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-sinh(x)/b/(a+b*cosh(x))-2*a*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b^2/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.59, size = 700, normalized size = 10.45 \[ \left [\frac {{\left (a^{2} b - b^{3}\right )} x \cosh \relax (x)^{2} + {\left (a^{2} b - b^{3}\right )} x \sinh \relax (x)^{2} + 2 \, a^{2} b - 2 \, b^{3} + {\left (a b \cosh \relax (x)^{2} + a b \sinh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) + a b + 2 \, {\left (a b \cosh \relax (x) + a^{2}\right )} \sinh \relax (x)\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + {\left (a^{2} b - b^{3}\right )} x + 2 \, {\left (a^{3} - a b^{2} + {\left (a^{3} - a b^{2}\right )} x\right )} \cosh \relax (x) + 2 \, {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} x \cosh \relax (x) + {\left (a^{3} - a b^{2}\right )} x\right )} \sinh \relax (x)}{a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{2} + {\left (a^{2} b^{3} - b^{5}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cosh \relax (x) + 2 \, {\left (a^{3} b^{2} - a b^{4} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}, \frac {{\left (a^{2} b - b^{3}\right )} x \cosh \relax (x)^{2} + {\left (a^{2} b - b^{3}\right )} x \sinh \relax (x)^{2} + 2 \, a^{2} b - 2 \, b^{3} + 2 \, {\left (a b \cosh \relax (x)^{2} + a b \sinh \relax (x)^{2} + 2 \, a^{2} \cosh \relax (x) + a b + 2 \, {\left (a b \cosh \relax (x) + a^{2}\right )} \sinh \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} b - b^{3}\right )} x + 2 \, {\left (a^{3} - a b^{2} + {\left (a^{3} - a b^{2}\right )} x\right )} \cosh \relax (x) + 2 \, {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} x \cosh \relax (x) + {\left (a^{3} - a b^{2}\right )} x\right )} \sinh \relax (x)}{a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)^{2} + {\left (a^{2} b^{3} - b^{5}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cosh \relax (x) + 2 \, {\left (a^{3} b^{2} - a b^{4} + {\left (a^{2} b^{3} - b^{5}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}\right ] \]

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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giac [A]  time = 0.15, size = 68, normalized size = 1.01 \[ -\frac {2 \, a \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} + \frac {2 \, {\left (a e^{x} + b\right )}}{{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )} b^{2}} \]

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)

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maple [A]  time = 0.07, size = 99, normalized size = 1.48 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {2 \tanh \left (\frac {x}{2}\right )}{b \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}-\frac {2 a \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.12, size = 139, normalized size = 2.07 \[ \frac {x}{b^2}+\frac {\frac {2}{b}+\frac {2\,a\,{\mathrm {e}}^x}{b^2}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}-\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}+\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b*cosh(x))^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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