3.124 \(\int \sqrt {a \cosh ^2(x)} \, dx\)

Optimal. Leaf size=13 \[ \tanh (x) \sqrt {a \cosh ^2(x)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3207, 2637} \[ \tanh (x) \sqrt {a \cosh ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Cosh[x]^2],x]

[Out]

Sqrt[a*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 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 {a \cosh ^2(x)} \, dx &=\left (\sqrt {a \cosh ^2(x)} \text {sech}(x)\right ) \int \cosh (x) \, dx\\ &=\sqrt {a \cosh ^2(x)} \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \tanh (x) \sqrt {a \cosh ^2(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Cosh[x]^2],x]

[Out]

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

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fricas [B]  time = 1.06, size = 69, normalized size = 5.31 \[ \frac {{\left (2 \, \cosh \relax (x) e^{x} \sinh \relax (x) + e^{x} \sinh \relax (x)^{2} + {\left (\cosh \relax (x)^{2} - 1\right )} e^{x}\right )} \sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + a} e^{\left (-x\right )}}{2 \, {\left (\cosh \relax (x) e^{\left (2 \, x\right )} + {\left (e^{\left (2 \, x\right )} + 1\right )} \sinh \relax (x) + \cosh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/2*(2*cosh(x)*e^x*sinh(x) + e^x*sinh(x)^2 + (cosh(x)^2 - 1)*e^x)*sqrt(a*e^(4*x) + 2*a*e^(2*x) + a)*e^(-x)/(co
sh(x)*e^(2*x) + (e^(2*x) + 1)*sinh(x) + cosh(x))

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giac [A]  time = 0.13, size = 14, normalized size = 1.08 \[ -\frac {1}{2} \, \sqrt {a} {\left (e^{\left (-x\right )} - e^{x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(a)*(e^(-x) - e^x)

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maple [A]  time = 0.15, size = 15, normalized size = 1.15 \[ \frac {a \cosh \relax (x ) \sinh \relax (x )}{\sqrt {a \left (\cosh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.42, size = 17, normalized size = 1.31 \[ -\frac {1}{2} \, \sqrt {a} e^{\left (-x\right )} + \frac {1}{2} \, \sqrt {a} e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(a)*e^(-x) + 1/2*sqrt(a)*e^x

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mupad [B]  time = 0.05, size = 17, normalized size = 1.31 \[ \sqrt {a}\,\mathrm {tanh}\relax (x)\,\left (\frac {{\mathrm {e}}^{-x}}{2}+\frac {{\mathrm {e}}^x}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x)^2)^(1/2),x)

[Out]

a^(1/2)*tanh(x)*(exp(-x)/2 + exp(x)/2)

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sympy [A]  time = 0.45, size = 19, normalized size = 1.46 \[ \frac {\sqrt {a} \sqrt {\cosh ^{2}{\relax (x )}} \sinh {\relax (x )}}{\cosh {\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)**2)**(1/2),x)

[Out]

sqrt(a)*sqrt(cosh(x)**2)*sinh(x)/cosh(x)

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