### 3.561 $$\int (\sinh (x) \tanh (x))^{3/2} \, dx$$

Optimal. Leaf size=31 $\frac{2}{3} \sinh (x) \sqrt{\sinh (x) \tanh (x)}+\frac{8}{3} \text{csch}(x) \sqrt{\sinh (x) \tanh (x)}$

[Out]

(8*Csch[x]*Sqrt[Sinh[x]*Tanh[x]])/3 + (2*Sinh[x]*Sqrt[Sinh[x]*Tanh[x]])/3

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Rubi [A]  time = 0.08793, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.444, Rules used = {4398, 4400, 2598, 2589} $\frac{2}{3} \sinh (x) \sqrt{\sinh (x) \tanh (x)}+\frac{8}{3} \text{csch}(x) \sqrt{\sinh (x) \tanh (x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sinh[x]*Tanh[x])^(3/2),x]

[Out]

(8*Csch[x]*Sqrt[Sinh[x]*Tanh[x]])/3 + (2*Sinh[x]*Sqrt[Sinh[x]*Tanh[x]])/3

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 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))^{3/2} \, dx &=-\frac{\sqrt{\sinh (x) \tanh (x)} \int (-\sinh (x) \tanh (x))^{3/2} \, dx}{\sqrt{-\sinh (x) \tanh (x)}}\\ &=-\frac{\sqrt{\sinh (x) \tanh (x)} \int (i \sinh (x))^{3/2} (i \tanh (x))^{3/2} \, dx}{\sqrt{i \sinh (x)} \sqrt{i \tanh (x)}}\\ &=\frac{2}{3} \sinh (x) \sqrt{\sinh (x) \tanh (x)}-\frac{\left (4 \sqrt{\sinh (x) \tanh (x)}\right ) \int \frac{(i \tanh (x))^{3/2}}{\sqrt{i \sinh (x)}} \, dx}{3 \sqrt{i \sinh (x)} \sqrt{i \tanh (x)}}\\ &=\frac{8}{3} \text{csch}(x) \sqrt{\sinh (x) \tanh (x)}+\frac{2}{3} \sinh (x) \sqrt{\sinh (x) \tanh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0673715, size = 23, normalized size = 0.74 $\frac{2}{3} \sinh (x) \left (4 \text{csch}^2(x)+1\right ) \sqrt{\sinh (x) \tanh (x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sinh[x]*Tanh[x])^(3/2),x]

[Out]

(2*(1 + 4*Csch[x]^2)*Sinh[x]*Sqrt[Sinh[x]*Tanh[x]])/3

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( x \right ) \tanh \left ( x \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(x)*tanh(x))^(3/2),x)

[Out]

int((sinh(x)*tanh(x))^(3/2),x)

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Maxima [B]  time = 1.62995, size = 93, normalized size = 3. \begin{align*} -\frac{\sqrt{2} e^{\left (\frac{3}{2} \, x\right )}}{6 \,{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}} - \frac{5 \, \sqrt{2} e^{\left (-\frac{1}{2} \, x\right )}}{2 \,{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}} - \frac{5 \, \sqrt{2} e^{\left (-\frac{5}{2} \, x\right )}}{2 \,{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}} - \frac{\sqrt{2} e^{\left (-\frac{9}{2} \, x\right )}}{6 \,{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sinh(x)*tanh(x))^(3/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(2)*e^(3/2*x)/(e^(-2*x) + 1)^(3/2) - 5/2*sqrt(2)*e^(-1/2*x)/(e^(-2*x) + 1)^(3/2) - 5/2*sqrt(2)*e^(-5/
2*x)/(e^(-2*x) + 1)^(3/2) - 1/6*sqrt(2)*e^(-9/2*x)/(e^(-2*x) + 1)^(3/2)

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Fricas [B]  time = 2.30302, size = 348, normalized size = 11.23 \begin{align*} \frac{\sqrt{\frac{1}{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} + 7\right )} \sinh \left (x\right )^{2} + 14 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + 7 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}}{3 \, \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 )}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sinh(x)*tanh(x))^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(1/2)*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 7)*sinh(x)^2 + 14*cosh(x)^2 + 4*
(cosh(x)^3 + 7*cosh(x))*sinh(x) + 1)/(sqrt(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*cosh(x)^2 + 1)*sin
h(x) + cosh(x))*(cosh(x) + sinh(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))**(3/2),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{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sinh(x)*tanh(x))^(3/2),x, algorithm="giac")

[Out]

integrate((sinh(x)*tanh(x))^(3/2), x)