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

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Rubi [A]  time = 0.27, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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 {\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*}

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Mathematica [A]  time = 0.41, 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]

-1/2*(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[Ta
nh[(c + d*x)/2]] + a*Tanh[(c + d*x)/2])/(a^2*d)

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fricas [B]  time = 0.49, size = 360, normalized size = 4.68 \[ \frac {\sqrt {a^{2} + b^{2}} {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) - 2 \, a}{a^{2} d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )^{2} - a^{2} d} \]

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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giac [B]  time = 1.11, size = 149, normalized size = 1.94 \[ \frac {\frac {{\left (a^{2} e^{c} + b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left (\frac {{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} - 2 \, \sqrt {a^{2} + b^{2}} e^{c} \right |}}{{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} + 2 \, \sqrt {a^{2} + b^{2}} e^{c} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} - \frac {b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \]

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

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maple [B]  time = 0.00, size = 147, normalized size = 1.91 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {2 \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \sqrt {a^{2}+b^{2}}}+\frac {2 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

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)+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))-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))

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maxima [A]  time = 0.41, size = 134, normalized size = 1.74 \[ \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} + \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{a^{2} d} + \frac {2}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} \]

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]

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

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mupad [B]  time = 0.64, size = 380, normalized size = 4.94 \[ \frac {2}{a\,d-a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}-\frac {b\,\ln \left (32\,a^2+32\,b^2-32\,a^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^2\,d}+\frac {b\,\ln \left (32\,a^2+32\,b^2+32\,a^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^2\,d}+\frac {\ln \left (128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a\,b^3-64\,a^3\,b-32\,b^3\,\sqrt {a^2+b^2}+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,d}-\frac {\ln \left (32\,b^3\,\sqrt {a^2+b^2}-64\,a\,b^3-64\,a^3\,b+128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a*d - a*d*exp(2*c + 2*d*x)) - (b*log(32*a^2 + 32*b^2 - 32*a^2*exp(d*x)*exp(c) - 32*b^2*exp(d*x)*exp(c)))/(a
^2*d) + (b*log(32*a^2 + 32*b^2 + 32*a^2*exp(d*x)*exp(c) + 32*b^2*exp(d*x)*exp(c)))/(a^2*d) + (log(128*a^4*exp(
d*x)*exp(c) - 64*a*b^3 - 64*a^3*b - 32*b^3*(a^2 + b^2)^(1/2) + 32*b^4*exp(d*x)*exp(c) - 64*a^2*b*(a^2 + b^2)^(
1/2) + 160*a^2*b^2*exp(d*x)*exp(c) + 128*a^3*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 96*a*b^2*exp(d*x)*exp(c)*(a^2
+ b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^2*d) - (log(32*b^3*(a^2 + b^2)^(1/2) - 64*a*b^3 - 64*a^3*b + 128*a^4*exp(
d*x)*exp(c) + 32*b^4*exp(d*x)*exp(c) + 64*a^2*b*(a^2 + b^2)^(1/2) + 160*a^2*b^2*exp(d*x)*exp(c) - 128*a^3*exp(
d*x)*exp(c)*(a^2 + b^2)^(1/2) - 96*a*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{2}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

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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