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

Optimal. Leaf size=67 \[ -\frac{2 a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2 \sqrt{a-b} \sqrt{a+b}}-\frac{\sinh (x)}{b (a+b \cosh (x))}+\frac{x}{b^2} \]

[Out]

x/b^2 - (2*a*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]) - Sinh[x]/(b*(a + b*C
osh[x]))

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Rubi [A]  time = 0.104158, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2693, 2735, 2659, 208} \[ -\frac{2 a \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2 \sqrt{a-b} \sqrt{a+b}}-\frac{\sinh (x)}{b (a+b \cosh (x))}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Cosh[x])^2,x]

[Out]

x/b^2 - (2*a*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]) - Sinh[x]/(b*(a + b*C
osh[x]))

Rule 2693

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

Rule 2735

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0990332, size = 61, normalized size = 0.91 \[ \frac{\frac{2 a \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-\frac{b \sinh (x)}{a+b \cosh (x)}+x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 99, normalized size = 1.5 \begin{align*} 2\,{\frac{\tanh \left ( x/2 \right ) }{b \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b-a-b \right ) }}-2\,{\frac{a}{{b}^{2}\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{1}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b*tanh(1/2*x)/(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b-a-b)-2/b^2*a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((
a+b)*(a-b))^(1/2))+1/b^2*ln(tanh(1/2*x)+1)-1/b^2*ln(tanh(1/2*x)-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(sinh(x)^2/(a+b*cosh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.1027, size = 1663, normalized size = 24.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((a^2*b - b^3)*x*cosh(x)^2 + (a^2*b - b^3)*x*sinh(x)^2 + 2*a^2*b - 2*b^3 + (a*b*cosh(x)^2 + a*b*sinh(x)^2 + 2
*a^2*cosh(x) + a*b + 2*(a*b*cosh(x) + a^2)*sinh(x))*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*cosh(x) + b*sinh(x) + a))/(b*cos
h(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) + (a^2*b - b^3)*x + 2*(a^3 - a*b^2 + (a^3
- a*b^2)*x)*cosh(x) + 2*(a^3 - a*b^2 + (a^2*b - b^3)*x*cosh(x) + (a^3 - a*b^2)*x)*sinh(x))/(a^2*b^3 - b^5 + (
a^2*b^3 - b^5)*cosh(x)^2 + (a^2*b^3 - b^5)*sinh(x)^2 + 2*(a^3*b^2 - a*b^4)*cosh(x) + 2*(a^3*b^2 - a*b^4 + (a^2
*b^3 - b^5)*cosh(x))*sinh(x)), ((a^2*b - b^3)*x*cosh(x)^2 + (a^2*b - b^3)*x*sinh(x)^2 + 2*a^2*b - 2*b^3 + 2*(a
*b*cosh(x)^2 + a*b*sinh(x)^2 + 2*a^2*cosh(x) + a*b + 2*(a*b*cosh(x) + a^2)*sinh(x))*sqrt(-a^2 + b^2)*arctan(-s
qrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (a^2*b - b^3)*x + 2*(a^3 - a*b^2 + (a^3 - a*b^2)*x)
*cosh(x) + 2*(a^3 - a*b^2 + (a^2*b - b^3)*x*cosh(x) + (a^3 - a*b^2)*x)*sinh(x))/(a^2*b^3 - b^5 + (a^2*b^3 - b^
5)*cosh(x)^2 + (a^2*b^3 - b^5)*sinh(x)^2 + 2*(a^3*b^2 - a*b^4)*cosh(x) + 2*(a^3*b^2 - a*b^4 + (a^2*b^3 - b^5)*
cosh(x))*sinh(x))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.16578, size = 92, 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}} b^{2}} + \frac{x}{b^{2}} + \frac{2 \,{\left (a e^{x} + b\right )}}{{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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