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

Optimal. Leaf size=54 \[ -\frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2}-\frac{a x}{b^2}+\frac{\cosh (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.108934, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2695, 2735, 2660, 618, 206} \[ -\frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2}-\frac{a x}{b^2}+\frac{\cosh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + b*Sinh[x]),x]

[Out]

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

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]

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

Mathematica [C]  time = 1.02959, size = 429, normalized size = 7.94 \[ \frac{\cosh (x) \left (2 \sqrt{b^2} (a-i b) \sqrt{1+i \sinh (x)} \tanh ^{-1}\left (\frac{\sqrt{a-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{a+i b} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}}}\right )+\sqrt{a+i b} \left (\sqrt{b^2} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}} \left (\sqrt{a-i b} \sqrt{1+i \sinh (x)} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}-2 (-1)^{3/4} \sqrt{b} \sin ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{a-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{b}}\right )\right )-2 i b \sqrt{a-i b} \sqrt{1+i \sinh (x)} \tan ^{-1}\left (\frac{\sqrt{-i b} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}}{\sqrt{i b} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}}}\right )\right )\right )}{b \sqrt{b^2} \sqrt{a-i b} \sqrt{a+i b} \sqrt{1+i \sinh (x)} \sqrt{-\frac{b (\sinh (x)-i)}{a+i b}} \sqrt{-\frac{b (\sinh (x)+i)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[x]*(2*(a - I*b)*Sqrt[b^2]*ArcTanh[(Sqrt[a - I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/(Sqrt[a + I*b]*Sq
rt[-((b*(-I + Sinh[x]))/(a + I*b))])]*Sqrt[1 + I*Sinh[x]] + Sqrt[a + I*b]*((-2*I)*Sqrt[a - I*b]*b*ArcTan[(Sqrt
[(-I)*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/(Sqrt[I*b]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))])]*Sqrt[1 + I*S
inh[x]] + Sqrt[b^2]*Sqrt[-((b*(-I + Sinh[x]))/(a + I*b))]*(-2*(-1)^(3/4)*Sqrt[b]*ArcSin[((1/2 + I/2)*Sqrt[a -
I*b]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])/Sqrt[b]] + Sqrt[a - I*b]*Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*(I + Sinh[x]
))/(a - I*b))]))))/(Sqrt[a - I*b]*Sqrt[a + I*b]*b*Sqrt[b^2]*Sqrt[1 + I*Sinh[x]]*Sqrt[-((b*(-I + Sinh[x]))/(a +
I*b))]*Sqrt[-((b*(I + Sinh[x]))/(a - I*b))])

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Maple [B]  time = 0.024, size = 126, normalized size = 2.3 \begin{align*}{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{2}}{{b}^{2}\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 ) }+2\,{\frac{1}{\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(cosh(x)^2/(a+b*sinh(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*a^2/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))+2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^
2+b^2)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.74153, size = 504, normalized size = 9.33 \begin{align*} -\frac{2 \, a x \cosh \left (x\right ) - b \cosh \left (x\right )^{2} - b \sinh \left (x\right )^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \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 x - b \cosh \left (x\right )\right )} \sinh \left (x\right ) - b}{2 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 177.406, size = 398, normalized size = 7.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(log(tanh(x/2))*tanh(x/2)**2/(tanh(x/2)**2 - 1) - log(tanh(x/2))/(tanh(x/2)**2 - 1) - 2/(tanh(x
/2)**2 - 1)), Eq(a, 0) & Eq(b, 0)), ((-x*sinh(x)**2/2 + x*cosh(x)**2/2 + sinh(x)*cosh(x)/2)/a, Eq(b, 0)), ((lo
g(tanh(x/2))*tanh(x/2)**2/(tanh(x/2)**2 - 1) - log(tanh(x/2))/(tanh(x/2)**2 - 1) - 2/(tanh(x/2)**2 - 1))/b, Eq
(a, 0)), (-a*x*tanh(x/2)**2/(b**2*tanh(x/2)**2 - b**2) + a*x/(b**2*tanh(x/2)**2 - b**2) - b*tanh(x/2)**2/(b**2
*tanh(x/2)**2 - b**2) - b/(b**2*tanh(x/2)**2 - b**2) - sqrt(a**2 + b**2)*log(tanh(x/2) - b/a - sqrt(a**2 + b**
2)/a)*tanh(x/2)**2/(b**2*tanh(x/2)**2 - b**2) + sqrt(a**2 + b**2)*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/(
b**2*tanh(x/2)**2 - b**2) + sqrt(a**2 + b**2)*log(tanh(x/2) - b/a + sqrt(a**2 + b**2)/a)*tanh(x/2)**2/(b**2*ta
nh(x/2)**2 - b**2) - sqrt(a**2 + b**2)*log(tanh(x/2) - b/a + sqrt(a**2 + b**2)/a)/(b**2*tanh(x/2)**2 - b**2),
True))

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Giac [A]  time = 1.26692, size = 112, normalized size = 2.07 \begin{align*} -\frac{a x}{b^{2}} + \frac{e^{\left (-x\right )}}{2 \, b} + \frac{e^{x}}{2 \, b} + \frac{\sqrt{a^{2} + b^{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 )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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