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

Optimal. Leaf size=56 \[ \frac {B x}{b}-\frac {2 B \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2735, 2659, 208} \[ \frac {B x}{b}-\frac {2 B \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 56, normalized size = 1.00 \[ \frac {B \left (\frac {2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{b}+\frac {a x}{b}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.80, size = 190, normalized size = 3.39 \[ \left [\frac {B a x + \sqrt {a^{2} - b^{2}} B \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right )}{a b}, \frac {B a x + 2 \, \sqrt {-a^{2} + b^{2}} B \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right )}{a b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(B*a*x + sqrt(a^2 - b^2)*B*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)))/(a*b), (B*a*x + 2*sqrt(-a^2 + b^2)*B*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sin
h(x) + a)/(a^2 - b^2)))/(a*b)]

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giac [A]  time = 0.14, size = 57, normalized size = 1.02 \[ \frac {B x}{b} - \frac {2 \, {\left (B a^{2} - B b^{2}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 107, normalized size = 1.91 \[ -\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}-\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) a B}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 B b \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*B/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/b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(
1/2))*a*B+2*B/a*b/((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((b*B/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 = 0.49, size = 205, normalized size = 3.66 \[ \frac {2\,\mathrm {atan}\left (\frac {b\,\sqrt {a^2\,b^2}\,\sqrt {B^2\,b^2-B^2\,a^2}}{B\,\left (b^4-a^2\,b^2\right )}+\frac {a\,b^2\,{\mathrm {e}}^x\,\left (\frac {2\,\sqrt {B^2\,b^2-B^2\,a^2}}{B\,b^2\,\left (b^4-a^2\,b^2\right )}-\frac {2\,\left (B\,a^2\,\sqrt {a^2\,b^2}-B\,b^2\,\sqrt {a^2\,b^2}\right )}{a^2\,b^4\,\sqrt {-B^2\,\left (a^2-b^2\right )}\,\sqrt {a^2\,b^2}}\right )\,\sqrt {a^2\,b^2}}{2}\right )\,\sqrt {B^2\,b^2-B^2\,a^2}}{\sqrt {a^2\,b^2}}+\frac {B\,x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 28.17, size = 170, normalized size = 3.04 \[ \begin {cases} \text {NaN} & \text {for}\: a = 0 \wedge b = 0 \\\frac {B x}{b} & \text {for}\: a = - b \\\frac {B \sinh {\relax (x )}}{a} & \text {for}\: b = 0 \\\frac {B x}{b} & \text {for}\: a = b \\\frac {B x}{b} + \frac {B \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {B \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} + \frac {B \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {B \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((nan, Eq(a, 0) & Eq(b, 0)), (B*x/b, Eq(a, -b)), (B*sinh(x)/a, Eq(b, 0)), (B*x/b, Eq(a, b)), (B*x/b +
B*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(b*sqrt(a/(a - b) + b/(a - b))) - B*log(sqrt(a/(a - b) + b/(a
- b)) + tanh(x/2))/(b*sqrt(a/(a - b) + b/(a - b))) + B*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*sqrt(
a/(a - b) + b/(a - b))) - B*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*sqrt(a/(a - b) + b/(a - b))), True
))

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