3.231 \(\int \frac{\tanh (x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=48 \[ -\frac{a \log (a+b \sinh (x))}{a^2+b^2}+\frac{b \tan ^{-1}(\sinh (x))}{a^2+b^2}+\frac{a \log (\cosh (x))}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.0688227, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {2721, 801, 635, 203, 260} \[ -\frac{a \log (a+b \sinh (x))}{a^2+b^2}+\frac{b \tan ^{-1}(\sinh (x))}{a^2+b^2}+\frac{a \log (\cosh (x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Sinh[x]),x]

[Out]

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

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]

Rule 635

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.0579387, size = 36, normalized size = 0.75 \[ \frac{-a \log (a+b \sinh (x))+a \log (\cosh (x))+2 b \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Sinh[x]),x]

[Out]

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

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Maple [A]  time = 0.024, size = 84, normalized size = 1.8 \begin{align*} 2\,{\frac{a\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }{2\,{a}^{2}+2\,{b}^{2}}}+4\,{\frac{b\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{2\,{a}^{2}+2\,{b}^{2}}}-2\,{\frac{a\ln \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }{2\,{a}^{2}+2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a+b*sinh(x)),x)

[Out]

2/(2*a^2+2*b^2)*a*ln(tanh(1/2*x)^2+1)+4/(2*a^2+2*b^2)*b*arctan(tanh(1/2*x))-2*a/(2*a^2+2*b^2)*ln(a*tanh(1/2*x)
^2-2*tanh(1/2*x)*b-a)

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Maxima [A]  time = 1.53292, size = 89, normalized size = 1.85 \begin{align*} -\frac{2 \, b \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} - \frac{a \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} + \frac{a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.12042, size = 177, normalized size = 3.69 \begin{align*} \frac{2 \, b \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - a \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + a \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

(2*b*arctan(cosh(x) + sinh(x)) - a*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + a*log(2*cosh(x)/(cosh(x) - sin
h(x))))/(a^2 + b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*sinh(x)),x)

[Out]

Integral(tanh(x)/(a + b*sinh(x)), x)

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Giac [A]  time = 1.13206, size = 120, normalized size = 2.5 \begin{align*} -\frac{a b \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} b}{2 \,{\left (a^{2} + b^{2}\right )}} + \frac{a \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{2} + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="giac")

[Out]

-a*b*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^2*b + b^3) + 1/2*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*b/(a^2 +
b^2) + 1/2*a*log((e^(-x) - e^x)^2 + 4)/(a^2 + b^2)