3.34 \(\int \frac{1}{(1+\cosh (c+d x))^3} \, dx\)

Optimal. Leaf size=70 \[ \frac{2 \sinh (c+d x)}{15 d (\cosh (c+d x)+1)}+\frac{2 \sinh (c+d x)}{15 d (\cosh (c+d x)+1)^2}+\frac{\sinh (c+d x)}{5 d (\cosh (c+d x)+1)^3} \]

[Out]

Sinh[c + d*x]/(5*d*(1 + Cosh[c + d*x])^3) + (2*Sinh[c + d*x])/(15*d*(1 + Cosh[c + d*x])^2) + (2*Sinh[c + d*x])
/(15*d*(1 + Cosh[c + d*x]))

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Rubi [A]  time = 0.036479, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2650, 2648} \[ \frac{2 \sinh (c+d x)}{15 d (\cosh (c+d x)+1)}+\frac{2 \sinh (c+d x)}{15 d (\cosh (c+d x)+1)^2}+\frac{\sinh (c+d x)}{5 d (\cosh (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cosh[c + d*x])^(-3),x]

[Out]

Sinh[c + d*x]/(5*d*(1 + Cosh[c + d*x])^3) + (2*Sinh[c + d*x])/(15*d*(1 + Cosh[c + d*x])^2) + (2*Sinh[c + d*x])
/(15*d*(1 + Cosh[c + d*x]))

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(1+\cosh (c+d x))^3} \, dx &=\frac{\sinh (c+d x)}{5 d (1+\cosh (c+d x))^3}+\frac{2}{5} \int \frac{1}{(1+\cosh (c+d x))^2} \, dx\\ &=\frac{\sinh (c+d x)}{5 d (1+\cosh (c+d x))^3}+\frac{2 \sinh (c+d x)}{15 d (1+\cosh (c+d x))^2}+\frac{2}{15} \int \frac{1}{1+\cosh (c+d x)} \, dx\\ &=\frac{\sinh (c+d x)}{5 d (1+\cosh (c+d x))^3}+\frac{2 \sinh (c+d x)}{15 d (1+\cosh (c+d x))^2}+\frac{2 \sinh (c+d x)}{15 d (1+\cosh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.055126, size = 44, normalized size = 0.63 \[ \frac{15 \sinh (c+d x)+6 \sinh (2 (c+d x))+\sinh (3 (c+d x))}{30 d (\cosh (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cosh[c + d*x])^(-3),x]

[Out]

(15*Sinh[c + d*x] + 6*Sinh[2*(c + d*x)] + Sinh[3*(c + d*x)])/(30*d*(1 + Cosh[c + d*x])^3)

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Maple [A]  time = 0.009, size = 43, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({\frac{1}{20} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{6} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{1}{4}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+cosh(d*x+c))^3,x)

[Out]

1/d*(1/20*tanh(1/2*d*x+1/2*c)^5-1/6*tanh(1/2*d*x+1/2*c)^3+1/4*tanh(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.05483, size = 277, normalized size = 3.96 \begin{align*} \frac{4 \, e^{\left (-d x - c\right )}}{3 \, d{\left (5 \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{3 \, d{\left (5 \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} + \frac{4}{15 \, d{\left (5 \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^3,x, algorithm="maxima")

[Out]

4/3*e^(-d*x - c)/(d*(5*e^(-d*x - c) + 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) + 5*e^(-4*d*x - 4*c) + e^(-5*d
*x - 5*c) + 1)) + 8/3*e^(-2*d*x - 2*c)/(d*(5*e^(-d*x - c) + 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) + 5*e^(-
4*d*x - 4*c) + e^(-5*d*x - 5*c) + 1)) + 4/15/(d*(5*e^(-d*x - c) + 10*e^(-2*d*x - 2*c) + 10*e^(-3*d*x - 3*c) +
5*e^(-4*d*x - 4*c) + e^(-5*d*x - 5*c) + 1))

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Fricas [B]  time = 1.79728, size = 486, normalized size = 6.94 \begin{align*} -\frac{4 \,{\left (11 \, \cosh \left (d x + c\right ) + 9 \, \sinh \left (d x + c\right ) + 5\right )}}{15 \,{\left (d \cosh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} +{\left (4 \, d \cosh \left (d x + c\right ) + 5 \, d\right )} \sinh \left (d x + c\right )^{3} + 10 \, d \cosh \left (d x + c\right )^{2} +{\left (6 \, d \cosh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) + 10 \, d\right )} \sinh \left (d x + c\right )^{2} + 11 \, d \cosh \left (d x + c\right ) +{\left (4 \, d \cosh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 20 \, d \cosh \left (d x + c\right ) + 9 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^3,x, algorithm="fricas")

[Out]

-4/15*(11*cosh(d*x + c) + 9*sinh(d*x + c) + 5)/(d*cosh(d*x + c)^4 + d*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 +
(4*d*cosh(d*x + c) + 5*d)*sinh(d*x + c)^3 + 10*d*cosh(d*x + c)^2 + (6*d*cosh(d*x + c)^2 + 15*d*cosh(d*x + c) +
10*d)*sinh(d*x + c)^2 + 11*d*cosh(d*x + c) + (4*d*cosh(d*x + c)^3 + 15*d*cosh(d*x + c)^2 + 20*d*cosh(d*x + c)
+ 9*d)*sinh(d*x + c) + 5*d)

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Sympy [A]  time = 3.49056, size = 51, normalized size = 0.73 \begin{align*} \begin{cases} \frac{\tanh ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 d} - \frac{\tanh ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{6 d} + \frac{\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 d} & \text{for}\: d \neq 0 \\\frac{x}{\left (\cosh{\left (c \right )} + 1\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))**3,x)

[Out]

Piecewise((tanh(c/2 + d*x/2)**5/(20*d) - tanh(c/2 + d*x/2)**3/(6*d) + tanh(c/2 + d*x/2)/(4*d), Ne(d, 0)), (x/(
cosh(c) + 1)**3, True))

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Giac [A]  time = 1.14721, size = 49, normalized size = 0.7 \begin{align*} -\frac{4 \,{\left (10 \, e^{\left (2 \, d x + 2 \, c\right )} + 5 \, e^{\left (d x + c\right )} + 1\right )}}{15 \, d{\left (e^{\left (d x + c\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^3,x, algorithm="giac")

[Out]

-4/15*(10*e^(2*d*x + 2*c) + 5*e^(d*x + c) + 1)/(d*(e^(d*x + c) + 1)^5)