\(\int \sqrt {a \sinh ^2(x)} \, dx\) [142]

Optimal result

Integrand size = 10, antiderivative size = 13 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\coth (x) \sqrt {a \sinh ^2(x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3286, 2718} \[ \int \sqrt {a \sinh ^2(x)} \, dx=\coth (x) \sqrt {a \sinh ^2(x)} \]

[In]

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

[Out]

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

Rule 2718

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

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
]])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\coth (x) \sqrt {a \sinh ^2(x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {a \sinh \left (x \right ) \cosh \left (x \right )}{\sqrt {a \sinh \left (x \right )^{2}}}\)	\(15\)
risch	\(\frac {\sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}\, {\mathrm e}^{2 x}}{2 \,{\mathrm e}^{2 x}-2}+\frac {\sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}{2 \,{\mathrm e}^{2 x}-2}\)	\(58\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (11) = 22\).

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 5.46 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\frac {{\left (2 \, \cosh \left (x\right ) e^{x} \sinh \left (x\right ) + e^{x} \sinh \left (x\right )^{2} + {\left (\cosh \left (x\right )^{2} + 1\right )} e^{x}\right )} \sqrt {a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a} e^{\left (-x\right )}}{2 \, {\left (\cosh \left (x\right ) e^{\left (2 \, x\right )} + {\left (e^{\left (2 \, x\right )} - 1\right )} \sinh \left (x\right ) - \cosh \left (x\right )\right )}} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\frac {\sqrt {a \sinh ^{2}{\left (x \right )}} \cosh {\left (x \right )}}{\sinh {\left (x \right )}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \sqrt {a \sinh ^2(x)} \, dx=-\frac {1}{2} \, \sqrt {a} e^{\left (-x\right )} - \frac {1}{2} \, \sqrt {a} e^{x} \]

[In]

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

[Out]

-1/2*sqrt(a)*e^(-x) - 1/2*sqrt(a)*e^x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (11) = 22\).

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.62 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\frac {1}{2} \, {\left (e^{\left (-x\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) + e^{x} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )\right )} \sqrt {a} \]

[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)

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.62 \[ \int \sqrt {a \sinh ^2(x)} \, dx=\sqrt {a}\,\mathrm {coth}\left (x\right )\,\sqrt {{\left (\frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^x}{2}\right )}^2} \]

[In]

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

[Out]

a^(1/2)*coth(x)*((exp(-x)/2 - exp(x)/2)^2)^(1/2)