3.565 \(\int (\cosh (x) \coth (x))^{5/2} \, dx\)

Optimal. Leaf size=50 \[ \frac{2}{5} \cosh ^2(x) \coth (x) \sqrt{\cosh (x) \coth (x)}-\frac{16}{15} \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{64}{15} \tanh (x) \sqrt{\cosh (x) \coth (x)} \]

[Out]

(-16*Coth[x]*Sqrt[Cosh[x]*Coth[x]])/15 + (2*Cosh[x]^2*Coth[x]*Sqrt[Cosh[x]*Coth[x]])/5 + (64*Sqrt[Cosh[x]*Coth
[x]]*Tanh[x])/15

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Rubi [A]  time = 0.131647, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4398, 4400, 2598, 2594, 2589} \[ \frac{2}{5} \cosh ^2(x) \coth (x) \sqrt{\cosh (x) \coth (x)}-\frac{16}{15} \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{64}{15} \tanh (x) \sqrt{\cosh (x) \coth (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[x]*Coth[x])^(5/2),x]

[Out]

(-16*Coth[x]*Sqrt[Cosh[x]*Coth[x]])/15 + (2*Cosh[x]^2*Coth[x]*Sqrt[Cosh[x]*Coth[x]])/5 + (64*Sqrt[Cosh[x]*Coth
[x]]*Tanh[x])/15

Rule 4398

Int[(u_.)*((a_)*(v_))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v]}, Dist[(a^IntPart[p]
*(a*vv)^FracPart[p])/vv^FracPart[p], Int[uu*vv^p, x], x]] /; FreeQ[{a, p}, x] &&  !IntegerQ[p] &&  !InertTrigF
reeQ[v]

Rule 4400

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])), Int[uu*vv^(m*p)*ww^(n*p)
, x], x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  !InertTrigFreeQ[w])

Rule 2598

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(b*(a*Sin[
e + f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2
*m, 2*n]

Rule 2594

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(n - 1)), x] - Dist[(b^2*(m + n - 1))/(n - 1), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

Rule 2589

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*(a*Sin[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rubi steps

\begin{align*} \int (\cosh (x) \coth (x))^{5/2} \, dx &=-\frac{\sqrt{\cosh (x) \coth (x)} \int (-i \cosh (x) \coth (x))^{5/2} \, dx}{\sqrt{-i \cosh (x) \coth (x)}}\\ &=-\frac{\sqrt{\cosh (x) \coth (x)} \int \cosh ^{\frac{5}{2}}(x) (-i \coth (x))^{5/2} \, dx}{\sqrt{\cosh (x)} \sqrt{-i \coth (x)}}\\ &=\frac{2}{5} \cosh ^2(x) \coth (x) \sqrt{\cosh (x) \coth (x)}-\frac{\left (8 \sqrt{\cosh (x) \coth (x)}\right ) \int \sqrt{\cosh (x)} (-i \coth (x))^{5/2} \, dx}{5 \sqrt{\cosh (x)} \sqrt{-i \coth (x)}}\\ &=-\frac{16}{15} \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{2}{5} \cosh ^2(x) \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{\left (32 \sqrt{\cosh (x) \coth (x)}\right ) \int \sqrt{\cosh (x)} \sqrt{-i \coth (x)} \, dx}{15 \sqrt{\cosh (x)} \sqrt{-i \coth (x)}}\\ &=-\frac{16}{15} \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{2}{5} \cosh ^2(x) \coth (x) \sqrt{\cosh (x) \coth (x)}+\frac{64}{15} \sqrt{\cosh (x) \coth (x)} \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.311567, size = 44, normalized size = 0.88 \[ \frac{1}{15} \sqrt{\cosh (x) \coth (x)} \left (64 \tanh (x)-10 \coth (x)+6 \sinh (x) \cosh (x)+57 \left (-\sinh ^2(x)\right )^{3/4} \text{csch}(x) \text{sech}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[x]*Coth[x])^(5/2),x]

[Out]

(Sqrt[Cosh[x]*Coth[x]]*(-10*Coth[x] + 6*Cosh[x]*Sinh[x] + 57*Csch[x]*Sech[x]*(-Sinh[x]^2)^(3/4) + 64*Tanh[x]))
/15

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Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int \left ( \cosh \left ( x \right ){\rm coth} \left (x\right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)*coth(x))^(5/2),x)

[Out]

int((cosh(x)*coth(x))^(5/2),x)

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Maxima [B]  time = 1.69129, size = 220, normalized size = 4.4 \begin{align*} \frac{\sqrt{2} e^{\left (\frac{5}{2} \, x\right )}}{20 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} + \frac{7 \, \sqrt{2} e^{\left (\frac{1}{2} \, x\right )}}{4 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} - \frac{41 \, \sqrt{2} e^{\left (-\frac{3}{2} \, x\right )}}{6 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} + \frac{41 \, \sqrt{2} e^{\left (-\frac{7}{2} \, x\right )}}{6 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} - \frac{7 \, \sqrt{2} e^{\left (-\frac{11}{2} \, x\right )}}{4 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} - \frac{\sqrt{2} e^{\left (-\frac{15}{2} \, x\right )}}{20 \,{\left (e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}{\left (-e^{\left (-x\right )} + 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(x)*coth(x))^(5/2),x, algorithm="maxima")

[Out]

1/20*sqrt(2)*e^(5/2*x)/((e^(-x) + 1)^(5/2)*(-e^(-x) + 1)^(5/2)) + 7/4*sqrt(2)*e^(1/2*x)/((e^(-x) + 1)^(5/2)*(-
e^(-x) + 1)^(5/2)) - 41/6*sqrt(2)*e^(-3/2*x)/((e^(-x) + 1)^(5/2)*(-e^(-x) + 1)^(5/2)) + 41/6*sqrt(2)*e^(-7/2*x
)/((e^(-x) + 1)^(5/2)*(-e^(-x) + 1)^(5/2)) - 7/4*sqrt(2)*e^(-11/2*x)/((e^(-x) + 1)^(5/2)*(-e^(-x) + 1)^(5/2))
- 1/20*sqrt(2)*e^(-15/2*x)/((e^(-x) + 1)^(5/2)*(-e^(-x) + 1)^(5/2))

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Fricas [B]  time = 2.46197, size = 883, normalized size = 17.66 \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (3 \, \cosh \left (x\right )^{8} + 24 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + 3 \, \sinh \left (x\right )^{8} + 12 \,{\left (7 \, \cosh \left (x\right )^{2} + 9\right )} \sinh \left (x\right )^{6} + 108 \, \cosh \left (x\right )^{6} + 24 \,{\left (7 \, \cosh \left (x\right )^{3} + 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + 2 \,{\left (105 \, \cosh \left (x\right )^{4} + 810 \, \cosh \left (x\right )^{2} - 151\right )} \sinh \left (x\right )^{4} - 302 \, \cosh \left (x\right )^{4} + 8 \,{\left (21 \, \cosh \left (x\right )^{5} + 270 \, \cosh \left (x\right )^{3} - 151 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 12 \,{\left (7 \, \cosh \left (x\right )^{6} + 135 \, \cosh \left (x\right )^{4} - 151 \, \cosh \left (x\right )^{2} + 9\right )} \sinh \left (x\right )^{2} + 108 \, \cosh \left (x\right )^{2} + 8 \,{\left (3 \, \cosh \left (x\right )^{7} + 81 \, \cosh \left (x\right )^{5} - 151 \, \cosh \left (x\right )^{3} + 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 3\right )}}{30 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} +{\left (6 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - \cosh \left (x\right )^{2} + 2 \,{\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt{\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} +{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right ) - \cosh \left (x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(x)*coth(x))^(5/2),x, algorithm="fricas")

[Out]

1/30*sqrt(1/2)*(3*cosh(x)^8 + 24*cosh(x)*sinh(x)^7 + 3*sinh(x)^8 + 12*(7*cosh(x)^2 + 9)*sinh(x)^6 + 108*cosh(x
)^6 + 24*(7*cosh(x)^3 + 27*cosh(x))*sinh(x)^5 + 2*(105*cosh(x)^4 + 810*cosh(x)^2 - 151)*sinh(x)^4 - 302*cosh(x
)^4 + 8*(21*cosh(x)^5 + 270*cosh(x)^3 - 151*cosh(x))*sinh(x)^3 + 12*(7*cosh(x)^6 + 135*cosh(x)^4 - 151*cosh(x)
^2 + 9)*sinh(x)^2 + 108*cosh(x)^2 + 8*(3*cosh(x)^7 + 81*cosh(x)^5 - 151*cosh(x)^3 + 27*cosh(x))*sinh(x) + 3)/(
(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 - 1)*sinh(x)^2 - cosh(x)^2 + 2*(2*cosh(x)^3 - cosh
(x))*sinh(x))*sqrt(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*cosh(x)^2 - 1)*sinh(x) - cosh(x)))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(x)*coth(x))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\cosh \left (x\right ) \coth \left (x\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(x)*coth(x))^(5/2),x, algorithm="giac")

[Out]

integrate((cosh(x)*coth(x))^(5/2), x)