\(\int (1+\text {csch}^2(x))^{3/2} \, dx\) [16]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 29 \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=-\frac {1}{2} \coth ^2(x)^{3/2} \tanh (x)+\sqrt {\coth ^2(x)} \log (\sinh (x)) \tanh (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4206, 3739, 3554, 3556} \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=\tanh (x) \sqrt {\coth ^2(x)} \log (\sinh (x))-\frac {1}{2} \tanh (x) \coth ^2(x)^{3/2} \]

[In]

Int[(1 + Csch[x]^2)^(3/2),x]

[Out]

-1/2*((Coth[x]^2)^(3/2)*Tanh[x]) + Sqrt[Coth[x]^2]*Log[Sinh[x]]*Tanh[x]

Rule 3554

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 3556

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

Rule 3739

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 4206

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=-\frac {1}{2} \sqrt {\coth ^2(x)} \left (\coth ^2(x)-2 (\log (\cosh (x))+\log (\tanh (x)))\right ) \tanh (x) \]

[In]

Integrate[(1 + Csch[x]^2)^(3/2),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method result size
default \(\operatorname {csgn}\left (\coth \left (x \right )\right ) \left (-\frac {\coth \left (x \right )^{2}}{2}-\frac {\ln \left (\coth \left (x \right )-1\right )}{2}-\frac {\ln \left (1+\coth \left (x \right )\right )}{2}\right )\) \(26\)
risch \(\frac {\sqrt {\frac {\left (1+{\mathrm e}^{2 x}\right )^{2}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \left ({\mathrm e}^{4 x} \ln \left ({\mathrm e}^{2 x}-1\right )-{\mathrm e}^{4 x} x -2 \,{\mathrm e}^{2 x} \ln \left ({\mathrm e}^{2 x}-1\right )+2 \,{\mathrm e}^{2 x} x -2 \,{\mathrm e}^{2 x}+\ln \left ({\mathrm e}^{2 x}-1\right )-x \right )}{\left (1+{\mathrm e}^{2 x}\right ) \left ({\mathrm e}^{2 x}-1\right )}\) \(93\)

[In]

int((1+csch(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

csgn(coth(x))*(-1/2*coth(x)^2-1/2*ln(coth(x)-1)-1/2*ln(1+coth(x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 6.55 \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=-\frac {x \cosh \left (x\right )^{4} + 4 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + x \sinh \left (x\right )^{4} - 2 \, {\left (x - 1\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, x \cosh \left (x\right )^{2} - x + 1\right )} \sinh \left (x\right )^{2} - {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left (x \cosh \left (x\right )^{3} - {\left (x - 1\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + x}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1} \]

[In]

integrate((1+csch(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*sinh(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)

Sympy [F]

\[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=\int \left (\operatorname {csch}^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

Integral((csch(x)**2 + 1)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=-x - \frac {2 \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (23) = 46\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52 \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=-x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \frac {3 \, e^{\left (4 \, x\right )} \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - 2 \, e^{\left (2 \, x\right )} \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + 3 \, \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right )}{2 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \]

[In]

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

[Out]

-x*sgn(e^(4*x) - 1) + log(abs(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

Mupad [F(-1)]

Timed out. \[ \int \left (1+\text {csch}^2(x)\right )^{3/2} \, dx=\int {\left (\frac {1}{{\mathrm {sinh}\left (x\right )}^2}+1\right )}^{3/2} \,d x \]

[In]

int((1/sinh(x)^2 + 1)^(3/2),x)

[Out]

int((1/sinh(x)^2 + 1)^(3/2), x)