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

Optimal. Leaf size=11 \[ \sqrt {\cosh ^2(x)} \tanh (x) \]

[Out]

(cosh(x)^2)^(1/2)*tanh(x)

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Rubi [A]  time = 0.02, antiderivative size = 11, 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 = {3176, 3207, 2637} \[ \sqrt {\cosh ^2(x)} \tanh (x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Sinh[x]^2],x]

[Out]

Sqrt[Cosh[x]^2]*Tanh[x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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
]])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \[ \sqrt {\cosh ^2(x)} \tanh (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Sinh[x]^2],x]

[Out]

Sqrt[Cosh[x]^2]*Tanh[x]

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fricas [A]  time = 3.47, size = 2, normalized size = 0.18 \[ \sinh \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sinh(x)

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giac [A]  time = 0.11, size = 11, normalized size = 1.00 \[ -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^(-x) + 1/2*e^x

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maple [A]  time = 0.08, size = 14, normalized size = 1.27 \[ \frac {\sqrt {\frac {\cosh \left (2 x \right )}{2}+\frac {1}{2}}\, \sinh \relax (x )}{\cosh \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 11, normalized size = 1.00 \[ -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^(-x) + 1/2*e^x

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mupad [B]  time = 0.07, size = 2, normalized size = 0.18 \[ \mathrm {sinh}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(x)^2 + 1)^(1/2),x)

[Out]

sinh(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sinh(x)**2 + 1), x)

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