3.68 \(\int \frac{1}{(a+b \cosh (c+d x))^2} \, dx\)

Optimal. Leaf size=86 \[ \frac{2 a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))} \]

[Out]

(2*a*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) - (b*Sinh[c + d*x])
/((a^2 - b^2)*d*(a + b*Cosh[c + d*x]))

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Rubi [A]  time = 0.0840867, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2664, 12, 2659, 205} \[ \frac{2 a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^(-2),x]

[Out]

(2*a*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) - (b*Sinh[c + d*x])
/((a^2 - b^2)*d*(a + b*Cosh[c + d*x]))

Rule 2664

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^2} \, dx &=-\frac{b \sinh (c+d x)}{\left (a^2-b^2\right ) d (a+b \cosh (c+d x))}-\frac{\int \frac{a}{a+b \cosh (c+d x)} \, dx}{-a^2+b^2}\\ &=-\frac{b \sinh (c+d x)}{\left (a^2-b^2\right ) d (a+b \cosh (c+d x))}+\frac{a \int \frac{1}{a+b \cosh (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac{b \sinh (c+d x)}{\left (a^2-b^2\right ) d (a+b \cosh (c+d x))}-\frac{(2 i a) \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 )}{\left (a^2-b^2\right ) d}\\ &=\frac{2 a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac{b \sinh (c+d x)}{\left (a^2-b^2\right ) d (a+b \cosh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.206597, size = 84, normalized size = 0.98 \[ \frac{\frac{2 a \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac{b \sinh (c+d x)}{(a-b) (a+b) (a+b \cosh (c+d x))}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*ArcTan[((a - b)*Tanh[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - (b*Sinh[c + d*x])/((a - b)*(a
+ b)*(a + b*Cosh[c + d*x])))/d

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Maple [A]  time = 0.02, size = 118, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( 2\,{\frac{b\tanh \left ( 1/2\,dx+c/2 \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \left ( a \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{a}{ \left ( a+b \right ) \left ( a-b \right ) \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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(2*b/(a^2-b^2)*tanh(1/2*d*x+1/2*c)/(a*tanh(1/2*d*x+1/2*c)^2-tanh(1/2*d*x+1/2*c)^2*b-a-b)+2*a/(a+b)/(a-b)/(
(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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.41934, size = 1787, normalized size = 20.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^2*b - 2*b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) + a*b + 2*(a*b*cosh(d*x +
c) + a^2)*sinh(d*x + c))*sqrt(a^2 - b^2)*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)) + 2*(a^3 - a*b^2)*cosh(d*x + c) + 2*(a^3 - a*b^2)*sinh(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) + (a^4*b - 2*
a^2*b^3 + b^5)*d + 2*((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c)),
2*(a^2*b - b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) + a*b + 2*(a*b*cosh(d*x + c
) + a^2)*sinh(d*x + c))*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^3 - a*b^2)*cosh(d*x + c) + (a^3 - a*b^2)*sinh(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x +
c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*d*sinh(d*x + c)^2 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) + (a^4*b - 2*
a^2*b^3 + b^5)*d + 2*((a^4*b - 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)*sinh(d*x + c))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.23238, size = 138, normalized size = 1.6 \begin{align*} \frac{2 \, a \arctan \left (\frac{b e^{\left (d x + c\right )} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{{\left (a^{2} d - b^{2} d\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} d - b^{2} d\right )}{\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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