3.144 \(\int \frac{1}{(a \sinh ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=42 \[ \frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{2 a \sqrt{a \sinh ^2(x)}}-\frac{\coth (x)}{2 a \sqrt{a \sinh ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.0262079, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3204, 3207, 3770} \[ \frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{2 a \sqrt{a \sinh ^2(x)}}-\frac{\coth (x)}{2 a \sqrt{a \sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^2)^(-3/2),x]

[Out]

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

Rule 3204

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^(p + 1))/(b*f*(
2*p + 1)), x] + Dist[(2*(p + 1))/(b*(2*p + 1)), Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x]
&&  !IntegerQ[p] && LtQ[p, -1]

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}{\left (a \sinh ^2(x)\right )^{3/2}} \, dx &=-\frac{\coth (x)}{2 a \sqrt{a \sinh ^2(x)}}-\frac{\int \frac{1}{\sqrt{a \sinh ^2(x)}} \, dx}{2 a}\\ &=-\frac{\coth (x)}{2 a \sqrt{a \sinh ^2(x)}}-\frac{\sinh (x) \int \text{csch}(x) \, dx}{2 a \sqrt{a \sinh ^2(x)}}\\ &=-\frac{\coth (x)}{2 a \sqrt{a \sinh ^2(x)}}+\frac{\tanh ^{-1}(\cosh (x)) \sinh (x)}{2 a \sqrt{a \sinh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0375964, size = 44, normalized size = 1.05 \[ -\frac{\sinh ^3(x) \left (\text{csch}^2\left (\frac{x}{2}\right )+\text{sech}^2\left (\frac{x}{2}\right )+4 \log \left (\tanh \left (\frac{x}{2}\right )\right )\right )}{8 \left (a \sinh ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^2)^(-3/2),x]

[Out]

-((Csch[x/2]^2 + 4*Log[Tanh[x/2]] + Sech[x/2]^2)*Sinh[x]^3)/(8*(a*Sinh[x]^2)^(3/2))

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Maple [B]  time = 0.05, size = 71, normalized size = 1.7 \begin{align*} -{\frac{1}{2\,\cosh \left ( x \right ) \sinh \left ( x \right ) }\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}} \left ( -\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}}+a}{\sinh \left ( x \right ) }} \right ) a \left ( \sinh \left ( x \right ) \right ) ^{2}+\sqrt{a}\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}} \right ){a}^{-{\frac{5}{2}}}{\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)^(3/2),x)

[Out]

-1/2/a^(5/2)/sinh(x)*(a*cosh(x)^2)^(1/2)*(-ln(2*(a^(1/2)*(a*cosh(x)^2)^(1/2)+a)/sinh(x))*a*sinh(x)^2+a^(1/2)*(
a*cosh(x)^2)^(1/2))/cosh(x)/(a*sinh(x)^2)^(1/2)

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Maxima [A]  time = 1.75467, size = 84, normalized size = 2. \begin{align*} -\frac{e^{\left (-x\right )} + e^{\left (-3 \, x\right )}}{2 \, a^{\frac{3}{2}} e^{\left (-2 \, x\right )} - a^{\frac{3}{2}} e^{\left (-4 \, x\right )} - a^{\frac{3}{2}}} - \frac{\log \left (e^{\left (-x\right )} + 1\right )}{2 \, a^{\frac{3}{2}}} + \frac{\log \left (e^{\left (-x\right )} - 1\right )}{2 \, a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

-(e^(-x) + e^(-3*x))/(2*a^(3/2)*e^(-2*x) - a^(3/2)*e^(-4*x) - a^(3/2)) - 1/2*log(e^(-x) + 1)/a^(3/2) + 1/2*log
(e^(-x) - 1)/a^(3/2)

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Fricas [B]  time = 1.81061, size = 927, normalized size = 22.07 \begin{align*} \frac{{\left (6 \, \cosh \left (x\right ) e^{x} \sinh \left (x\right )^{2} + 2 \, e^{x} \sinh \left (x\right )^{3} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} e^{x} \sinh \left (x\right ) + 2 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} e^{x} -{\left (4 \, \cosh \left (x\right ) e^{x} \sinh \left (x\right )^{3} + e^{x} \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} e^{x} \sinh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} e^{x} \sinh \left (x\right ) +{\left (\cosh \left (x\right )^{4} - 2 \, \cosh \left (x\right )^{2} + 1\right )} e^{x}\right )} \log \left (\frac{\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right )\right )} \sqrt{a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a} e^{\left (-x\right )}}{2 \,{\left (a^{2} \cosh \left (x\right )^{4} -{\left (a^{2} e^{\left (2 \, x\right )} - a^{2}\right )} \sinh \left (x\right )^{4} - 2 \, a^{2} \cosh \left (x\right )^{2} - 4 \,{\left (a^{2} \cosh \left (x\right ) e^{\left (2 \, x\right )} - a^{2} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 2 \,{\left (3 \, a^{2} \cosh \left (x\right )^{2} - a^{2} -{\left (3 \, a^{2} \cosh \left (x\right )^{2} - a^{2}\right )} e^{\left (2 \, x\right )}\right )} \sinh \left (x\right )^{2} + a^{2} -{\left (a^{2} \cosh \left (x\right )^{4} - 2 \, a^{2} \cosh \left (x\right )^{2} + a^{2}\right )} e^{\left (2 \, x\right )} + 4 \,{\left (a^{2} \cosh \left (x\right )^{3} - a^{2} \cosh \left (x\right ) -{\left (a^{2} \cosh \left (x\right )^{3} - a^{2} \cosh \left (x\right )\right )} e^{\left (2 \, x\right )}\right )} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(6*cosh(x)*e^x*sinh(x)^2 + 2*e^x*sinh(x)^3 + 2*(3*cosh(x)^2 + 1)*e^x*sinh(x) + 2*(cosh(x)^3 + cosh(x))*e^x
- (4*cosh(x)*e^x*sinh(x)^3 + e^x*sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*e^x*sinh(x)^2 + 4*(cosh(x)^3 - cosh(x))*e^x*
sinh(x) + (cosh(x)^4 - 2*cosh(x)^2 + 1)*e^x)*log((cosh(x) + sinh(x) + 1)/(cosh(x) + sinh(x) - 1)))*sqrt(a*e^(4
*x) - 2*a*e^(2*x) + a)*e^(-x)/(a^2*cosh(x)^4 - (a^2*e^(2*x) - a^2)*sinh(x)^4 - 2*a^2*cosh(x)^2 - 4*(a^2*cosh(x
)*e^(2*x) - a^2*cosh(x))*sinh(x)^3 + 2*(3*a^2*cosh(x)^2 - a^2 - (3*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^2 + a
^2 - (a^2*cosh(x)^4 - 2*a^2*cosh(x)^2 + a^2)*e^(2*x) + 4*(a^2*cosh(x)^3 - a^2*cosh(x) - (a^2*cosh(x)^3 - a^2*c
osh(x))*e^(2*x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sinh ^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)**2)**(3/2),x)

[Out]

Integral((a*sinh(x)**2)**(-3/2), x)

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Giac [B]  time = 1.33836, size = 127, normalized size = 3.02 \begin{align*} \frac{\frac{\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{\sqrt{a} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} - \frac{\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{\sqrt{a} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} - \frac{4 \,{\left (e^{\left (-x\right )} + e^{x}\right )}}{{\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )} \sqrt{a} \mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )}}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*(log(e^(-x) + e^x + 2)/(sqrt(a)*sgn(e^(3*x) - e^x)) - log(e^(-x) + e^x - 2)/(sqrt(a)*sgn(e^(3*x) - e^x)) -
4*(e^(-x) + e^x)/(((e^(-x) + e^x)^2 - 4)*sqrt(a)*sgn(e^(3*x) - e^x)))/a