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*ln(cosh(x))/(a^2+b^2)-a*ln(a+b*sinh(x))/(a^2+b^2)

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Rubi [A]  time = 0.07, antiderivative size = 48, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.63, size = 57, normalized size = 1.19 \[ \frac {2 \, b \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - a \log \left (\frac {2 \, {\left (b \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + a \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} + b^{2}} \]

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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giac [A]  time = 0.27, size = 89, normalized size = 1.85 \[ -\frac {a b \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} b}{2 \, {\left (a^{2} + b^{2}\right )}} + \frac {a \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]

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)

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maple [A]  time = 0.05, size = 84, normalized size = 1.75 \[ -\frac {2 a \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}{2 a^{2}+2 b^{2}}+\frac {2 a \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{2 a^{2}+2 b^{2}}+\frac {4 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}+2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 66, normalized size = 1.38 \[ -\frac {2 \, b \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} - \frac {a \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} + \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \]

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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mupad [B]  time = 1.40, size = 95, normalized size = 1.98 \[ \frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{a-b\,1{}\mathrm {i}}-\frac {a\,\ln \left (b^3\,{\mathrm {e}}^{2\,x}-4\,a^2\,b-b^3+8\,a^3\,{\mathrm {e}}^x+2\,a\,b^2\,{\mathrm {e}}^x+4\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2+b^2}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{-b+a\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b*sinh(x)),x)

[Out]

(log(exp(x)*1i + 1)*1i)/(a*1i - b) + log(exp(x) + 1i)/(a - b*1i) - (a*log(b^3*exp(2*x) - 4*a^2*b - b^3 + 8*a^3
*exp(x) + 2*a*b^2*exp(x) + 4*a^2*b*exp(2*x)))/(a^2 + b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\relax (x )}}{a + b \sinh {\relax (x )}}\, dx \]

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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