\(\int (\sinh (x) \tanh (x))^{5/2} \, dx\) [562]

Optimal result

Integrand size = 9, antiderivative size = 50 \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=-\frac {64}{15} \coth (x) \sqrt {\sinh (x) \tanh (x)}+\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)} \]

[Out]

-64/15*coth(x)*(sinh(x)*tanh(x))^(1/2)+16/15*(sinh(x)*tanh(x))^(1/2)*tanh(x)+2/5*sinh(x)^2*(sinh(x)*tanh(x))^(
1/2)*tanh(x)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4483, 4485, 2678, 2674, 2669} \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=\frac {2}{5} \sinh ^2(x) \tanh (x) \sqrt {\sinh (x) \tanh (x)}+\frac {16}{15} \tanh (x) \sqrt {\sinh (x) \tanh (x)}-\frac {64}{15} \coth (x) \sqrt {\sinh (x) \tanh (x)} \]

[In]

Int[(Sinh[x]*Tanh[x])^(5/2),x]

[Out]

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

Rule 2669

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]

Rule 2674

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 2678

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*
Tan[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 4483

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 4485

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])

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58 \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=-\frac {2}{15} \left (-5-3 \cosh ^2(x)+32 \coth ^2(x)\right ) \tanh (x) \sqrt {\sinh (x) \tanh (x)} \]

[In]

Integrate[(Sinh[x]*Tanh[x])^(5/2),x]

[Out]

(-2*(-5 - 3*Cosh[x]^2 + 32*Coth[x]^2)*Tanh[x]*Sqrt[Sinh[x]*Tanh[x]])/15

Maple [F]

\[\int \left (\sinh \left (x \right ) \tanh \left (x \right )\right )^{\frac {5}{2}}d x\]

[In]

int((sinh(x)*tanh(x))^(5/2),x)

[Out]

int((sinh(x)*tanh(x))^(5/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (38) = 76\).

Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 5.06 \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (3 \, \cosh \left (x\right )^{8} + 24 \, \cosh \left (x\right ) \sinh \left (x\right )^{7} + 3 \, \sinh \left (x\right )^{8} + 12 \, {\left (7 \, \cosh \left (x\right )^{2} - 9\right )} \sinh \left (x\right )^{6} - 108 \, \cosh \left (x\right )^{6} + 24 \, {\left (7 \, \cosh \left (x\right )^{3} - 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + 2 \, {\left (105 \, \cosh \left (x\right )^{4} - 810 \, \cosh \left (x\right )^{2} - 151\right )} \sinh \left (x\right )^{4} - 302 \, \cosh \left (x\right )^{4} + 8 \, {\left (21 \, \cosh \left (x\right )^{5} - 270 \, \cosh \left (x\right )^{3} - 151 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 12 \, {\left (7 \, \cosh \left (x\right )^{6} - 135 \, \cosh \left (x\right )^{4} - 151 \, \cosh \left (x\right )^{2} - 9\right )} \sinh \left (x\right )^{2} - 108 \, \cosh \left (x\right )^{2} + 8 \, {\left (3 \, \cosh \left (x\right )^{7} - 81 \, \cosh \left (x\right )^{5} - 151 \, \cosh \left (x\right )^{3} - 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 3\right )}}{30 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} + {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right )}} \]

[In]

integrate((sinh(x)*tanh(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)))

Sympy [F(-1)]

Timed out. \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((sinh(x)*tanh(x))**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (38) = 76\).

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.06 \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=-\frac {\sqrt {2} e^{\left (\frac {5}{2} \, x\right )}}{20 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {7 \, \sqrt {2} e^{\left (\frac {1}{2} \, x\right )}}{4 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {41 \, \sqrt {2} e^{\left (-\frac {3}{2} \, x\right )}}{6 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {41 \, \sqrt {2} e^{\left (-\frac {7}{2} \, x\right )}}{6 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} + \frac {7 \, \sqrt {2} e^{\left (-\frac {11}{2} \, x\right )}}{4 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} - \frac {\sqrt {2} e^{\left (-\frac {15}{2} \, x\right )}}{20 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} \]

[In]

integrate((sinh(x)*tanh(x))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=\int { \left (\sinh \left (x\right ) \tanh \left (x\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((sinh(x)*tanh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((sinh(x)*tanh(x))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (\sinh (x) \tanh (x))^{5/2} \, dx=\int {\left (\mathrm {sinh}\left (x\right )\,\mathrm {tanh}\left (x\right )\right )}^{5/2} \,d x \]

[In]

int((sinh(x)*tanh(x))^(5/2),x)

[Out]

int((sinh(x)*tanh(x))^(5/2), x)