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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.00 \[ \frac {1}{2 d (a \cosh (c+d x)-a \sinh (c+d x))^2} \]

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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fricas [A]  time = 0.83, size = 41, normalized size = 1.52 \[ \frac {\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )}{2 \, {\left (a^{2} d \cosh \left (d x + c\right ) - a^{2} d \sinh \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.12, size = 17, normalized size = 0.63 \[ \frac {e^{\left (2 \, d x + 2 \, c\right )}}{2 \, a^{2} d} \]

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)

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maple [A]  time = 0.02, size = 26, normalized size = 0.96 \[ \frac {1}{2 d \,a^{2} \left (\cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )^{2}} \]

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 = 0.35, size = 17, normalized size = 0.63 \[ \frac {e^{\left (2 \, d x + 2 \, c\right )}}{2 \, a^{2} d} \]

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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mupad [B]  time = 0.10, size = 17, normalized size = 0.63 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*c + 2*d*x)/(2*a^2*d)

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sympy [A]  time = 0.62, size = 65, normalized size = 2.41 \[ \begin {cases} \frac {1}{2 a^{2} d \sinh ^{2}{\left (c + d x \right )} - 4 a^{2} d \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )} + 2 a^{2} d \cosh ^{2}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x}{\left (- a \sinh {\relax (c )} + a \cosh {\relax (c )}\right )^{2}} & \text {otherwise} \end {cases} \]

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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