3.676 \(\int (\text{csch}(x)+\sinh (x))^3 \, dx\)

Optimal. Leaf size=34 \[ \frac{5 \cosh ^3(x)}{6}+\frac{5 \cosh (x)}{2}-\frac{1}{2} \cosh ^3(x) \coth ^2(x)-\frac{5}{2} \tanh ^{-1}(\cosh (x)) \]

[Out]

(-5*ArcTanh[Cosh[x]])/2 + (5*Cosh[x])/2 + (5*Cosh[x]^3)/6 - (Cosh[x]^3*Coth[x]^2)/2

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Rubi [A]  time = 0.0517341, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {4397, 2592, 288, 302, 206} \[ \frac{5 \cosh ^3(x)}{6}+\frac{5 \cosh (x)}{2}-\frac{1}{2} \cosh ^3(x) \coth ^2(x)-\frac{5}{2} \tanh ^{-1}(\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Csch[x] + Sinh[x])^3,x]

[Out]

(-5*ArcTanh[Cosh[x]])/2 + (5*Cosh[x])/2 + (5*Cosh[x]^3)/6 - (Cosh[x]^3*Coth[x]^2)/2

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (\text{csch}(x)+\sinh (x))^3 \, dx &=\int \cosh ^3(x) \coth ^3(x) \, dx\\ &=\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cosh (x)\right )\\ &=-\frac{1}{2} \cosh ^3(x) \coth ^2(x)-\frac{5}{2} \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cosh (x)\right )\\ &=-\frac{1}{2} \cosh ^3(x) \coth ^2(x)-\frac{5}{2} \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cosh (x)\right )\\ &=\frac{5 \cosh (x)}{2}+\frac{5 \cosh ^3(x)}{6}-\frac{1}{2} \cosh ^3(x) \coth ^2(x)-\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cosh (x)\right )\\ &=-\frac{5}{2} \tanh ^{-1}(\cosh (x))+\frac{5 \cosh (x)}{2}+\frac{5 \cosh ^3(x)}{6}-\frac{1}{2} \cosh ^3(x) \coth ^2(x)\\ \end{align*}

Mathematica [A]  time = 0.0682244, size = 45, normalized size = 1.32 \[ \frac{1}{48} \text{csch}^2(x) \left (-50 \cosh (x)+25 \cosh (3 x)+\cosh (5 x)-60 \log \left (\tanh \left (\frac{x}{2}\right )\right )+60 \cosh (2 x) \log \left (\tanh \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Csch[x] + Sinh[x])^3,x]

[Out]

(Csch[x]^2*(-50*Cosh[x] + 25*Cosh[3*x] + Cosh[5*x] - 60*Log[Tanh[x/2]] + 60*Cosh[2*x]*Log[Tanh[x/2]]))/48

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Maple [A]  time = 0.017, size = 28, normalized size = 0.8 \begin{align*} -{\frac{{\rm csch} \left (x\right ){\rm coth} \left (x\right )}{2}}-5\,{\it Artanh} \left ({{\rm e}^{x}} \right ) +3\,\cosh \left ( x \right ) + \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((csch(x)+sinh(x))^3,x)

[Out]

-1/2*csch(x)*coth(x)-5*arctanh(exp(x))+3*cosh(x)+(-2/3+1/3*sinh(x)^2)*cosh(x)

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Maxima [B]  time = 1.13593, size = 90, normalized size = 2.65 \begin{align*} \frac{e^{\left (-x\right )} + e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + \frac{1}{24} \, e^{\left (3 \, x\right )} + \frac{9}{8} \, e^{\left (-x\right )} + \frac{1}{24} \, e^{\left (-3 \, x\right )} + \frac{9}{8} \, e^{x} - \frac{5}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{5}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((csch(x)+sinh(x))^3,x, algorithm="maxima")

[Out]

(e^(-x) + e^(-3*x))/(2*e^(-2*x) - e^(-4*x) - 1) + 1/24*e^(3*x) + 9/8*e^(-x) + 1/24*e^(-3*x) + 9/8*e^x - 5/2*lo
g(e^(-x) + 1) + 5/2*log(e^(-x) - 1)

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Fricas [B]  time = 1.89613, size = 2090, normalized size = 61.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((csch(x)+sinh(x))^3,x, algorithm="fricas")

[Out]

1/24*(cosh(x)^10 + 10*cosh(x)*sinh(x)^9 + sinh(x)^10 + 5*(9*cosh(x)^2 + 5)*sinh(x)^8 + 25*cosh(x)^8 + 40*(3*co
sh(x)^3 + 5*cosh(x))*sinh(x)^7 + 10*(21*cosh(x)^4 + 70*cosh(x)^2 - 5)*sinh(x)^6 - 50*cosh(x)^6 + 4*(63*cosh(x)
^5 + 350*cosh(x)^3 - 75*cosh(x))*sinh(x)^5 + 10*(21*cosh(x)^6 + 175*cosh(x)^4 - 75*cosh(x)^2 - 5)*sinh(x)^4 -
50*cosh(x)^4 + 40*(3*cosh(x)^7 + 35*cosh(x)^5 - 25*cosh(x)^3 - 5*cosh(x))*sinh(x)^3 + 5*(9*cosh(x)^8 + 140*cos
h(x)^6 - 150*cosh(x)^4 - 60*cosh(x)^2 + 5)*sinh(x)^2 + 25*cosh(x)^2 - 60*(cosh(x)^7 + 7*cosh(x)*sinh(x)^6 + si
nh(x)^7 + (21*cosh(x)^2 - 2)*sinh(x)^5 - 2*cosh(x)^5 + 5*(7*cosh(x)^3 - 2*cosh(x))*sinh(x)^4 + (35*cosh(x)^4 -
20*cosh(x)^2 + 1)*sinh(x)^3 + cosh(x)^3 + (21*cosh(x)^5 - 20*cosh(x)^3 + 3*cosh(x))*sinh(x)^2 + (7*cosh(x)^6
- 10*cosh(x)^4 + 3*cosh(x)^2)*sinh(x))*log(cosh(x) + sinh(x) + 1) + 60*(cosh(x)^7 + 7*cosh(x)*sinh(x)^6 + sinh
(x)^7 + (21*cosh(x)^2 - 2)*sinh(x)^5 - 2*cosh(x)^5 + 5*(7*cosh(x)^3 - 2*cosh(x))*sinh(x)^4 + (35*cosh(x)^4 - 2
0*cosh(x)^2 + 1)*sinh(x)^3 + cosh(x)^3 + (21*cosh(x)^5 - 20*cosh(x)^3 + 3*cosh(x))*sinh(x)^2 + (7*cosh(x)^6 -
10*cosh(x)^4 + 3*cosh(x)^2)*sinh(x))*log(cosh(x) + sinh(x) - 1) + 10*(cosh(x)^9 + 20*cosh(x)^7 - 30*cosh(x)^5
- 20*cosh(x)^3 + 5*cosh(x))*sinh(x) + 1)/(cosh(x)^7 + 7*cosh(x)*sinh(x)^6 + sinh(x)^7 + (21*cosh(x)^2 - 2)*sin
h(x)^5 - 2*cosh(x)^5 + 5*(7*cosh(x)^3 - 2*cosh(x))*sinh(x)^4 + (35*cosh(x)^4 - 20*cosh(x)^2 + 1)*sinh(x)^3 + c
osh(x)^3 + (21*cosh(x)^5 - 20*cosh(x)^3 + 3*cosh(x))*sinh(x)^2 + (7*cosh(x)^6 - 10*cosh(x)^4 + 3*cosh(x)^2)*si
nh(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((csch(x)+sinh(x))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.12487, size = 84, normalized size = 2.47 \begin{align*} \frac{1}{24} \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - \frac{e^{\left (-x\right )} + e^{x}}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4} + e^{\left (-x\right )} + e^{x} - \frac{5}{4} \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \frac{5}{4} \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((csch(x)+sinh(x))^3,x, algorithm="giac")

[Out]

1/24*(e^(-x) + e^x)^3 - (e^(-x) + e^x)/((e^(-x) + e^x)^2 - 4) + e^(-x) + e^x - 5/4*log(e^(-x) + e^x + 2) + 5/4
*log(e^(-x) + e^x - 2)