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

Optimal. Leaf size=59 \[ -\frac {a x}{b^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}+\frac {\sinh (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 59, 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 = {2695, 2735, 2659, 208} \[ -\frac {a x}{b^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}+\frac {\sinh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Cosh[x]),x]

[Out]

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

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 2695

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 + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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)} \, dx &=\frac {\sinh (x)}{b}+\frac {\int \frac {-b-a \cosh (x)}{a+b \cosh (x)} \, dx}{b}\\ &=-\frac {a x}{b^2}+\frac {\sinh (x)}{b}-\left (1-\frac {a^2}{b^2}\right ) \int \frac {1}{a+b \cosh (x)} \, dx\\ &=-\frac {a x}{b^2}+\frac {\sinh (x)}{b}-\left (2 \left (1-\frac {a^2}{b^2}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )\\ &=-\frac {a x}{b^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^2}+\frac {\sinh (x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 54, normalized size = 0.92 \[ \frac {2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-a x+b \sinh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.29, size = 279, normalized size = 4.73 \[ \left [-\frac {2 \, a x \cosh \relax (x) - b \cosh \relax (x)^{2} - b \sinh \relax (x)^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left (a x - b \cosh \relax (x)\right )} \sinh \relax (x) + b}{2 \, {\left (b^{2} \cosh \relax (x) + b^{2} \sinh \relax (x)\right )}}, -\frac {2 \, a x \cosh \relax (x) - b \cosh \relax (x)^{2} - b \sinh \relax (x)^{2} + 4 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + 2 \, {\left (a x - b \cosh \relax (x)\right )} \sinh \relax (x) + b}{2 \, {\left (b^{2} \cosh \relax (x) + b^{2} \sinh \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*a*x*cosh(x) - b*cosh(x)^2 - b*sinh(x)^2 - 2*sqrt(a^2 - b^2)*(cosh(x) + sinh(x))*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*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) + 2*(a*x - b*cosh(x)
)*sinh(x) + b)/(b^2*cosh(x) + b^2*sinh(x)), -1/2*(2*a*x*cosh(x) - b*cosh(x)^2 - b*sinh(x)^2 + 4*sqrt(-a^2 + b^
2)*(cosh(x) + sinh(x))*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 2*(a*x - b*cosh(x))
*sinh(x) + b)/(b^2*cosh(x) + b^2*sinh(x))]

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giac [A]  time = 0.13, size = 68, normalized size = 1.15 \[ -\frac {a x}{b^{2}} - \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*x/b^2 - 1/2*e^(-x)/b + 1/2*e^x/b + 2*(a^2 - b^2)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b^2
)

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maple [B]  time = 0.06, size = 129, normalized size = 2.19 \[ -\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a^{2}}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/(tanh(1/2*x)-1)+a/b^2*ln(tanh(1/2*x)-1)-1/b/(tanh(1/2*x)+1)-a/b^2*ln(tanh(1/2*x)+1)+2/b^2/((a+b)*(a-b))^(
1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*a^2-2/((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)),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 = 1.04, size = 139, normalized size = 2.36 \[ \frac {{\mathrm {e}}^x}{2\,b}-\frac {{\mathrm {e}}^{-x}}{2\,b}-\frac {a\,x}{b^2}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (a^2-b^2\right )}{b^3}-\frac {2\,\sqrt {a+b}\,\left (b+a\,{\mathrm {e}}^x\right )\,\sqrt {a-b}}{b^3}\right )\,\sqrt {a+b}\,\sqrt {a-b}}{b^2}-\frac {\ln \left (\frac {2\,\sqrt {a+b}\,\left (b+a\,{\mathrm {e}}^x\right )\,\sqrt {a-b}}{b^3}-\frac {2\,{\mathrm {e}}^x\,\left (a^2-b^2\right )}{b^3}\right )\,\sqrt {a+b}\,\sqrt {a-b}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b*cosh(x)),x)

[Out]

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

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sympy [A]  time = 93.14, size = 892, normalized size = 15.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*tanh(x/2)**2*atan(tanh(x/2))/(tanh(x/2)**2 - 1) - 2*tanh(x/2)/(tanh(x/2)**2 - 1) + 2*atan(t
anh(x/2))/(tanh(x/2)**2 - 1)), Eq(a, 0) & Eq(b, 0)), (x*tanh(x/2)**2/(b*tanh(x/2)**2 - b) - x/(b*tanh(x/2)**2
- b) - 2*tanh(x/2)/(b*tanh(x/2)**2 - b), Eq(a, -b)), ((x*sinh(x)**2/2 - x*cosh(x)**2/2 + sinh(x)*cosh(x)/2)/a,
Eq(b, 0)), (-x*tanh(x/2)**2/(b*tanh(x/2)**2 - b) + x/(b*tanh(x/2)**2 - b) - 2*tanh(x/2)/(b*tanh(x/2)**2 - b),
Eq(a, b)), (-a*x*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b*
*2*sqrt(a/(a - b) + b/(a - b))) + a*x*sqrt(a/(a - b) + b/(a - b))/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)*
*2 - b**2*sqrt(a/(a - b) + b/(a - b))) - a*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))*tanh(x/2)**2/(b**2*sq
rt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) + a*log(-sqrt(a/(a - b) + b/(a - b)
) + tanh(x/2))/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) + a*log(sqrt
(a/(a - b) + b/(a - b)) + tanh(x/2))*tanh(x/2)**2/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a
/(a - b) + b/(a - b))) - a*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh
(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) - 2*b*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)/(b**2*sqrt(a/(a - b)
+ b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) - b*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2)
)*tanh(x/2)**2/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) + b*log(-sqr
t(a/(a - b) + b/(a - b)) + tanh(x/2))/(b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b
/(a - b))) + b*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))*tanh(x/2)**2/(b**2*sqrt(a/(a - b) + b/(a - b))*tan
h(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))) - b*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(b**2*sqrt(a/(a
- b) + b/(a - b))*tanh(x/2)**2 - b**2*sqrt(a/(a - b) + b/(a - b))), True))

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