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]

-5/2*arctanh(cosh(x))+5/2*cosh(x)+5/6*cosh(x)^3-1/2*cosh(x)^3*coth(x)^2

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, 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 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])

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 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 4397

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

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*}

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Mathematica [A]  time = 0.07, 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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fricas [B]  time = 0.43, size = 616, normalized size = 18.12 \[ \text {result too large to display} \]

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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giac [B]  time = 0.11, size = 62, normalized size = 1.82 \[ \frac {1}{24} \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - \frac {e^{\left (-x\right )} + e^{x}}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4} + e^{\left (-x\right )} + e^{x} - \frac {5}{4} \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \frac {5}{4} \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) \]

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)

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maple [A]  time = 0.47, size = 28, normalized size = 0.82 \[ -\frac {\coth \relax (x ) \mathrm {csch}\relax (x )}{2}-5 \arctanh \left ({\mathrm e}^{x}\right )+3 \cosh \relax (x )+\left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\relax (x )\right )}{3}\right ) \cosh \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.37, size = 67, normalized size = 1.97 \[ \frac {e^{\left (-x\right )} + e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + \frac {1}{24} \, e^{\left (3 \, x\right )} + \frac {9}{8} \, e^{\left (-x\right )} + \frac {1}{24} \, e^{\left (-3 \, x\right )} + \frac {9}{8} \, e^{x} - \frac {5}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {5}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \]

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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mupad [B]  time = 0.06, size = 71, normalized size = 2.09 \[ \frac {5\,\ln \left (5-5\,{\mathrm {e}}^x\right )}{2}-\frac {5\,\ln \left (-5\,{\mathrm {e}}^x-5\right )}{2}+\frac {9\,{\mathrm {e}}^{-x}}{8}+\frac {{\mathrm {e}}^{-3\,x}}{24}+\frac {{\mathrm {e}}^{3\,x}}{24}+\frac {9\,{\mathrm {e}}^x}{8}-\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1}-\frac {2\,{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(x) + 1/sinh(x))^3,x)

[Out]

(5*log(5 - 5*exp(x)))/2 - (5*log(- 5*exp(x) - 5))/2 + (9*exp(-x))/8 + exp(-3*x)/24 + exp(3*x)/24 + (9*exp(x))/
8 - exp(x)/(exp(2*x) - 1) - (2*exp(x))/(exp(4*x) - 2*exp(2*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\sinh {\relax (x )} + \operatorname {csch}{\relax (x )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((csch(x)+sinh(x))**3,x)

[Out]

Integral((sinh(x) + csch(x))**3, x)

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