3.457 \(\int \frac{\coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=77 \[ -\frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d}+\frac{b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac{\coth (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.266282, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2723, 3056, 3001, 3770, 2660, 618, 204} \[ -\frac{2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d}+\frac{b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac{\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^2/(a + b*Sinh[c + d*x]),x]

[Out]

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

Rule 2723

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rule 3770

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

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])

Rubi steps

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

Mathematica [A]  time = 0.441296, size = 98, normalized size = 1.27 \[ -\frac{4 \sqrt{-a^2-b^2} \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )+a \tanh \left (\frac{1}{2} (c+d x)\right )+a \coth \left (\frac{1}{2} (c+d x)\right )+2 b \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^2/(a + b*Sinh[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.003, size = 147, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{1}{d\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}^{2}}{d{a}^{2}\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(coth(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

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

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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(coth(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25519, size = 975, normalized size = 12.66 \begin{align*} \frac{\sqrt{a^{2} + b^{2}}{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \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 ) +{\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) -{\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) - 2 \, a}{a^{2} d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )^{2} - a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{2}{\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(coth(d*x+c)**2/(a+b*sinh(d*x+c)),x)

[Out]

Integral(coth(c + d*x)**2/(a + b*sinh(c + d*x)), x)

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Giac [B]  time = 1.83974, size = 201, normalized size = 2.61 \begin{align*} \frac{\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^{2}} + \frac{b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} - \frac{b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac{2}{a{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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