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

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Rubi [A]  time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.01, 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]

-1/2*((2*Coth[x]*Log[Cosh[x]] + Csch[x]*Sech[x])*Sqrt[-Tanh[x]^2])

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fricas [A]  time = 0.68, size = 1, normalized size = 0.03 \[ 0 \]

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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giac [C]  time = 0.14, size = 83, normalized size = 2.44 \[ -i \, x \mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + i \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) - \frac {i \, {\left (3 \, e^{\left (4 \, x\right )} \mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + 2 \, e^{\left (2 \, x\right )} \mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) + 3 \, \mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right )\right )}}{2 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

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

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maple [B]  time = 0.32, size = 123, normalized size = 3.62 \[ \frac {\left (1+{\mathrm e}^{2 x}\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, x}{{\mathrm e}^{2 x}-1}-\frac {2 \sqrt {-\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right ) \left (1+{\mathrm e}^{2 x}\right )}-\frac {\left (1+{\mathrm e}^{2 x}\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \ln \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{2 x}-1} \]

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

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maxima [C]  time = 0.49, size = 33, normalized size = 0.97 \[ i \, x + \frac {2 i \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + i \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (\frac {1}{{\mathrm {cosh}\relax (x)}^2}-1\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(x)^2 - 1)^(3/2),x)

[Out]

int((1/cosh(x)^2 - 1)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\operatorname {sech}^{2}{\relax (x )} - 1\right )^{\frac {3}{2}}\, dx \]

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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