### 3.610 $$\int \frac{1}{(a \cosh (c+d x)-a \sinh (c+d x))^2} \, dx$$

Optimal. Leaf size=27 $\frac{1}{2 d (a \cosh (c+d x)-a \sinh (c+d x))^2}$

[Out]

1/(2*d*(a*Cosh[c + d*x] - a*Sinh[c + d*x])^2)

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Rubi [A]  time = 0.0153704, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.05, Rules used = {3071} $\frac{1}{2 d (a \cosh (c+d x)-a \sinh (c+d x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Cosh[c + d*x] - a*Sinh[c + d*x])^(-2),x]

[Out]

1/(2*d*(a*Cosh[c + d*x] - a*Sinh[c + d*x])^2)

Rule 3071

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

Rubi steps

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

Mathematica [A]  time = 0.044446, size = 27, normalized size = 1. $\frac{1}{2 d (a \cosh (c+d x)-a \sinh (c+d x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Cosh[c + d*x] - a*Sinh[c + d*x])^(-2),x]

[Out]

1/(2*d*(a*Cosh[c + d*x] - a*Sinh[c + d*x])^2)

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Maple [A]  time = 0.001, size = 26, normalized size = 1. \begin{align*}{\frac{1}{2\,d{a}^{2} \left ( \cosh \left ( dx+c \right ) -\sinh \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(d*x+c)-a*sinh(d*x+c))^2,x)

[Out]

1/2/d/a^2/(cosh(d*x+c)-sinh(d*x+c))^2

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Maxima [A]  time = 1.06942, size = 23, normalized size = 0.85 \begin{align*} \frac{e^{\left (2 \, d x + 2 \, c\right )}}{2 \, a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(d*x+c)-a*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*e^(2*d*x + 2*c)/(a^2*d)

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Fricas [A]  time = 2.17896, size = 109, normalized size = 4.04 \begin{align*} \frac{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}{2 \,{\left (a^{2} d \cosh \left (d x + c\right ) - a^{2} d \sinh \left (d x + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(d*x+c)-a*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(cosh(d*x + c) + sinh(d*x + c))/(a^2*d*cosh(d*x + c) - a^2*d*sinh(d*x + c))

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Sympy [A]  time = 0.923225, size = 65, normalized size = 2.41 \begin{align*} \begin{cases} \frac{1}{2 a^{2} d \sinh ^{2}{\left (c + d x \right )} - 4 a^{2} d \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )} + 2 a^{2} d \cosh ^{2}{\left (c + d x \right )}} & \text{for}\: d \neq 0 \\\frac{x}{\left (- a \sinh{\left (c \right )} + a \cosh{\left (c \right )}\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(d*x+c)-a*sinh(d*x+c))**2,x)

[Out]

Piecewise((1/(2*a**2*d*sinh(c + d*x)**2 - 4*a**2*d*sinh(c + d*x)*cosh(c + d*x) + 2*a**2*d*cosh(c + d*x)**2), N
e(d, 0)), (x/(-a*sinh(c) + a*cosh(c))**2, True))

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Giac [A]  time = 1.11292, size = 23, normalized size = 0.85 \begin{align*} \frac{e^{\left (2 \, d x + 2 \, c\right )}}{2 \, a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(d*x+c)-a*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*e^(2*d*x + 2*c)/(a^2*d)