3.205 \(\int \frac{A+B \text{csch}(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=99 \[ \frac{b B \log (a+b \cosh (x))}{a^2-b^2}+\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cosh (x))}{2 (a+b)}-\frac{B \log (\cosh (x)+1)}{2 (a-b)} \]

[Out]

(2*A*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cosh[x]])/(2*(a + b)
) - (B*Log[1 + Cosh[x]])/(2*(a - b)) + (b*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.305051, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {4225, 4401, 2659, 208, 2668, 706, 31, 633} \[ \frac{b B \log (a+b \cosh (x))}{a^2-b^2}+\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cosh (x))}{2 (a+b)}-\frac{B \log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Csch[x])/(a + b*Cosh[x]),x]

[Out]

(2*A*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cosh[x]])/(2*(a + b)
) - (B*Log[1 + Cosh[x]])/(2*(a - b)) + (b*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

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 208

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rubi steps

\begin{align*} \int \frac{A+B \text{csch}(x)}{a+b \cosh (x)} \, dx &=-\left (i \int \frac{\text{csch}(x) (i B+i A \sinh (x))}{a+b \cosh (x)} \, dx\right )\\ &=\int \left (\frac{A}{a+b \cosh (x)}+\frac{B \text{csch}(x)}{a+b \cosh (x)}\right ) \, dx\\ &=A \int \frac{1}{a+b \cosh (x)} \, dx+B \int \frac{\text{csch}(x)}{a+b \cosh (x)} \, dx\\ &=(2 A) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )-(b B) \operatorname{Subst}\left (\int \frac{1}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \cosh (x)\right )\\ &=\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \cosh (x)\right )}{a^2-b^2}+\frac{(b B) \operatorname{Subst}\left (\int \frac{-a+x}{b^2-x^2} \, dx,x,b \cosh (x)\right )}{a^2-b^2}\\ &=\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{b B \log (a+b \cosh (x))}{a^2-b^2}+\frac{B \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \cosh (x)\right )}{2 (a-b)}-\frac{B \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \cosh (x)\right )}{2 (a+b)}\\ &=\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cosh (x))}{2 (a+b)}-\frac{B \log (1+\cosh (x))}{2 (a-b)}+\frac{b B \log (a+b \cosh (x))}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.187558, size = 81, normalized size = 0.82 \[ \frac{B \left (b \log (a+b \cosh (x))+a \log \left (\tanh \left (\frac{x}{2}\right )\right )-b \log (\sinh (x))\right )}{a^2-b^2}-\frac{2 A \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Csch[x])/(a + b*Cosh[x]),x]

[Out]

(-2*A*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (B*(b*Log[a + b*Cosh[x]] - b*Log[Sinh[x
]] + a*Log[Tanh[x/2]]))/(a^2 - b^2)

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Maple [A]  time = 0.025, size = 138, normalized size = 1.4 \begin{align*}{\frac{Bb}{ \left ( a+b \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-a-b \right ) }+2\,{\frac{Aa}{ \left ( a+b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{Ab}{ \left ( a+b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{B}{a+b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*csch(x))/(a+b*cosh(x)),x)

[Out]

1/(a+b)*B*b/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)+2/(a+b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x
)/((a+b)*(a-b))^(1/2))*A*a+2/(a+b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*A*b+B/(a
+b)*ln(tanh(1/2*x))

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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((A+B*csch(x))/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 22.6808, size = 840, normalized size = 8.48 \begin{align*} \left [\frac{B b \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \sqrt{a^{2} - b^{2}} A \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} - b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b}\right ) -{\left (B a + B b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\left (B a - B b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}}, \frac{B b \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - 2 \, \sqrt{-a^{2} + b^{2}} A \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2}}\right ) -{\left (B a + B b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\left (B a - B b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*csch(x))/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[(B*b*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + sqrt(a^2 - b^2)*A*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*
b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*co
sh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) - (B*a + B*b)*log(cosh(x) + sinh(x) + 1)
+ (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2), (B*b*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - 2*sq
rt(-a^2 + b^2)*A*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - (B*a + B*b)*log(cosh(x) +
sinh(x) + 1) + (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \operatorname{csch}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*csch(x))/(a+b*cosh(x)),x)

[Out]

Integral((A + B*csch(x))/(a + b*cosh(x)), x)

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Giac [A]  time = 1.23717, size = 122, normalized size = 1.23 \begin{align*} \frac{B b \log \left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}{a^{2} - b^{2}} + \frac{2 \, A \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}}} - \frac{B \log \left (e^{x} + 1\right )}{a - b} + \frac{B \log \left ({\left | e^{x} - 1 \right |}\right )}{a + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*csch(x))/(a+b*cosh(x)),x, algorithm="giac")

[Out]

B*b*log(b*e^(2*x) + 2*a*e^x + b)/(a^2 - b^2) + 2*A*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/sqrt(-a^2 + b^2) - B*l
og(e^x + 1)/(a - b) + B*log(abs(e^x - 1))/(a + b)