3.113 \(\int \frac {\cosh (x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=73 \[ \frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh (x)}{a^2-b^2}-\frac {b^2 \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {3511, 3486, 2637, 3509, 206} \[ \frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh (x)}{a^2-b^2}-\frac {b^2 \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a + b*Tanh[x]),x]

[Out]

-((b^2*ArcTan[(Cosh[x]*(b + a*Tanh[x]))/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2)) - (b*Cosh[x])/(a^2 - b^2) + (a*Si
nh[x])/(a^2 - b^2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0]

Rule 3511

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^
2), Int[(d*Sec[e + f*x])^m*(a - b*Tan[e + f*x]), x], x] + Dist[b^2/(d^2*(a^2 + b^2)), Int[(d*Sec[e + f*x])^(m
+ 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\cosh (x)}{a+b \tanh (x)} \, dx &=\frac {\int \cosh (x) (a-b \tanh (x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {\text {sech}(x)}{a+b \tanh (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \cosh (x)}{a^2-b^2}+\frac {a \int \cosh (x) \, dx}{a^2-b^2}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,\cosh (x) (-i b-i a \tanh (x))\right )}{a^2-b^2}\\ &=-\frac {b^2 \tan ^{-1}\left (\frac {\cosh (x) (b+a \tanh (x))}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {b \cosh (x)}{a^2-b^2}+\frac {a \sinh (x)}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 80, normalized size = 1.10 \[ \frac {a \sinh (x)}{a^2-b^2}+\frac {b \cosh (x)}{b^2-a^2}-\frac {2 b^2 \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(a + b*Tanh[x]),x]

[Out]

(-2*b^2*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])])/((a - b)^(3/2)*(a + b)^(3/2)) + (b*Cosh[x])/(-a^2
+ b^2) + (a*Sinh[x])/(a^2 - b^2)

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fricas [B]  time = 0.71, size = 435, normalized size = 5.96 \[ \left [-\frac {a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{2} - 2 \, {\left (b^{2} \cosh \relax (x) + b^{2} \sinh \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - a + b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} + a - b}\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)\right )}}, -\frac {a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{2} - 4 \, {\left (b^{2} \cosh \relax (x) + b^{2} \sinh \relax (x)\right )} \sqrt {a^{2} - b^{2}} \arctan \left (\frac {\sqrt {a^{2} - b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^3 + a^2*b - a*b^2 - b^3 - (a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2 - 2*(a^3 - a^2*b - a*b^2 + b^3)*cosh(
x)*sinh(x) - (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^2 - 2*(b^2*cosh(x) + b^2*sinh(x))*sqrt(-a^2 + b^2)*log(((a +
b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x)) - a + b)
/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x
) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)), -1/2*(a^3 + a^2*b - a*b^2 - b^3 - (a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2
- 2*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x) - (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^2 - 4*(b^2*cosh(x) + b^2
*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))))/((a^4 - 2*a^2*b^2 + b^4
)*cosh(x) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x))]

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giac [A]  time = 0.13, size = 61, normalized size = 0.84 \[ -\frac {2 \, b^{2} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {e^{\left (-x\right )}}{2 \, {\left (a - b\right )}} + \frac {e^{x}}{2 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^2*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/(a^2 - b^2)^(3/2) - 1/2*e^(-x)/(a - b) + 1/2*e^x/(a + b)

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maple [A]  time = 0.12, size = 93, normalized size = 1.27 \[ -\frac {2 b^{2} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {2}{\left (2 a -2 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2}{\left (2 b +2 a \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^2/(a-b)/(a+b)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))-2/(2*a-2*b)/(tanh(1/2*x)+
1)-2/(2*b+2*a)/(tanh(1/2*x)-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(cosh(x)/(a+b*tanh(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.36, size = 157, normalized size = 2.15 \[ \frac {{\mathrm {e}}^x}{2\,a+2\,b}-\frac {{\mathrm {e}}^{-x}}{2\,a-2\,b}-\frac {b^2\,\ln \left (-\frac {2\,b^2}{{\left (a+b\right )}^{5/2}\,\sqrt {b-a}}-\frac {2\,b^2\,{\mathrm {e}}^x}{-a^3-a^2\,b+a\,b^2+b^3}\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}+\frac {b^2\,\ln \left (\frac {2\,b^2}{{\left (a+b\right )}^{5/2}\,\sqrt {b-a}}-\frac {2\,b^2\,{\mathrm {e}}^x}{-a^3-a^2\,b+a\,b^2+b^3}\right )}{{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a + b*tanh(x)),x)

[Out]

exp(x)/(2*a + 2*b) - exp(-x)/(2*a - 2*b) - (b^2*log(- (2*b^2)/((a + b)^(5/2)*(b - a)^(1/2)) - (2*b^2*exp(x))/(
a*b^2 - a^2*b - a^3 + b^3)))/((a + b)^(3/2)*(b - a)^(3/2)) + (b^2*log((2*b^2)/((a + b)^(5/2)*(b - a)^(1/2)) -
(2*b^2*exp(x))/(a*b^2 - a^2*b - a^3 + b^3)))/((a + b)^(3/2)*(b - a)^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)/(a + b*tanh(x)), x)

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