### 3.562 $$\int (\sinh (x) \tanh (x))^{5/2} \, dx$$

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

[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

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Rubi [A]  time = 0.117818, 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} $\frac{2}{5} \sinh ^2(x) \tanh (x) \sqrt{\sinh (x) \tanh (x)}+\frac{16}{15} \tanh (x) \sqrt{\sinh (x) \tanh (x)}-\frac{64}{15} \coth (x) \sqrt{\sinh (x) \tanh (x)}$

Antiderivative was successfully veriﬁed.

[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 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 (\sinh (x) \tanh (x))^{5/2} \, dx &=\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]  time = 0.113306, size = 29, normalized size = 0.58 $-\frac{2}{15} \tanh (x) \sqrt{\sinh (x) \tanh (x)} \left (-3 \cosh ^2(x)+32 \coth ^2(x)-5\right )$

Antiderivative was successfully veriﬁed.

[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

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Maple [F]  time = 0.103, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( x \right ) \tanh \left ( x \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.75192, size = 139, normalized size = 2.78 \begin{align*} -\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}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 2.39637, 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((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)))

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((sinh(x)*tanh(x))^(5/2), x)