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

Optimal. Leaf size=29 \[ \sqrt {\tanh ^2(x)} \coth (x) \log (\cosh (x))-\frac {1}{2} \tanh ^2(x)^{3/2} \coth (x) \]

[Out]

coth(x)*ln(cosh(x))*(tanh(x)^2)^(1/2)-1/2*coth(x)*(tanh(x)^2)^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, 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} \[ \sqrt {\tanh ^2(x)} \coth (x) \log (\cosh (x))-\frac {1}{2} \tanh ^2(x)^{3/2} \coth (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sech[x]^2)^(3/2),x]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[Tanh[x]^2] - (Coth[x]*(Tanh[x]^2)^(3/2))/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 \tanh ^2(x)^{3/2} \, dx\\ &=\left (\coth (x) \sqrt {\tanh ^2(x)}\right ) \int \tanh ^3(x) \, dx\\ &=-\frac {1}{2} \coth (x) \tanh ^2(x)^{3/2}+\left (\coth (x) \sqrt {\tanh ^2(x)}\right ) \int \tanh (x) \, dx\\ &=\coth (x) \log (\cosh (x)) \sqrt {\tanh ^2(x)}-\frac {1}{2} \coth (x) \tanh ^2(x)^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.86 \[ \frac {1}{2} \sqrt {\tanh ^2(x)} (\text {csch}(x) \text {sech}(x)+2 \coth (x) \log (\cosh (x))) \]

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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fricas [B]  time = 0.67, size = 183, normalized size = 6.31 \[ -\frac {x \cosh \relax (x)^{4} + 4 \, x \cosh \relax (x) \sinh \relax (x)^{3} + x \sinh \relax (x)^{4} + 2 \, {\left (x - 1\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, x \cosh \relax (x)^{2} + x - 1\right )} \sinh \relax (x)^{2} - {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 2 \, {\left (3 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 2 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left (x \cosh \relax (x)^{3} + {\left (x - 1\right )} \cosh \relax (x)\right )} \sinh \relax (x) + x}{\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 2 \, {\left (3 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 2 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, size = 72, normalized size = 2.48 \[ -x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \frac {3 \, e^{\left (4 \, x\right )} \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + 2 \, e^{\left (2 \, x\right )} \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + 3 \, \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right )}{2 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*sgn(e^(4*x) - 1) + log(e^(2*x) + 1)*sgn(e^(4*x) - 1) - 1/2*(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 = 0.28, size = 120, normalized size = 4.14 \[ -\frac {\left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, x}{{\mathrm e}^{2 x}-1}+\frac {2 \sqrt {\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right ) \left (1+{\mathrm e}^{2 x}\right )}+\frac {\left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \ln \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{2 x}-1} \]

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 [A]  time = 0.46, size = 33, normalized size = 1.14 \[ -x - \frac {2 \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} - \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 1/cosh(x)^2)^(3/2),x)

[Out]

int((1 - 1/cosh(x)^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((1 - sech(x)**2)**(3/2), x)

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