3.203 \(\int \frac{A+B \coth (x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac{a B \log (a+b \cosh (x))}{a^2-b^2}+\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cosh (x))}{2 (a+b)}+\frac{B \log (\cosh (x)+1)}{2 (a-b)} \]

[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)) - (a*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.166308, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4401, 2659, 208, 2721, 801} \[ -\frac{a B \log (a+b \cosh (x))}{a^2-b^2}+\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cosh (x))}{2 (a+b)}+\frac{B \log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Coth[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)) - (a*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.243717, size = 81, normalized size = 0.81 \[ \frac{B \left (a \log (a+b \cosh (x))-a \log (\sinh (x))+b \log \left (\tanh \left (\frac{x}{2}\right )\right )\right )}{b^2-a^2}-\frac{2 A \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.026, size = 139, normalized size = 1.4 \begin{align*} -{\frac{aB}{ \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*coth(x))/(a+b*cosh(x)),x)

[Out]

-1/(a+b)*a*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*coth(x))/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 14.0187, size = 842, normalized size = 8.42 \begin{align*} \left [-\frac{B a \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 a \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*coth(x))/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[-(B*a*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*c
osh(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*a*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + 2*
sqrt(-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 \coth{\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*coth(x))/(a+b*cosh(x)),x)

[Out]

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

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Giac [A]  time = 1.1764, size = 122, normalized size = 1.22 \begin{align*} -\frac{B a \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*coth(x))/(a+b*cosh(x)),x, algorithm="giac")

[Out]

-B*a*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*
log(e^x + 1)/(a - b) + B*log(abs(e^x - 1))/(a + b)