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]))/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)

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Rubi [A]  time = 0.0900475, antiderivative size = 73, normalized size of antiderivative = 1., 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 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]

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 2637

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

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 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])

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*}

Mathematica [A]  time = 0.224738, size = 80, normalized size = 1.1 \[ \frac{a \sinh (x)}{a^2-b^2}+\frac{b \cosh (x)}{b^2-a^2}-\frac{2 b^2 \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \]

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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Maple [A]  time = 0.038, size = 93, normalized size = 1.3 \begin{align*} -2\,{\frac{{b}^{2}}{ \left ( a+b \right ) \left ( a-b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{1}{ \left ( 2\,b+2\,a \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) }}-2\,{\frac{1}{ \left ( 2\,a-2\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) }} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.49567, size = 1106, normalized size = 15.15 \begin{align*} \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 \left (x\right )^{2} - 2 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - a + b}{{\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )^{2} + a - b}\right )}{2 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )\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 \left (x\right )^{2} - 2 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) -{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{2} - 4 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (\frac{\sqrt{a^{2} - b^{2}}}{{\left (a + b\right )} \cosh \left (x\right ) +{\left (a + b\right )} \sinh \left (x\right )}\right )}{2 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )\right )}}\right ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh{\left (x \right )}}{a + b \tanh{\left (x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.21068, size = 82, normalized size = 1.12 \begin{align*} -\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 )}} \end{align*}

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)