3.142 \(\int \sqrt{a \sinh ^2(x)} \, dx\)

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

[Out]

Coth[x]*Sqrt[a*Sinh[x]^2]

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Rubi [A]  time = 0.0128127, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3207, 2638} \[ \coth (x) \sqrt{a \sinh ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Coth[x]*Sqrt[a*Sinh[x]^2]

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 2638

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

Rubi steps

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

Mathematica [A]  time = 0.004513, size = 13, normalized size = 1. \[ \coth (x) \sqrt{a \sinh ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Coth[x]*Sqrt[a*Sinh[x]^2]

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Maple [A]  time = 0.03, size = 15, normalized size = 1.2 \begin{align*}{a\sinh \left ( x \right ) \cosh \left ( x \right ){\frac{1}{\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.82899, size = 23, normalized size = 1.77 \begin{align*} -\frac{1}{2} \, \sqrt{a} e^{\left (-x\right )} - \frac{1}{2} \, \sqrt{a} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a)*e^(-x) - 1/2*sqrt(a)*e^x

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Fricas [B]  time = 1.74253, size = 216, normalized size = 16.62 \begin{align*} \frac{{\left (2 \, \cosh \left (x\right ) e^{x} \sinh \left (x\right ) + e^{x} \sinh \left (x\right )^{2} +{\left (\cosh \left (x\right )^{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 \left (x\right ) e^{\left (2 \, x\right )} +{\left (e^{\left (2 \, x\right )} - 1\right )} \sinh \left (x\right ) - \cosh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(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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Sympy [A]  time = 0.609714, size = 19, normalized size = 1.46 \begin{align*} \frac{\sqrt{a} \sqrt{\sinh ^{2}{\left (x \right )}} \cosh{\left (x \right )}}{\sinh{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19944, size = 46, normalized size = 3.54 \begin{align*} \frac{1}{2} \,{\left (e^{\left (-x\right )} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + e^{x} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^(-x)*sgn(e^(3*x) - e^x) + e^x*sgn(e^(3*x) - e^x))*sqrt(a)