3.173 \(\int (-1+\text{sech}^2(x))^{3/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0288396, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4121, 3658, 3473, 3475} \[ \frac{1}{2} \tanh (x) \sqrt{-\tanh ^2(x)}-\sqrt{-\tanh ^2(x)} \coth (x) \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Sech[x]^2)^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0154117, size = 27, normalized size = 0.79 \[ -\frac{1}{2} \sqrt{-\tanh ^2(x)} (\text{csch}(x) \text{sech}(x)+2 \coth (x) \log (\cosh (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Sech[x]^2)^(3/2),x]

[Out]

-((2*Coth[x]*Log[Cosh[x]] + Csch[x]*Sech[x])*Sqrt[-Tanh[x]^2])/2

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Maple [B]  time = 0.102, 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+sech(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 = 1.66048, size = 45, normalized size = 1.32 \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 (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+sech(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^(-2*x) + 1)

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Fricas [A]  time = 2.05415, size = 4, normalized size = 0.12 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\operatorname{sech}^{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+sech(x)**2)**(3/2),x)

[Out]

Integral((sech(x)**2 - 1)**(3/2), x)

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Giac [C]  time = 1.13726, size = 112, normalized size = 3.29 \begin{align*} -i \, x \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + i \, \log \left (e^{\left (2 \, x\right )} + 1\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+sech(x)^2)^(3/2),x, algorithm="giac")

[Out]

-I*x*sgn(-e^(4*x) + 1) + I*log(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