3.143 \(\int \frac{1}{\sqrt{a \sinh ^2(x)}} \, dx\)

Optimal. Leaf size=17 \[ -\frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{\sqrt{a \sinh ^2(x)}} \]

[Out]

-((ArcTanh[Cosh[x]]*Sinh[x])/Sqrt[a*Sinh[x]^2])

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Rubi [A]  time = 0.0133405, antiderivative size = 17, 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, 3770} \[ -\frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{\sqrt{a \sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*Sinh[x]^2],x]

[Out]

-((ArcTanh[Cosh[x]]*Sinh[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 3770

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{a \sinh ^2(x)}} \, dx &=\frac{\sinh (x) \int \text{csch}(x) \, dx}{\sqrt{a \sinh ^2(x)}}\\ &=-\frac{\tanh ^{-1}(\cosh (x)) \sinh (x)}{\sqrt{a \sinh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0056344, size = 20, normalized size = 1.18 \[ \frac{\sinh (x) \log \left (\tanh \left (\frac{x}{2}\right )\right )}{\sqrt{a \sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*Sinh[x]^2],x]

[Out]

(Log[Tanh[x/2]]*Sinh[x])/Sqrt[a*Sinh[x]^2]

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Maple [B]  time = 0.053, size = 49, normalized size = 2.9 \begin{align*} -{\frac{\sinh \left ( x \right ) }{\cosh \left ( x \right ) }\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}}+a}{\sinh \left ( x \right ) }} \right ){\frac{1}{\sqrt{a}}}{\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(1/(a*sinh(x)^2)^(1/2),x)

[Out]

-sinh(x)*(a*cosh(x)^2)^(1/2)/a^(1/2)*ln(2*(a^(1/2)*(a*cosh(x)^2)^(1/2)+a)/sinh(x))/cosh(x)/(a*sinh(x)^2)^(1/2)

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Maxima [A]  time = 1.80917, size = 32, normalized size = 1.88 \begin{align*} \frac{\log \left (e^{\left (-x\right )} + 1\right )}{\sqrt{a}} - \frac{\log \left (e^{\left (-x\right )} - 1\right )}{\sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(-x) + 1)/sqrt(a) - log(e^(-x) - 1)/sqrt(a)

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Fricas [B]  time = 1.71426, size = 312, normalized size = 18.35 \begin{align*} \left [\frac{\sqrt{a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a} \log \left (\frac{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}{\cosh \left (x\right ) + \sinh \left (x\right ) + 1}\right )}{a e^{\left (2 \, x\right )} - a}, \frac{2 \, \sqrt{-a} \arctan \left (\frac{\sqrt{a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a} \sqrt{-a}}{a \cosh \left (x\right ) e^{\left (2 \, x\right )} - a \cosh \left (x\right ) +{\left (a e^{\left (2 \, x\right )} - a\right )} \sinh \left (x\right )}\right )}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[sqrt(a*e^(4*x) - 2*a*e^(2*x) + a)*log((cosh(x) + sinh(x) - 1)/(cosh(x) + sinh(x) + 1))/(a*e^(2*x) - a), 2*sqr
t(-a)*arctan(sqrt(a*e^(4*x) - 2*a*e^(2*x) + a)*sqrt(-a)/(a*cosh(x)*e^(2*x) - a*cosh(x) + (a*e^(2*x) - a)*sinh(
x)))/a]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a*sinh(x)**2), x)

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Giac [B]  time = 1.26213, size = 61, normalized size = 3.59 \begin{align*} -\frac{\log \left (e^{x} + 1\right )}{\sqrt{a} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} + \frac{\log \left ({\left | e^{x} - 1 \right |}\right )}{\sqrt{a} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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