 1) - dilog(-e^(2*x)))*n + x*log(a*cosh(x)^n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \ln \left (a\,{\mathrm {cosh}\relax (x)}^n\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a*cosh(x)^n),x)

[Out]

int(log(a*cosh(x)^n), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \cosh ^{n}{\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*cosh(x)**n),x)

[Out]

Integral(log(a*cosh(x)**n), x)

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