3.2.11 \(\int \frac {\text {sech}^3(x)}{1+\coth (x)} \, dx\) [111]

Optimal. Leaf size=20 \[ -\frac {1}{2} \text {ArcTan}(\sinh (x))-\text {sech}(x)+\frac {1}{2} \text {sech}(x) \tanh (x) \]

[Out]

-1/2*arctan(sinh(x))-sech(x)+1/2*sech(x)*tanh(x)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3599, 3187, 3186, 2686, 8, 2691, 3855} \begin {gather*} -\frac {1}{2} \text {ArcTan}(\sinh (x))-\text {sech}(x)+\frac {1}{2} \tanh (x) \text {sech}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(1 + Coth[x]),x]

[Out]

-1/2*ArcTan[Sinh[x]] - Sech[x] + (Sech[x]*Tanh[x])/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3186

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 3187

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3599

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Sin[e + f*x]^
m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/Cos[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3855

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 {\text {sech}^3(x)}{1+\coth (x)} \, dx &=-\left (i \int \frac {\text {sech}^2(x) \tanh (x)}{-i \cosh (x)-i \sinh (x)} \, dx\right )\\ &=-\int \text {sech}^2(x) (-\cosh (x)+\sinh (x)) \tanh (x) \, dx\\ &=i \int \left (-i \text {sech}(x) \tanh (x)+i \text {sech}(x) \tanh ^2(x)\right ) \, dx\\ &=\int \text {sech}(x) \tanh (x) \, dx-\int \text {sech}(x) \tanh ^2(x) \, dx\\ &=\frac {1}{2} \text {sech}(x) \tanh (x)-\frac {1}{2} \int \text {sech}(x) \, dx-\text {Subst}(\int 1 \, dx,x,\text {sech}(x))\\ &=-\frac {1}{2} \tan ^{-1}(\sinh (x))-\text {sech}(x)+\frac {1}{2} \text {sech}(x) \tanh (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 20, normalized size = 1.00 \begin {gather*} -\text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{2} \text {sech}(x) (-2+\tanh (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(1 + Coth[x]),x]

[Out]

-ArcTan[Tanh[x/2]] + (Sech[x]*(-2 + Tanh[x]))/2

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(44\) vs. \(2(16)=32\).
time = 0.59, size = 45, normalized size = 2.25

method	result	size
risch	\(-\frac {{\mathrm e}^{x} \left ({\mathrm e}^{2 x}+3\right )}{\left (1+{\mathrm e}^{2 x}\right )^{2}}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}\)	\(38\)
default	\(\frac {-\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )-2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\tanh \left (\frac {x}{2}\right )-2}{\left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\arctan \left (\tanh \left (\frac {x}{2}\right )\right )\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(1+coth(x)),x,method=_RETURNVERBOSE)

[Out]

4*(-1/4*tanh(1/2*x)^3-1/2*tanh(1/2*x)^2+1/4*tanh(1/2*x)-1/2)/(tanh(1/2*x)^2+1)^2-arctan(tanh(1/2*x))

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
time = 0.48, size = 33, normalized size = 1.65 \begin {gather*} -\frac {e^{\left (-x\right )} + 3 \, e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + \arctan \left (e^{\left (-x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(1+coth(x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (16) = 32\).
time = 0.35, size = 140, normalized size = 7.00 \begin {gather*} -\frac {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} + {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 3 \, {\left (\cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + 3 \, \cosh \left (x\right )}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (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)*arctan(cosh(x) + sinh(x)) + 3*(cosh(x)^2 +
1)*sinh(x) + 3*cosh(x))/(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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{3}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(1+coth(x)),x)

[Out]

Integral(sech(x)**3/(coth(x) + 1), x)

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 25, normalized size = 1.25 \begin {gather*} -\frac {e^{\left (3 \, x\right )} + 3 \, e^{x}}{{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} - \arctan \left (e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(1+coth(x)),x, algorithm="giac")

[Out]

-(e^(3*x) + 3*e^x)/(e^(2*x) + 1)^2 - arctan(e^x)

________________________________________________________________________________________

Mupad [B]
time = 1.25, size = 22, normalized size = 1.10 \begin {gather*} -\mathrm {atan}\left ({\mathrm {e}}^x\right )-\frac {1}{2\,\mathrm {cosh}\left (x\right )}-\frac {{\mathrm {e}}^{-x}}{2\,{\mathrm {cosh}\left (x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(coth(x) + 1)),x)

[Out]

- atan(exp(x)) - 1/(2*cosh(x)) - exp(-x)/(2*cosh(x)^2)

________________________________________________________________________________________