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

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

[Out]

1/2*(a^2+2*b^2)*x/a^3-1/2*cosh(x)*(2*b-a*sinh(x))/a^2+2*b*arctanh((a-b*tanh(1/2*x))/(a^2+b^2)^(1/2))*(a^2+b^2)
^(1/2)/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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]

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 = 0.22, 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)

________________________________________________________________________________________

fricas [B]  time = 0.76, size = 304, normalized size = 3.95 \[ \frac {a^{2} \cosh \relax (x)^{4} + a^{2} \sinh \relax (x)^{4} - 4 \, a b \cosh \relax (x)^{3} + 4 \, {\left (a^{2} + 2 \, b^{2}\right )} x \cosh \relax (x)^{2} + 4 \, {\left (a^{2} \cosh \relax (x) - a b\right )} \sinh \relax (x)^{3} - 4 \, a b \cosh \relax (x) + 2 \, {\left (3 \, a^{2} \cosh \relax (x)^{2} - 6 \, a b \cosh \relax (x) + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} x\right )} \sinh \relax (x)^{2} + 8 \, {\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) - a^{2} + 4 \, {\left (a^{2} \cosh \relax (x)^{3} - 3 \, a b \cosh \relax (x)^{2} + 2 \, {\left (a^{2} + 2 \, b^{2}\right )} x \cosh \relax (x) - a b\right )} \sinh \relax (x)}{8 \, {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) \sinh \relax (x) + a^{3} \sinh \relax (x)^{2}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 121, normalized size = 1.57 \[ \frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {{\left (4 \, a b e^{x} + a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.14, size = 172, normalized size = 2.23 \[ \frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{a^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}}-\frac {2 b \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}} \]

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^2-1/2/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))

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 122, normalized size = 1.58 \[ -\frac {{\left (4 \, b e^{\left (-x\right )} - a\right )} e^{\left (2 \, x\right )}}{8 \, a^{2}} - \frac {4 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.69, size = 159, normalized size = 2.06 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}-\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}+\frac {x\,\left (a^2+2\,b^2\right )}{2\,a^3}-\frac {b\,\ln \left (\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{a^4}-\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{a^4}\right )\,\sqrt {a^2+b^2}}{a^3}+\frac {b\,\ln \left (\frac {2\,b\,\left (a-b\,{\mathrm {e}}^x\right )\,\sqrt {a^2+b^2}}{a^4}+\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2+b^2\right )}{a^4}\right )\,\sqrt {a^2+b^2}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(a + b/sinh(x)),x)

[Out]

exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) - (b*exp(-x))/(2*a^2) + (x*(a^2 + 2*b^2))/(2*a^3) - (b*l
og((2*b*exp(x)*(a^2 + b^2))/a^4 - (2*b*(a - b*exp(x))*(a^2 + b^2)^(1/2))/a^4)*(a^2 + b^2)^(1/2))/a^3 + (b*log(
(2*b*(a - b*exp(x))*(a^2 + b^2)^(1/2))/a^4 + (2*b*exp(x)*(a^2 + b^2))/a^4)*(a^2 + b^2)^(1/2))/a^3

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{2}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

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)

________________________________________________________________________________________