3.95 \(\int \frac{\cosh ^2(x)}{a+b \text{csch}(x)} \, dx\)

Optimal. Leaf size=77 \[ \frac{x \left (a^2+2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2} \]

[Out]

((a^2 + 2*b^2)*x)/(2*a^3) + (2*b*Sqrt[a^2 + b^2]*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/a^3 - (Cosh[x]*(2
*b - a*Sinh[x]))/(2*a^2)

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Rubi [A]  time = 0.206218, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2865, 2735, 2660, 618, 204} \[ \frac{x \left (a^2+2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + b*Csch[x]),x]

[Out]

((a^2 + 2*b^2)*x)/(2*a^3) + (2*b*Sqrt[a^2 + b^2]*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/a^3 - (Cosh[x]*(2
*b - a*Sinh[x]))/(2*a^2)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

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

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cosh ^2(x)}{a+b \text{csch}(x)} \, dx &=i \int \frac{\cosh ^2(x) \sinh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2}+\frac{\int \frac{-i a b+i \left (a^2+2 b^2\right ) \sinh (x)}{i b+i a \sinh (x)} \, dx}{2 a^2}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2}-\frac{\left (i b \left (a^2+b^2\right )\right ) \int \frac{1}{i b+i a \sinh (x)} \, dx}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2}-\frac{\left (2 i b \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2}+\frac{\left (4 i b \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 i a-2 i b \tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}+\frac{2 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3}-\frac{\cosh (x) (2 b-a \sinh (x))}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.227557, size = 80, normalized size = 1.04 \[ \frac{8 b \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )+2 a^2 x+a^2 \sinh (2 x)-4 a b \cosh (x)+4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*x + 4*b^2*x + 8*b*Sqrt[-a^2 - b^2]*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]] - 4*a*b*Cosh[x] + a^2*Sin
h[2*x])/(4*a^3)

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Maple [B]  time = 0.03, size = 172, normalized size = 2.2 \begin{align*} -{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-2\,{\frac{b\sqrt{{a}^{2}+{b}^{2}}}{{a}^{3}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/(tanh(1/2*x)+1)^2+1/2/a/(tanh(1/2*x)+1)-1/a^2/(tanh(1/2*x)+1)*b+1/2/a*ln(tanh(1/2*x)+1)+1/a^3*ln(tanh(1
/2*x)+1)*b^2-2*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))+1/2/a/(tanh(1/2*x)-1)^
2+1/2/a/(tanh(1/2*x)-1)+1/a^2/(tanh(1/2*x)-1)*b-1/2/a*ln(tanh(1/2*x)-1)-1/a^3*ln(tanh(1/2*x)-1)*b^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*csch(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.57327, size = 859, normalized size = 11.16 \begin{align*} \frac{a^{2} \cosh \left (x\right )^{4} + a^{2} \sinh \left (x\right )^{4} - 4 \, a b \cosh \left (x\right )^{3} + 4 \,{\left (a^{2} + 2 \, b^{2}\right )} x \cosh \left (x\right )^{2} + 4 \,{\left (a^{2} \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )^{3} - 4 \, a b \cosh \left (x\right ) + 2 \,{\left (3 \, a^{2} \cosh \left (x\right )^{2} - 6 \, a b \cosh \left (x\right ) + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \,{\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2}\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) - a}\right ) - a^{2} + 4 \,{\left (a^{2} \cosh \left (x\right )^{3} - 3 \, a b \cosh \left (x\right )^{2} + 2 \,{\left (a^{2} + 2 \, b^{2}\right )} x \cosh \left (x\right ) - a b\right )} \sinh \left (x\right )}{8 \,{\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*cosh(x)^4 + a^2*sinh(x)^4 - 4*a*b*cosh(x)^3 + 4*(a^2 + 2*b^2)*x*cosh(x)^2 + 4*(a^2*cosh(x) - a*b)*sin
h(x)^3 - 4*a*b*cosh(x) + 2*(3*a^2*cosh(x)^2 - 6*a*b*cosh(x) + 2*(a^2 + 2*b^2)*x)*sinh(x)^2 + 8*(b*cosh(x)^2 +
2*b*cosh(x)*sinh(x) + b*sinh(x)^2)*sqrt(a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 +
2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x
)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a)) - a^2 + 4*(a^2*cosh(x)^3 - 3*a*b*cosh(x)^2 + 2*(a^2 + 2*b^
2)*x*cosh(x) - a*b)*sinh(x))/(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)**2/(a + b*csch(x)), x)

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Giac [A]  time = 1.1684, size = 163, normalized size = 2.12 \begin{align*} \frac{a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} + \frac{{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac{{\left (4 \, a b e^{x} + a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} - \frac{{\left (a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a*e^(2*x) - 4*b*e^x)/a^2 + 1/2*(a^2 + 2*b^2)*x/a^3 - 1/8*(4*a*b*e^x + a^2)*e^(-2*x)/a^3 - (a^2*b + b^3)*l
og(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3)