3.230 \(\int \frac{\tanh ^2(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=69 \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{a \tanh (x)}{a^2+b^2}-\frac{b \text{sech}(x)}{a^2+b^2} \]

[Out]

(-2*a^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) - (b*Sech[x])/(a^2 + b^2) - (a*Tanh[x])/
(a^2 + b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0914194, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2727, 3767, 8, 2606, 2660, 618, 206} \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{a \tanh (x)}{a^2+b^2}-\frac{b \text{sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Sinh[x]),x]

[Out]

(-2*a^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) - (b*Sech[x])/(a^2 + b^2) - (a*Tanh[x])/
(a^2 + b^2)

Rule 2727

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\tanh ^2(x)}{a+b \sinh (x)} \, dx &=-\frac{a \int \text{sech}^2(x) \, dx}{a^2+b^2}+\frac{a^2 \int \frac{1}{a+b \sinh (x)} \, dx}{a^2+b^2}+\frac{b \int \text{sech}(x) \tanh (x) \, dx}{a^2+b^2}\\ &=-\frac{(i a) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))}{a^2+b^2}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^2+b^2}-\frac{b \operatorname{Subst}(\int 1 \, dx,x,\text{sech}(x))}{a^2+b^2}\\ &=-\frac{b \text{sech}(x)}{a^2+b^2}-\frac{a \tanh (x)}{a^2+b^2}-\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{a^2+b^2}\\ &=-\frac{2 a^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{b \text{sech}(x)}{a^2+b^2}-\frac{a \tanh (x)}{a^2+b^2}\\ \end{align*}

Mathematica [A]  time = 0.188303, size = 69, normalized size = 1. \[ \frac{a \left (\frac{2 a \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-\tanh (x)\right )-b \text{sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/(a + b*Sinh[x]),x]

[Out]

(-(b*Sech[x]) + a*((2*a*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - Tanh[x]))/(a^2 + b^2)

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 84, normalized size = 1.2 \begin{align*} 2\,{\frac{-a\tanh \left ( x/2 \right ) -b}{ \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+8\,{\frac{{a}^{2}}{ \left ( 4\,{a}^{2}+4\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a+b*sinh(x)),x)

[Out]

2/(a^2+b^2)*(-a*tanh(1/2*x)-b)/(tanh(1/2*x)^2+1)+8*a^2/(4*a^2+4*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2
*x)-2*b)/(a^2+b^2)^(1/2))

________________________________________________________________________________________

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(tanh(x)^2/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.05387, size = 691, normalized size = 10.01 \begin{align*} \frac{2 \, a^{3} + 2 \, a b^{2} +{\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} + a^{2}\right )} \sqrt{a^{2} + b^{2}} \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 ) - 2 \,{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

(2*a^3 + 2*a*b^2 + (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2)*sqrt(a^2 + b^2)*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*cos
h(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)) - 2*(a^2*b +
b^3)*cosh(x) - 2*(a^2*b + b^3)*sinh(x))/(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 +
2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{2}{\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)**2/(a+b*sinh(x)),x)

[Out]

Integral(tanh(x)**2/(a + b*sinh(x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.19777, size = 117, normalized size = 1.7 \begin{align*} \frac{a^{2} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (b e^{x} - a\right )}}{{\left (a^{2} + b^{2}\right )}{\left (e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*sinh(x)),x, algorithm="giac")

[Out]

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