3.682 \(\int (-\cosh (x)+\text {sech}(x))^3 \, dx\)

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

[Out]

-5/2*arctan(sinh(x))+5/2*sinh(x)-5/6*sinh(x)^3+1/2*sinh(x)^3*tanh(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 = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4397, 2592, 288, 302, 203} \[ -\frac {5 \sinh ^3(x)}{6}+\frac {5 \sinh (x)}{2}+\frac {1}{2} \sinh ^3(x) \tanh ^2(x)-\frac {5}{2} \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(-Cosh[x] + Sech[x])^3,x]

[Out]

(-5*ArcTan[Sinh[x]])/2 + (5*Sinh[x])/2 - (5*Sinh[x]^3)/6 + (Sinh[x]^3*Tanh[x]^2)/2

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 (-\cosh (x)+\text {sech}(x))^3 \, dx &=-\int \sinh ^3(x) \tanh ^3(x) \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2} \, dx,x,\sinh (x)\right )\\ &=\frac {1}{2} \sinh ^3(x) \tanh ^2(x)-\frac {5}{2} \operatorname {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=\frac {1}{2} \sinh ^3(x) \tanh ^2(x)-\frac {5}{2} \operatorname {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\sinh (x)\right )\\ &=\frac {5 \sinh (x)}{2}-\frac {5 \sinh ^3(x)}{6}+\frac {1}{2} \sinh ^3(x) \tanh ^2(x)-\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac {5}{2} \tan ^{-1}(\sinh (x))+\frac {5 \sinh (x)}{2}-\frac {5 \sinh ^3(x)}{6}+\frac {1}{2} \sinh ^3(x) \tanh ^2(x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 37, normalized size = 1.09 \[ -\frac {1}{48} \text {sech}^2(x) \left (-50 \sinh (x)-25 \sinh (3 x)+\sinh (5 x)+60 \tan ^{-1}(\sinh (x))+60 \cosh (2 x) \tan ^{-1}(\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-Cosh[x] + Sech[x])^3,x]

[Out]

-1/48*(Sech[x]^2*(60*ArcTan[Sinh[x]] + 60*ArcTan[Sinh[x]]*Cosh[2*x] - 50*Sinh[x] - 25*Sinh[3*x] + Sinh[5*x]))

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fricas [B]  time = 0.45, size = 486, normalized size = 14.29 \[ -\frac {\cosh \relax (x)^{10} + 10 \, \cosh \relax (x) \sinh \relax (x)^{9} + \sinh \relax (x)^{10} + 5 \, {\left (9 \, \cosh \relax (x)^{2} - 5\right )} \sinh \relax (x)^{8} - 25 \, \cosh \relax (x)^{8} + 40 \, {\left (3 \, \cosh \relax (x)^{3} - 5 \, \cosh \relax (x)\right )} \sinh \relax (x)^{7} + 10 \, {\left (21 \, \cosh \relax (x)^{4} - 70 \, \cosh \relax (x)^{2} - 5\right )} \sinh \relax (x)^{6} - 50 \, \cosh \relax (x)^{6} + 4 \, {\left (63 \, \cosh \relax (x)^{5} - 350 \, \cosh \relax (x)^{3} - 75 \, \cosh \relax (x)\right )} \sinh \relax (x)^{5} + 10 \, {\left (21 \, \cosh \relax (x)^{6} - 175 \, \cosh \relax (x)^{4} - 75 \, \cosh \relax (x)^{2} + 5\right )} \sinh \relax (x)^{4} + 50 \, \cosh \relax (x)^{4} + 40 \, {\left (3 \, \cosh \relax (x)^{7} - 35 \, \cosh \relax (x)^{5} - 25 \, \cosh \relax (x)^{3} + 5 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 5 \, {\left (9 \, \cosh \relax (x)^{8} - 140 \, \cosh \relax (x)^{6} - 150 \, \cosh \relax (x)^{4} + 60 \, \cosh \relax (x)^{2} + 5\right )} \sinh \relax (x)^{2} + 120 \, {\left (\cosh \relax (x)^{7} + 7 \, \cosh \relax (x) \sinh \relax (x)^{6} + \sinh \relax (x)^{7} + {\left (21 \, \cosh \relax (x)^{2} + 2\right )} \sinh \relax (x)^{5} + 2 \, \cosh \relax (x)^{5} + 5 \, {\left (7 \, \cosh \relax (x)^{3} + 2 \, \cosh \relax (x)\right )} \sinh \relax (x)^{4} + {\left (35 \, \cosh \relax (x)^{4} + 20 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{3} + \cosh \relax (x)^{3} + {\left (21 \, \cosh \relax (x)^{5} + 20 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{2} + {\left (7 \, \cosh \relax (x)^{6} + 10 \, \cosh \relax (x)^{4} + 3 \, \cosh \relax (x)^{2}\right )} \sinh \relax (x)\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + 25 \, \cosh \relax (x)^{2} + 10 \, {\left (\cosh \relax (x)^{9} - 20 \, \cosh \relax (x)^{7} - 30 \, \cosh \relax (x)^{5} + 20 \, \cosh \relax (x)^{3} + 5 \, \cosh \relax (x)\right )} \sinh \relax (x) - 1}{24 \, {\left (\cosh \relax (x)^{7} + 7 \, \cosh \relax (x) \sinh \relax (x)^{6} + \sinh \relax (x)^{7} + {\left (21 \, \cosh \relax (x)^{2} + 2\right )} \sinh \relax (x)^{5} + 2 \, \cosh \relax (x)^{5} + 5 \, {\left (7 \, \cosh \relax (x)^{3} + 2 \, \cosh \relax (x)\right )} \sinh \relax (x)^{4} + {\left (35 \, \cosh \relax (x)^{4} + 20 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{3} + \cosh \relax (x)^{3} + {\left (21 \, \cosh \relax (x)^{5} + 20 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{2} + {\left (7 \, \cosh \relax (x)^{6} + 10 \, \cosh \relax (x)^{4} + 3 \, \cosh \relax (x)^{2}\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cosh(x)+sech(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*c
osh(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*co
sh(x)^6 - 150*cosh(x)^4 + 60*cosh(x)^2 + 5)*sinh(x)^2 + 120*(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 + 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))*arctan(cosh(x) + sinh(x)) + 25*cosh(x)^2 + 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)*
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))

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giac [B]  time = 0.12, size = 66, normalized size = 1.94 \[ -\frac {5}{4} \, \pi + \frac {1}{24} \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - \frac {e^{\left (-x\right )} - e^{x}}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4} - \frac {5}{2} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - e^{\left (-x\right )} + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cosh(x)+sech(x))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.46, size = 29, normalized size = 0.85 \[ -\left (\frac {2}{3}+\frac {\left (\cosh ^{2}\relax (x )\right )}{3}\right ) \sinh \relax (x )+3 \sinh \relax (x )-5 \arctan \left ({\mathrm e}^{x}\right )+\frac {\mathrm {sech}\relax (x ) \tanh \relax (x )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-cosh(x)+sech(x))^3,x)

[Out]

-(2/3+1/3*cosh(x)^2)*sinh(x)+3*sinh(x)-5*arctan(exp(x))+1/2*sech(x)*tanh(x)

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maxima [B]  time = 0.99, size = 56, normalized size = 1.65 \[ \frac {e^{\left (-x\right )} - e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + 5 \, \arctan \left (e^{\left (-x\right )}\right ) - \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {9}{8} \, e^{\left (-x\right )} + \frac {1}{24} \, e^{\left (-3 \, x\right )} + \frac {9}{8} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cosh(x)+sech(x))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 57, normalized size = 1.68 \[ \frac {{\mathrm {e}}^{-3\,x}}{24}-\frac {9\,{\mathrm {e}}^{-x}}{8}-\frac {{\mathrm {e}}^{3\,x}}{24}-5\,\mathrm {atan}\left ({\mathrm {e}}^x\right )+\frac {9\,{\mathrm {e}}^x}{8}+\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}+1}-\frac {2\,{\mathrm {e}}^x}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(cosh(x) - 1/cosh(x))^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int 3 \cosh {\relax (x )} \operatorname {sech}^{2}{\relax (x )}\, dx - \int \left (- 3 \cosh ^{2}{\relax (x )} \operatorname {sech}{\relax (x )}\right )\, dx - \int \cosh ^{3}{\relax (x )}\, dx - \int \left (- \operatorname {sech}^{3}{\relax (x )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-cosh(x)+sech(x))**3,x)

[Out]

-Integral(3*cosh(x)*sech(x)**2, x) - Integral(-3*cosh(x)**2*sech(x), x) - Integral(cosh(x)**3, x) - Integral(-
sech(x)**3, x)

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