3.17 \(\int \sqrt {1+\text {csch}^2(x)} \, dx\)

Optimal. Leaf size=14 \[ \tanh (x) \sqrt {\coth ^2(x)} \log (\sinh (x)) \]

[Out]

ln(sinh(x))*(coth(x)^2)^(1/2)*tanh(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 = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4121, 3658, 3475} \[ \tanh (x) \sqrt {\coth ^2(x)} \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Csch[x]^2],x]

[Out]

Sqrt[Coth[x]^2]*Log[Sinh[x]]*Tanh[x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 4121

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(b*tan[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rubi steps

\begin {align*} \int \sqrt {1+\text {csch}^2(x)} \, dx &=\int \sqrt {\coth ^2(x)} \, dx\\ &=\left (\sqrt {\coth ^2(x)} \tanh (x)\right ) \int \coth (x) \, dx\\ &=\sqrt {\coth ^2(x)} \log (\sinh (x)) \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \tanh (x) \sqrt {\coth ^2(x)} \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Csch[x]^2],x]

[Out]

Sqrt[Coth[x]^2]*Log[Sinh[x]]*Tanh[x]

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fricas [A]  time = 0.45, size = 18, normalized size = 1.29 \[ -x + \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+csch(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-x + log(2*sinh(x)/(cosh(x) - sinh(x)))

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giac [B]  time = 0.12, size = 27, normalized size = 1.93 \[ -x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+csch(x)^2)^(1/2),x, algorithm="giac")

[Out]

-x*sgn(e^(4*x) - 1) + log(abs(e^(2*x) - 1))*sgn(e^(4*x) - 1)

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maple [B]  time = 0.32, size = 79, normalized size = 5.64 \[ -\frac {\left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 x}\right )^{2}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, x}{1+{\mathrm e}^{2 x}}+\frac {\left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 x}\right )^{2}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+csch(x)^2)^(1/2),x)

[Out]

-1/(1+exp(2*x))*(exp(2*x)-1)*((1+exp(2*x))^2/(exp(2*x)-1)^2)^(1/2)*x+1/(1+exp(2*x))*(exp(2*x)-1)*((1+exp(2*x))
^2/(exp(2*x)-1)^2)^(1/2)*ln(exp(2*x)-1)

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maxima [A]  time = 0.51, size = 22, normalized size = 1.57 \[ -x - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+csch(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-x - log(e^(-x) + 1) - log(e^(-x) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.07 \[ \int \sqrt {\frac {1}{{\mathrm {sinh}\relax (x)}^2}+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sinh(x)^2 + 1)^(1/2),x)

[Out]

int((1/sinh(x)^2 + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\operatorname {csch}^{2}{\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+csch(x)**2)**(1/2),x)

[Out]

Integral(sqrt(csch(x)**2 + 1), x)

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