\(\int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx\) [600]

Optimal result

Integrand size = 19, antiderivative size = 24 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {1}{d (a \cosh (c+d x)+a \sinh (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {3150} \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {1}{d (a \sinh (c+d x)+a \cosh (c+d x))} \]

[In]

Int[(a*Cosh[c + d*x] + a*Sinh[c + d*x])^(-1),x]

[Out]

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

Rule 3150

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*} \text {integral}& = -\frac {1}{d (a \cosh (c+d x)+a \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {1}{d (a \cosh (c+d x)+a \sinh (c+d x))} \]

[In]

Integrate[(a*Cosh[c + d*x] + a*Sinh[c + d*x])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {{\mathrm e}^{-d x -c}}{a d}\)	\(18\)
gosper	\(-\frac {1}{d a \left (\cosh \left (d x +c \right )+\sinh \left (d x +c \right )\right )}\)	\(24\)
derivativedivides	\(-\frac {1}{d \left (a \cosh \left (d x +c \right )+a \sinh \left (d x +c \right )\right )}\)	\(25\)
default	\(-\frac {1}{d \left (a \cosh \left (d x +c \right )+a \sinh \left (d x +c \right )\right )}\)	\(25\)

[In]

int(1/(a*cosh(d*x+c)+a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/a/d*exp(-d*x-c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {1}{a d \cosh \left (d x + c\right ) + a d \sinh \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=\begin {cases} - \frac {1}{a d \sinh {\left (c + d x \right )} + a d \cosh {\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x}{a \sinh {\left (c \right )} + a \cosh {\left (c \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-1/(a*d*sinh(c + d*x) + a*d*cosh(c + d*x)), Ne(d, 0)), (x/(a*sinh(c) + a*cosh(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {e^{\left (-d x - c\right )}}{a d} \]

[In]

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

[Out]

-e^(-d*x - c)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {e^{\left (-d x - c\right )}}{a d} \]

[In]

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

[Out]

-e^(-d*x - c)/(a*d)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1}{a \cosh (c+d x)+a \sinh (c+d x)} \, dx=-\frac {{\mathrm {e}}^{-c-d\,x}}{a\,d} \]

[In]

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

[Out]

-exp(- c - d*x)/(a*d)