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

Optimal. Leaf size=100 \[ -\frac {a B \log (a+b \cosh (x))}{a^2-b^2}+\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cosh (x))}{2 (a+b)}+\frac {B \log (\cosh (x)+1)}{2 (a-b)} \]

[Out]

1/2*B*ln(1-cosh(x))/(a+b)+1/2*B*ln(1+cosh(x))/(a-b)-a*B*ln(a+b*cosh(x))/(a^2-b^2)+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 = 0.17, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4401, 2659, 208, 2721, 801} \[ -\frac {a B \log (a+b \cosh (x))}{a^2-b^2}+\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (1-\cosh (x))}{2 (a+b)}+\frac {B \log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Coth[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[1 - Cosh[x]])/(2*(a + b)
) + (B*Log[1 + Cosh[x]])/(2*(a - b)) - (a*B*Log[a + b*Cosh[x]])/(a^2 - b^2)

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

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Mathematica [A]  time = 0.26, size = 81, normalized size = 0.81 \[ \frac {B \left (a \log (a+b \cosh (x))-a \log (\sinh (x))+b \log \left (\tanh \left (\frac {x}{2}\right )\right )\right )}{b^2-a^2}-\frac {2 A \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Coth[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*(a*Log[a + b*Cosh[x]] - a*Log[Sinh[x
]] + b*Log[Tanh[x/2]]))/(-a^2 + b^2)

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fricas [A]  time = 2.75, size = 303, normalized size = 3.03 \[ \left [-\frac {B a \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - \sqrt {a^{2} - b^{2}} A \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) - {\left (B a + B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (B a - B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}}, -\frac {B a \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 2 \, \sqrt {-a^{2} + b^{2}} A \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) - {\left (B a + B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (B a - B b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(B*a*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - sqrt(a^2 - b^2)*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*c
osh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b)) - (B*a + B*b)*log(cosh(x) + sinh(x) + 1
) - (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2), -(B*a*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + 2*
sqrt(-a^2 + b^2)*A*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) - (B*a + B*b)*log(cosh(x)
+ sinh(x) + 1) - (B*a - B*b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)]

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giac [A]  time = 0.13, size = 90, normalized size = 0.90 \[ -\frac {B a \log \left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}{a^{2} - b^{2}} + \frac {2 \, A \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}}} + \frac {B \log \left (e^{x} + 1\right )}{a - b} + \frac {B \log \left ({\left | e^{x} - 1 \right |}\right )}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 139, normalized size = 1.39 \[ -\frac {a B \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A a}{\left (a +b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right ) A b}{\left (a +b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*coth(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 = 3.68, size = 974, normalized size = 9.74 \[ \frac {B\,\ln \left ({\mathrm {e}}^x+1\right )}{a-b}+\frac {\ln \left (\frac {\left (\frac {32\,\left (A^2\,a^2\,b+2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3+8\,{\mathrm {e}}^x\,A\,B\,a^3+4\,A\,B\,a^2\,b-2\,{\mathrm {e}}^x\,A\,B\,a\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^3+3\,B^2\,a^2\,b+5\,{\mathrm {e}}^x\,B^2\,a\,b^2+B^2\,b^3\right )}{b^5}+\frac {\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )\,\left (128\,{\mathrm {e}}^x\,{\left (a^2-b^2\right )}^3\,\left (A-2\,B\right )+a\,b^5\,\left (64\,A-128\,B\right )+a^5\,b\,\left (64\,A-128\,B\right )+96\,b^6\,{\mathrm {e}}^x\,\left (A-3\,B\right )-a^3\,b^3\,\left (128\,A-256\,B\right )-192\,a^2\,b^4\,{\mathrm {e}}^x\,\left (A-3\,B\right )+96\,a^4\,b^2\,{\mathrm {e}}^x\,\left (A-3\,B\right )+128\,A\,a^3\,{\mathrm {e}}^x\,\sqrt {{\left (a^2-b^2\right )}^3}+96\,A\,a^2\,b\,\sqrt {{\left (a^2-b^2\right )}^3}-32\,A\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {{\left (a^2-b^2\right )}^3}\right )}{\left (b^7-a^2\,b^5\right )\,{\left (a^2-b^2\right )}^2}\right )\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )}{{\left (a^2-b^2\right )}^2}-\frac {32\,B\,\left (A^2\,a\,b+{\mathrm {e}}^x\,A^2\,b^2+4\,{\mathrm {e}}^x\,A\,B\,a^2+2\,A\,B\,a\,b-{\mathrm {e}}^x\,A\,B\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^2+B^2\,a\,b\right )}{b^5}\right )\,\left (A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^3+B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (-\frac {32\,B\,\left (A^2\,a\,b+{\mathrm {e}}^x\,A^2\,b^2+4\,{\mathrm {e}}^x\,A\,B\,a^2+2\,A\,B\,a\,b-{\mathrm {e}}^x\,A\,B\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^2+B^2\,a\,b\right )}{b^5}-\frac {\left (\frac {32\,\left (A^2\,a^2\,b+2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3+8\,{\mathrm {e}}^x\,A\,B\,a^3+4\,A\,B\,a^2\,b-2\,{\mathrm {e}}^x\,A\,B\,a\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^3+3\,B^2\,a^2\,b+5\,{\mathrm {e}}^x\,B^2\,a\,b^2+B^2\,b^3\right )}{b^5}-\frac {\left (B\,a^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )\,\left (128\,{\mathrm {e}}^x\,{\left (a^2-b^2\right )}^3\,\left (A-2\,B\right )+a\,b^5\,\left (64\,A-128\,B\right )+a^5\,b\,\left (64\,A-128\,B\right )+96\,b^6\,{\mathrm {e}}^x\,\left (A-3\,B\right )-a^3\,b^3\,\left (128\,A-256\,B\right )-192\,a^2\,b^4\,{\mathrm {e}}^x\,\left (A-3\,B\right )+96\,a^4\,b^2\,{\mathrm {e}}^x\,\left (A-3\,B\right )-128\,A\,a^3\,{\mathrm {e}}^x\,\sqrt {{\left (a^2-b^2\right )}^3}-96\,A\,a^2\,b\,\sqrt {{\left (a^2-b^2\right )}^3}+32\,A\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {{\left (a^2-b^2\right )}^3}\right )}{\left (b^7-a^2\,b^5\right )\,{\left (a^2-b^2\right )}^2}\right )\,\left (B\,a^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )}{{\left (a^2-b^2\right )}^2}\right )\,\left (B\,a^3+A\,\sqrt {{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {B\,\ln \left ({\mathrm {e}}^x-1\right )}{a+b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*log(exp(x) + 1))/(a - b) + (log((((32*(A^2*b^3 + B^2*b^3 + A^2*a^2*b + 3*B^2*a^2*b + 4*B^2*a^3*exp(x) + 5*B
^2*a*b^2*exp(x) + 4*A*B*a^2*b + 8*A*B*a^3*exp(x) + 2*A^2*a*b^2*exp(x) - 2*A*B*a*b^2*exp(x)))/b^5 + ((A*((a + b
)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2)*(128*exp(x)*(a^2 - b^2)^3*(A - 2*B) + a*b^5*(64*A - 128*B) + a^5*b*(64
*A - 128*B) + 96*b^6*exp(x)*(A - 3*B) - a^3*b^3*(128*A - 256*B) - 192*a^2*b^4*exp(x)*(A - 3*B) + 96*a^4*b^2*ex
p(x)*(A - 3*B) + 128*A*a^3*exp(x)*((a^2 - b^2)^3)^(1/2) + 96*A*a^2*b*((a^2 - b^2)^3)^(1/2) - 32*A*a*b^2*exp(x)
*((a^2 - b^2)^3)^(1/2)))/((b^7 - a^2*b^5)*(a^2 - b^2)^2))*(A*((a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2))/(
a^2 - b^2)^2 - (32*B*(A^2*b^2*exp(x) + 4*B^2*a^2*exp(x) + A^2*a*b + B^2*a*b + 4*A*B*a^2*exp(x) - A*B*b^2*exp(x
) + 2*A*B*a*b))/b^5)*(A*((a + b)^3*(a - b)^3)^(1/2) - B*a^3 + B*a*b^2))/(a^4 + b^4 - 2*a^2*b^2) - (log(- (32*B
*(A^2*b^2*exp(x) + 4*B^2*a^2*exp(x) + A^2*a*b + B^2*a*b + 4*A*B*a^2*exp(x) - A*B*b^2*exp(x) + 2*A*B*a*b))/b^5
- (((32*(A^2*b^3 + B^2*b^3 + A^2*a^2*b + 3*B^2*a^2*b + 4*B^2*a^3*exp(x) + 5*B^2*a*b^2*exp(x) + 4*A*B*a^2*b + 8
*A*B*a^3*exp(x) + 2*A^2*a*b^2*exp(x) - 2*A*B*a*b^2*exp(x)))/b^5 - ((B*a^3 + A*((a + b)^3*(a - b)^3)^(1/2) - B*
a*b^2)*(128*exp(x)*(a^2 - b^2)^3*(A - 2*B) + a*b^5*(64*A - 128*B) + a^5*b*(64*A - 128*B) + 96*b^6*exp(x)*(A -
3*B) - a^3*b^3*(128*A - 256*B) - 192*a^2*b^4*exp(x)*(A - 3*B) + 96*a^4*b^2*exp(x)*(A - 3*B) - 128*A*a^3*exp(x)
*((a^2 - b^2)^3)^(1/2) - 96*A*a^2*b*((a^2 - b^2)^3)^(1/2) + 32*A*a*b^2*exp(x)*((a^2 - b^2)^3)^(1/2)))/((b^7 -
a^2*b^5)*(a^2 - b^2)^2))*(B*a^3 + A*((a + b)^3*(a - b)^3)^(1/2) - B*a*b^2))/(a^2 - b^2)^2)*(B*a^3 + A*((a + b)
^3*(a - b)^3)^(1/2) - B*a*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (B*log(exp(x) - 1))/(a + b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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