3.2.25 \(\int \frac {1}{\sqrt {1+\sinh ^2(x)}} \, dx\) [125]

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

[Out]

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

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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 = {3255, 3286, 3855} \begin {gather*} \frac {\cosh (x) \text {ArcTan}(\sinh (x))}{\sqrt {\cosh ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Sinh[x]^2],x]

[Out]

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

Rule 3255

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 3286

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 3855

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*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 1.36 \begin {gather*} \frac {2 \text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right ) \cosh (x)}{\sqrt {\cosh ^2(x)}} \end {gather*}

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]

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Maple [A]
time = 0.79, size = 15, normalized size = 1.07

method	result	size
default	\(\frac {\sqrt {\frac {1}{2}+\frac {\cosh \left (2 x \right )}{2}}\, \arctan \left (\sinh \left (x \right )\right )}{\cosh \left (x \right )}\)	\(15\)
risch	\(\frac {i {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}+i\right )}{\sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}-\frac {i {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}-i\right )}{\sqrt {\left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sinh(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 5, normalized size = 0.36 \begin {gather*} 2 \, \arctan \left (e^{x}\right ) \end {gather*}

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)

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Fricas [A]
time = 0.45, size = 8, normalized size = 0.57 \begin {gather*} 2 \, \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) \end {gather*}

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))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\sinh ^{2}{\left (x \right )} + 1}}\, dx \end {gather*}

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)

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Giac [A]
time = 0.41, size = 5, normalized size = 0.36 \begin {gather*} 2 \, \arctan \left (e^{x}\right ) \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.07 \begin {gather*} \int \frac {1}{\sqrt {{\mathrm {sinh}\left (x\right )}^2+1}} \,d x \end {gather*}

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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