3.25 \(\int (-1-\text{csch}^2(x))^{3/2} \, dx\)

Optimal. Leaf size=34 \[ \frac{1}{2} \coth (x) \sqrt{-\coth ^2(x)}-\tanh (x) \sqrt{-\coth ^2(x)} \log (\sinh (x)) \]

[Out]

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

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Rubi [A]  time = 0.0373094, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4121, 3658, 3473, 3475} \[ \frac{1}{2} \coth (x) \sqrt{-\coth ^2(x)}-\tanh (x) \sqrt{-\coth ^2(x)} \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(-1 - Csch[x]^2)^(3/2),x]

[Out]

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

Rule 4121

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(b*tan[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

Rubi steps

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

Mathematica [A]  time = 0.0156181, size = 26, normalized size = 0.76 \[ \frac{1}{2} \tanh (x) \sqrt{-\coth ^2(x)} \left (\text{csch}^2(x)-2 \log (\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 - Csch[x]^2)^(3/2),x]

[Out]

(Sqrt[-Coth[x]^2]*(Csch[x]^2 - 2*Log[Sinh[x]])*Tanh[x])/2

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Maple [B]  time = 0.098, size = 123, normalized size = 3.6 \begin{align*}{\frac{ \left ({{\rm e}^{2\,x}}-1 \right ) x}{{{\rm e}^{2\,x}}+1}\sqrt{-{\frac{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}}}}+2\,{\frac{{{\rm e}^{2\,x}}}{ \left ({{\rm e}^{2\,x}}+1 \right ) \left ({{\rm e}^{2\,x}}-1 \right ) }\sqrt{-{\frac{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}}}}-{\frac{ \left ({{\rm e}^{2\,x}}-1 \right ) \ln \left ({{\rm e}^{2\,x}}-1 \right ) }{{{\rm e}^{2\,x}}+1}\sqrt{-{\frac{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1-csch(x)^2)^(3/2),x)

[Out]

1/(exp(2*x)+1)*(exp(2*x)-1)*(-(exp(2*x)+1)^2/(exp(2*x)-1)^2)^(1/2)*x+2/(exp(2*x)+1)/(exp(2*x)-1)*(-(exp(2*x)+1
)^2/(exp(2*x)-1)^2)^(1/2)*exp(2*x)-1/(exp(2*x)+1)*(exp(2*x)-1)*(-(exp(2*x)+1)^2/(exp(2*x)-1)^2)^(1/2)*ln(exp(2
*x)-1)

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Maxima [C]  time = 1.51645, size = 59, normalized size = 1.74 \begin{align*} i \, x + \frac{2 i \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-csch(x)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.11553, size = 165, normalized size = 4.85 \begin{align*} \frac{i \, x e^{\left (4 \, x\right )} +{\left (-2 i \, x + 2 i\right )} e^{\left (2 \, x\right )} +{\left (-i \, e^{\left (4 \, x\right )} + 2 i \, e^{\left (2 \, x\right )} - i\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) + i \, x}{e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-csch(x)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-csch(x)**2)**(3/2),x)

[Out]

Integral((-csch(x)**2 - 1)**(3/2), x)

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Giac [C]  time = 1.16153, size = 113, normalized size = 3.32 \begin{align*} -i \, x \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + i \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) - \frac{i \,{\left (3 \, e^{\left (4 \, x\right )} \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) - 2 \, e^{\left (2 \, x\right )} \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + 3 \, \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right )\right )}}{2 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-csch(x)^2)^(3/2),x, algorithm="giac")

[Out]

-I*x*sgn(-e^(4*x) + 1) + I*log(abs(e^(2*x) - 1))*sgn(-e^(4*x) + 1) - 1/2*I*(3*e^(4*x)*sgn(-e^(4*x) + 1) - 2*e^
(2*x)*sgn(-e^(4*x) + 1) + 3*sgn(-e^(4*x) + 1))/(e^(2*x) - 1)^2