3.110 \(\int \frac {A+B \cosh (x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}+\frac {B x}{b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2735, 2659, 208} \[ \frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}+\frac {B x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(a + b*Cosh[x]),x]

[Out]

(B*x)/b + (2*(A*b - a*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + 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 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 {A+B \cosh (x)}{a+b \cosh (x)} \, dx &=\frac {B x}{b}-\frac {(-A b+a B) \int \frac {1}{a+b \cosh (x)} \, dx}{b}\\ &=\frac {B x}{b}-\frac {(2 (-A b+a B)) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b}\\ &=\frac {B x}{b}+\frac {2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 59, normalized size = 0.98 \[ \frac {2 (a B-A b) \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{b \sqrt {b^2-a^2}}+\frac {B x}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 0.77, size = 240, normalized size = 4.00 \[ \left [-\frac {{\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) - {\left (B a^{2} - B b^{2}\right )} x}{a^{2} b - b^{3}}, \frac {2 \, {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (B a^{2} - B b^{2}\right )} x}{a^{2} b - b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[-((B*a - A*b)*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*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) +
2*(b*cosh(x) + a)*sinh(x) + b)) - (B*a^2 - B*b^2)*x)/(a^2*b - b^3), (2*(B*a - A*b)*sqrt(-a^2 + b^2)*arctan(-sq
rt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (B*a^2 - B*b^2)*x)/(a^2*b - b^3)]

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giac [A]  time = 0.12, size = 50, normalized size = 0.83 \[ \frac {B x}{b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 103, normalized size = 1.72 \[ -\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A}{\sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a B}{b \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x))/(a+b*cosh(x)),x)

[Out]

-B/b*ln(tanh(1/2*x)-1)+B/b*ln(tanh(1/2*x)+1)+2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/
2))*A-2/b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*a*B

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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((A+B*cosh(x))/(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 = 1.12, size = 242, normalized size = 4.03 \[ \frac {2\,\mathrm {atan}\left (\frac {b^2\,{\mathrm {e}}^x\,\sqrt {b^4-a^2\,b^2}\,\left (\frac {2\,\left (A\,b\,\sqrt {b^4-a^2\,b^2}-B\,a\,\sqrt {b^4-a^2\,b^2}\right )}{b^4\,\sqrt {b^4-a^2\,b^2}\,\sqrt {{\left (A\,b-B\,a\right )}^2}}+\frac {2\,a^2\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{b^2\,\left (b^4-a^2\,b^2\right )\,\left (A\,b-B\,a\right )}\right )}{2}+\frac {a\,b\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{\sqrt {b^4-a^2\,b^2}\,\left (A\,b-B\,a\right )}\right )\,\sqrt {A^2\,b^2-2\,A\,B\,a\,b+B^2\,a^2}}{\sqrt {b^4-a^2\,b^2}}+\frac {B\,x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x))/(a + b*cosh(x)),x)

[Out]

(2*atan((b^2*exp(x)*(b^4 - a^2*b^2)^(1/2)*((2*(A*b*(b^4 - a^2*b^2)^(1/2) - B*a*(b^4 - a^2*b^2)^(1/2)))/(b^4*(b
^4 - a^2*b^2)^(1/2)*((A*b - B*a)^2)^(1/2)) + (2*a^2*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/2))/(b^2*(b^4 - a^2*b^2
)*(A*b - B*a))))/2 + (a*b*(A^2*b^2 + B^2*a^2 - 2*A*B*a*b)^(1/2))/((b^4 - a^2*b^2)^(1/2)*(A*b - B*a)))*(A^2*b^2
+ B^2*a^2 - 2*A*B*a*b)^(1/2))/(b^4 - a^2*b^2)^(1/2) + (B*x)/b

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sympy [A]  time = 27.03, size = 403, normalized size = 6.72 \[ \begin {cases} \tilde {\infty } \left (2 A \operatorname {atan}{\left (\tanh {\left (\frac {x}{2} \right )} \right )} + B x\right ) & \text {for}\: a = 0 \wedge b = 0 \\- \frac {A}{b \tanh {\left (\frac {x}{2} \right )}} + \frac {B x}{b} - \frac {B}{b \tanh {\left (\frac {x}{2} \right )}} & \text {for}\: a = - b \\\frac {A x + B \sinh {\relax (x )}}{a} & \text {for}\: b = 0 \\\frac {A \tanh {\left (\frac {x}{2} \right )}}{b} + \frac {B x}{b} - \frac {B \tanh {\left (\frac {x}{2} \right )}}{b} & \text {for}\: a = b \\- \frac {A b \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} + \frac {A b \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} + \frac {B a x \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} + \frac {B a \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {B a \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {B b x \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(a+b*cosh(x)),x)

[Out]

Piecewise((zoo*(2*A*atan(tanh(x/2)) + B*x), Eq(a, 0) & Eq(b, 0)), (-A/(b*tanh(x/2)) + B*x/b - B/(b*tanh(x/2)),
Eq(a, -b)), ((A*x + B*sinh(x))/a, Eq(b, 0)), (A*tanh(x/2)/b + B*x/b - B*tanh(x/2)/b, Eq(a, b)), (-A*b*log(-sq
rt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a - b) + b/(a - b))) +
A*b*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a - b) + b/(a
- b))) + B*a*x*sqrt(a/(a - b) + b/(a - b))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a - b) + b/(a - b)
)) + B*a*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a - b)
+ b/(a - b))) - B*a*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(
a/(a - b) + b/(a - b))) - B*b*x*sqrt(a/(a - b) + b/(a - b))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a
- b) + b/(a - b))), True))

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