3.67 \(\int \frac{1}{a+b \cosh (c+d x)} \, dx\)

Optimal. Leaf size=49 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d \sqrt{a-b} \sqrt{a+b}} \]

[Out]

(2*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d)

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Rubi [A]  time = 0.035172, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2659, 205} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^(-1),x]

[Out]

(2*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d)

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{a+b \cosh (c+d x)} \, dx &=-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b} d}\\ \end{align*}

Mathematica [A]  time = 0.0482518, size = 48, normalized size = 0.98 \[ -\frac{2 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{d \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[((a - b)*Tanh[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(Sqrt[-a^2 + b^2]*d)

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Maple [A]  time = 0.01, size = 44, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.25557, size = 585, normalized size = 11.94 \begin{align*} \left [\frac{\log \left (\frac{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} - b^{2} + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \,{\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) + b}\right )}{\sqrt{a^{2} - b^{2}} d}, -\frac{2 \, \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{a^{2} - b^{2}}\right )}{{\left (a^{2} - b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 - b^2 + 2*(b^2*cosh(d*x + c) + a
*b)*sinh(d*x + c) - 2*sqrt(a^2 - b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x
+ c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) + b))/(sqrt(a^2 - b^2)*d), -2*sqrt(-a^2 +
b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a)/(a^2 - b^2))/((a^2 - b^2)*d)]

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Sympy [A]  time = 15.5237, size = 163, normalized size = 3.33 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\cosh{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{b d} & \text{for}\: a = b \\- \frac{1}{b d \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}} & \text{for}\: a = - b \\\frac{x}{a + b \cosh{\left (c \right )}} & \text{for}\: d = 0 \\- \frac{\log{\left (- \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} + \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a d \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} - b d \sqrt{\frac{a}{a - b} + \frac{b}{a - b}}} + \frac{\log{\left (\sqrt{\frac{a}{a - b} + \frac{b}{a - b}} + \tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )} \right )}}{a d \sqrt{\frac{a}{a - b} + \frac{b}{a - b}} - b d \sqrt{\frac{a}{a - b} + \frac{b}{a - b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/cosh(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (tanh(c/2 + d*x/2)/(b*d), Eq(a, b)), (-1/(b*d*tanh(
c/2 + d*x/2)), Eq(a, -b)), (x/(a + b*cosh(c)), Eq(d, 0)), (-log(-sqrt(a/(a - b) + b/(a - b)) + tanh(c/2 + d*x/
2))/(a*d*sqrt(a/(a - b) + b/(a - b)) - b*d*sqrt(a/(a - b) + b/(a - b))) + log(sqrt(a/(a - b) + b/(a - b)) + ta
nh(c/2 + d*x/2))/(a*d*sqrt(a/(a - b) + b/(a - b)) - b*d*sqrt(a/(a - b) + b/(a - b))), True))

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Giac [A]  time = 1.22184, size = 53, normalized size = 1.08 \begin{align*} \frac{2 \, \arctan \left (\frac{b e^{\left (d x + c\right )} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan((b*e^(d*x + c) + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*d)