3.428 \(\int \frac{\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=71 \[ \frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a b d}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac{x}{b} \]

[Out]

x/b - ArcTanh[Cosh[c + d*x]]/(a*d) + (2*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a
*b*d)

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Rubi [A]  time = 0.212942, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2889, 3058, 2660, 618, 204, 3770} \[ \frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a b d}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[c + d*x]*Coth[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

x/b - ArcTanh[Cosh[c + d*x]]/(a*d) + (2*Sqrt[a^2 + b^2]*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a
*b*d)

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3058

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 + a^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[
e + f*x]), x], x] - Dist[(c^2*C + A*d^2)/(d*(b*c - a*d)), Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b,
c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx &=\int \frac{\text{csch}(c+d x) \left (1+\sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx\\ &=\frac{x}{b}+\frac{\int \text{csch}(c+d x) \, dx}{a}-\frac{\left (a^2+b^2\right ) \int \frac{1}{a+b \sinh (c+d x)} \, dx}{a b}\\ &=\frac{x}{b}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac{\left (2 i \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a b d}\\ &=\frac{x}{b}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d}-\frac{\left (4 i \left (a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a b d}\\ &=\frac{x}{b}-\frac{\tanh ^{-1}(\cosh (c+d x))}{a d}+\frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a b d}\\ \end{align*}

Mathematica [A]  time = 0.146125, size = 80, normalized size = 1.13 \[ \frac{2 \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )+a (c+d x)+b \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{a b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[c + d*x]*Coth[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.002, size = 150, normalized size = 2.1 \begin{align*}{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{a}{bd\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{b}{da\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

1/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/b*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/a*ln(tanh(1/2*d*x+1/2*c))-2/d*a/b/(a^2+b^2
)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-2/d*b/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*ta
nh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))

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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(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.58144, size = 559, normalized size = 7.87 \begin{align*} \frac{a d x - b \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + b \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) + \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \,{\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right )}{a b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(a*d*x - b*log(cosh(d*x + c) + sinh(d*x + c) + 1) + b*log(cosh(d*x + c) + sinh(d*x + c) - 1) + sqrt(a^2 + b^2)
*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a
*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x
+ c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)))/(a*b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral(cosh(c + d*x)*coth(c + d*x)/(a + b*sinh(c + d*x)), x)

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Giac [B]  time = 2.39724, size = 188, normalized size = 2.65 \begin{align*} \frac{\frac{d x}{b} - \frac{{\left (a^{2} e^{c} + b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left (\frac{{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} - 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}{{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} + 2 \, \sqrt{a^{2} + b^{2}} e^{c} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a b} - \frac{\log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac{\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

(d*x/b - (a^2*e^c + b^2*e^c)*e^(-c)*log(abs(2*b*e^(d*x + 2*c) + 2*a*e^c - 2*sqrt(a^2 + b^2)*e^c)/abs(2*b*e^(d*
x + 2*c) + 2*a*e^c + 2*sqrt(a^2 + b^2)*e^c))/(sqrt(a^2 + b^2)*a*b) - log(e^(d*x + c) + 1)/a + log(abs(e^(d*x +
c) - 1))/a)/d