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

Optimal. Leaf size=65 \[ \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}-\frac {B \log (a+b \cosh (x))}{a}+\frac {B \log (\cosh (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4401, 2659, 208, 2721, 36, 29, 31} \[ \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}-\frac {B \log (a+b \cosh (x))}{a}+\frac {B \log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 61, normalized size = 0.94 \[ \frac {B (\log (\cosh (x))-\log (a+b \cosh (x)))}{a}-\frac {2 A \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.58, size = 315, normalized size = 4.85 \[ \left [\frac {\sqrt {a^{2} - b^{2}} A a \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{3} - a b^{2}}, -\frac {2 \, \sqrt {-a^{2} + b^{2}} A a \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{3} - a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a^2 - b^2)*A*a*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)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + (B*a^2 - B*b^2)*log(2*co
sh(x)/(cosh(x) - sinh(x))))/(a^3 - a*b^2), -(2*sqrt(-a^2 + b^2)*A*a*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*si
nh(x) + a)/(a^2 - b^2)) + (B*a^2 - B*b^2)*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - (B*a^2 - B*b^2)*log(2*c
osh(x)/(cosh(x) - sinh(x))))/(a^3 - a*b^2)]

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giac [A]  time = 0.12, size = 66, normalized size = 1.02 \[ \frac {2 \, A \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}}} - \frac {B \log \left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}{a} + \frac {B \log \left (e^{\left (2 \, x\right )} + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 125, normalized size = 1.92 \[ -\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) B}{a -b}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right ) B b}{a \left (a -b \right )}+\frac {2 A \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {B \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*tanh(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 = 12.14, size = 160, normalized size = 2.46 \[ \frac {B\,\ln \left (16\,B^2\,b^2-16\,B^2\,a^2-16\,B^2\,a^2\,{\mathrm {e}}^{2\,x}+16\,B^2\,b^2\,{\mathrm {e}}^{2\,x}\right )}{a}-\frac {B\,\ln \left (16\,B^2\,b+32\,B^2\,a\,{\mathrm {e}}^x+16\,B^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a}-\frac {2\,\mathrm {atan}\left (\frac {A^2\,b^2\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}+A^2\,a\,b\,\sqrt {b^2-a^2}}{A\,b\,\left (a^2-b^2\right )\,\sqrt {A^2}}\right )\,\sqrt {A^2}}{\sqrt {b^2-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*log(16*B^2*b^2 - 16*B^2*a^2 - 16*B^2*a^2*exp(2*x) + 16*B^2*b^2*exp(2*x)))/a - (B*log(16*B^2*b + 32*B^2*a*ex
p(x) + 16*B^2*b*exp(2*x)))/a - (2*atan((A^2*b^2*exp(x)*(b^2 - a^2)^(1/2) + A^2*a*b*(b^2 - a^2)^(1/2))/(A*b*(a^
2 - b^2)*(A^2)^(1/2)))*(A^2)^(1/2))/(b^2 - a^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tanh {\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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