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

Optimal. Leaf size=65 \[ \frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}-\frac{B \log (a+b \cosh (x))}{a}+\frac{B \log (\cosh (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.147995, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {4401, 2659, 208, 2721, 36, 29, 31} \[ \frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}-\frac{B \log (a+b \cosh (x))}{a}+\frac{B \log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

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

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 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.142697, size = 61, normalized size = 0.94 \[ \frac{B (\log (\cosh (x))-\log (a+b \cosh (x)))}{a}-\frac{2 A \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.029, size = 125, normalized size = 1.9 \begin{align*} -{\frac{B}{a-b}\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 ) }+{\frac{Bb}{a \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{A}{\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}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*tanh(x))/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.09463, size = 806, normalized size = 12.4 \begin{align*} \left [\frac{\sqrt{a^{2} - b^{2}} A 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^{2} - B b^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) +{\left (B a^{2} - B b^{2}\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a b^{2}}, -\frac{2 \, \sqrt{-a^{2} + b^{2}} A 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^{2} - B b^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left (B a^{2} - B b^{2}\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a^2 - b^2)*A*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*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh
(x) + a)*sinh(x) + b)) - (B*a^2 - B*b^2)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + (B*a^2 - B*b^2)*log(2*co
sh(x)/(cosh(x) - sinh(x))))/(a^3 - a*b^2), -(2*sqrt(-a^2 + b^2)*A*a*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*si
nh(x) + a)/(a^2 - b^2)) + (B*a^2 - B*b^2)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - (B*a^2 - B*b^2)*log(2*c
osh(x)/(cosh(x) - sinh(x))))/(a^3 - a*b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \tanh{\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*tanh(x))/(a+b*cosh(x)),x)

[Out]

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

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Giac [A]  time = 1.19342, size = 89, normalized size = 1.37 \begin{align*} \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 (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}{a} + \frac{B \log \left (e^{\left (2 \, x\right )} + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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