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

Optimal. Leaf size=85 \[ \frac{a}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\left (a^2-b^2\right ) \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{2 a b \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cosh (x))}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.103188, antiderivative size = 85, 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} \[ \frac{a}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\left (a^2-b^2\right ) \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac{2 a b \tan ^{-1}(\sinh (x))}{\left (a^2+b^2\right )^2}+\frac{\left (a^2-b^2\right ) \log (\cosh (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Sinh[x])^2,x]

[Out]

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

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

Mathematica [C]  time = 0.225366, size = 146, normalized size = 1.72 \[ \frac{a \left (2 \left (\left (b^2-a^2\right ) \log (a+b \sinh (x))+a^2+b^2\right )+(a-i b)^2 \log (-\sinh (x)+i)+(a+i b)^2 \log (\sinh (x)+i)\right )+b \sinh (x) \left (2 \left (b^2-a^2\right ) \log (a+b \sinh (x))+(a-i b)^2 \log (-\sinh (x)+i)+(a+i b)^2 \log (\sinh (x)+i)\right )}{2 \left (a^2+b^2\right )^2 (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((a - I*b)^2*Log[I - Sinh[x]] + (a + I*b)^2*Log[I + Sinh[x]] + 2*(a^2 + b^2 + (-a^2 + b^2)*Log[a + b*Sinh[x
]])) + b*((a - I*b)^2*Log[I - Sinh[x]] + (a + I*b)^2*Log[I + Sinh[x]] + 2*(-a^2 + b^2)*Log[a + b*Sinh[x]])*Sin
h[x])/(2*(a^2 + b^2)^2*(a + b*Sinh[x]))

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Maple [B]  time = 0.042, size = 248, normalized size = 2.9 \begin{align*} 2\,{\frac{\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ){a}^{2}}{2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4}}}-2\,{\frac{\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ){b}^{2}}{2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4}}}+8\,{\frac{ab\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4}}}+2\,{\frac{\tanh \left ( x/2 \right ){a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{\tanh \left ( x/2 \right ){b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-{\frac{{a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }+{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(2*a^4+4*a^2*b^2+2*b^4)*ln(tanh(1/2*x)^2+1)*a^2-2/(2*a^4+4*a^2*b^2+2*b^4)*ln(tanh(1/2*x)^2+1)*b^2+8/(2*a^4+4
*a^2*b^2+2*b^4)*a*b*arctan(tanh(1/2*x))+2/(a^2+b^2)^2*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*a^2*b+2/
(a^2+b^2)^2*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b^3-1/(a^2+b^2)^2*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)
*b-a)*a^2+1/(a^2+b^2)^2*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b^2

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Maxima [A]  time = 1.57067, size = 209, normalized size = 2.46 \begin{align*} -\frac{4 \, a b \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \, a e^{\left (-x\right )}}{a^{2} b + b^{3} + 2 \,{\left (a^{3} + a b^{2}\right )} e^{\left (-x\right )} -{\left (a^{2} b + b^{3}\right )} e^{\left (-2 \, x\right )}} - \frac{{\left (a^{2} - b^{2}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.19909, size = 1029, normalized size = 12.11 \begin{align*} -\frac{4 \,{\left (a b^{2} \cosh \left (x\right )^{2} + a b^{2} \sinh \left (x\right )^{2} + 2 \, a^{2} b \cosh \left (x\right ) - a b^{2} + 2 \,{\left (a b^{2} \cosh \left (x\right ) + a^{2} b\right )} \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \cosh \left (x\right ) +{\left (a^{2} b - b^{3} -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left (a^{2} b - b^{3} -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \sinh \left (x\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{2} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(a*b^2*cosh(x)^2 + a*b^2*sinh(x)^2 + 2*a^2*b*cosh(x) - a*b^2 + 2*(a*b^2*cosh(x) + a^2*b)*sinh(x))*arctan(c
osh(x) + sinh(x)) + 2*(a^3 + a*b^2)*cosh(x) + (a^2*b - b^3 - (a^2*b - b^3)*cosh(x)^2 - (a^2*b - b^3)*sinh(x)^2
- 2*(a^3 - a*b^2)*cosh(x) - 2*(a^3 - a*b^2 + (a^2*b - b^3)*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) -
sinh(x))) - (a^2*b - b^3 - (a^2*b - b^3)*cosh(x)^2 - (a^2*b - b^3)*sinh(x)^2 - 2*(a^3 - a*b^2)*cosh(x) - 2*(a
^3 - a*b^2 + (a^2*b - b^3)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 2*(a^3 + a*b^2)*sinh(x))/(a^
4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^2 - (a^4*b + 2*a^2*b^3 + b^5)*sinh(x)^2 - 2*(a^5 + 2
*a^3*b^2 + a*b^4)*cosh(x) - 2*(a^5 + 2*a^3*b^2 + a*b^4 + (a^4*b + 2*a^2*b^3 + b^5)*cosh(x))*sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17057, size = 269, normalized size = 3.16 \begin{align*} \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{{\left (a^{2} b - b^{3}\right )} \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac{a^{2} b{\left (e^{\left (-x\right )} - e^{x}\right )} - b^{3}{\left (e^{\left (-x\right )} - e^{x}\right )} - 4 \, a^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*a*b/(a^4 + 2*a^2*b^2 + b^4) + 1/2*(a^2 - b^2)*log((e^(-x) - e^x)^2 +
4)/(a^4 + 2*a^2*b^2 + b^4) - (a^2*b - b^3)*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^4*b + 2*a^2*b^3 + b^5) + (a^2
*b*(e^(-x) - e^x) - b^3*(e^(-x) - e^x) - 4*a^3)/((a^4 + 2*a^2*b^2 + b^4)*(b*(e^(-x) - e^x) - 2*a))