3.125 \(\int \frac{1}{\sqrt{1+\sinh ^2(x)}} \, dx\)

Optimal. Leaf size=14 \[ \frac{\cosh (x) \tan ^{-1}(\sinh (x))}{\sqrt{\cosh ^2(x)}} \]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[Cosh[x]^2]

________________________________________________________________________________________

Rubi [A]  time = 0.0240236, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3176, 3207, 3770} \[ \frac{\cosh (x) \tan ^{-1}(\sinh (x))}{\sqrt{\cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Sinh[x]^2],x]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[Cosh[x]^2]

Rule 3176

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

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[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 3770

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{1+\sinh ^2(x)}} \, dx &=\int \frac{1}{\sqrt{\cosh ^2(x)}} \, dx\\ &=\frac{\cosh (x) \int \text{sech}(x) \, dx}{\sqrt{\cosh ^2(x)}}\\ &=\frac{\tan ^{-1}(\sinh (x)) \cosh (x)}{\sqrt{\cosh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0108901, size = 19, normalized size = 1.36 \[ \frac{2 \cosh (x) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )}{\sqrt{\cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Sinh[x]^2],x]

[Out]

(2*ArcTan[Tanh[x/2]]*Cosh[x])/Sqrt[Cosh[x]^2]

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 15, normalized size = 1.1 \begin{align*}{\frac{\arctan \left ( \sinh \left ( x \right ) \right ) }{\cosh \left ( x \right ) }\sqrt{ \left ( \cosh \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(x)^2)^(1/2)*arctan(sinh(x))/cosh(x)

________________________________________________________________________________________

Maxima [A]  time = 1.65678, size = 7, normalized size = 0.5 \begin{align*} 2 \, \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(e^x)

________________________________________________________________________________________

Fricas [A]  time = 2.03552, size = 39, normalized size = 2.79 \begin{align*} 2 \, \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(cosh(x) + sinh(x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\sinh ^{2}{\left (x \right )} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(sinh(x)**2 + 1), x)

________________________________________________________________________________________

Giac [A]  time = 1.14593, size = 7, normalized size = 0.5 \begin{align*} 2 \, \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(e^x)