3.241 \(\int \frac{\coth ^2(x)}{(a+b \sinh (x))^2} \, dx\)
Optimal. Leaf size=80 \[ -\frac{2 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2}}+\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3}-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))} \]
[Out]
(2*b*ArcTanh[Cosh[x]])/a^3 - (2*(a^2 + 2*b^2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 + b^2]
) - (2*Coth[x])/a^2 + Coth[x]/(a*(a + b*Sinh[x]))
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Rubi [A] time = 0.402169, antiderivative size = 80, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used =
{2723, 3056, 3001, 3770, 2660, 618, 206} \[ -\frac{2 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2}}+\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3}-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))} \]
Antiderivative was successfully verified.
[In]
Int[Coth[x]^2/(a + b*Sinh[x])^2,x]
[Out]
(2*b*ArcTanh[Cosh[x]])/a^3 - (2*(a^2 + 2*b^2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 + b^2]
) - (2*Coth[x])/a^2 + Coth[x]/(a*(a + b*Sinh[x]))
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 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{\coth ^2(x)}{(a+b \sinh (x))^2} \, dx &=\int \frac{\text{csch}^2(x) \left (1+\sinh ^2(x)\right )}{(a+b \sinh (x))^2} \, dx\\ &=\frac{\coth (x)}{a (a+b \sinh (x))}+\frac{\int \frac{\text{csch}^2(x) \left (2 \left (a^2+b^2\right )+\left (a^2+b^2\right ) \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))}+\frac{i \int \frac{\text{csch}(x) \left (2 i b \left (a^2+b^2\right )-i a \left (a^2+b^2\right ) \sinh (x)\right )}{a+b \sinh (x)} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))}-\frac{(2 b) \int \text{csch}(x) \, dx}{a^3}+\frac{\left (a^2+2 b^2\right ) \int \frac{1}{a+b \sinh (x)} \, dx}{a^3}\\ &=\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3}-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))}+\frac{\left (2 \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3}-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))}-\frac{\left (4 \left (a^2+2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{2 b \tanh ^{-1}(\cosh (x))}{a^3}-\frac{2 \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2}}-\frac{2 \coth (x)}{a^2}+\frac{\coth (x)}{a (a+b \sinh (x))}\\ \end{align*}
Mathematica [A] time = 0.477892, size = 102, normalized size = 1.27 \[ -\frac{-\frac{4 \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+\frac{2 a b \cosh (x)}{a+b \sinh (x)}+a \tanh \left (\frac{x}{2}\right )+a \coth \left (\frac{x}{2}\right )+4 b \log \left (\tanh \left (\frac{x}{2}\right )\right )}{2 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]^2/(a + b*Sinh[x])^2,x]
[Out]
-((-4*(a^2 + 2*b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + a*Coth[x/2] + 4*b*Log[Tanh[
x/2]] + (2*a*b*Cosh[x])/(a + b*Sinh[x]) + a*Tanh[x/2])/(2*a^3)
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Maple [B] time = 0.041, size = 170, normalized size = 2.1 \begin{align*} -{\frac{1}{2\,{a}^{2}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{2\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-2\,{\frac{b\ln \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{3}}}+2\,{\frac{{b}^{2}\tanh \left ( x/2 \right ) }{{a}^{3} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{b}{{a}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{1}{a\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+4\,{\frac{{b}^{2}}{{a}^{3}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/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(x)^2/(a+b*sinh(x))^2,x)
[Out]
-1/2/a^2*tanh(1/2*x)-1/2/a^2/tanh(1/2*x)-2/a^3*b*ln(tanh(1/2*x))+2/a^3/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b^2
*tanh(1/2*x)+2/a^2/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b+2/a/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)
/(a^2+b^2)^(1/2))+4/a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))*b^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(x)^2/(a+b*sinh(x))^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.74811, size = 3102, normalized size = 38.78 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sinh(x))^2,x, algorithm="fricas")
[Out]
(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2)*sinh(x)^3 - 4*(a^3*b + a*b^3)*cosh(x)^2 -
2*(2*a^3*b + 2*a*b^3 - 3*(a^4 + a^2*b^2)*cosh(x))*sinh(x)^2 + ((a^2*b + 2*b^3)*cosh(x)^4 + (a^2*b + 2*b^3)*si
nh(x)^4 + 2*(a^3 + 2*a*b^2)*cosh(x)^3 + 2*(a^3 + 2*a*b^2 + 2*(a^2*b + 2*b^3)*cosh(x))*sinh(x)^3 + a^2*b + 2*b^
3 - 2*(a^2*b + 2*b^3)*cosh(x)^2 - 2*(a^2*b + 2*b^3 - 3*(a^2*b + 2*b^3)*cosh(x)^2 - 3*(a^3 + 2*a*b^2)*cosh(x))*
sinh(x)^2 - 2*(a^3 + 2*a*b^2)*cosh(x) + 2*(2*(a^2*b + 2*b^3)*cosh(x)^3 - a^3 - 2*a*b^2 + 3*(a^3 + 2*a*b^2)*cos
h(x)^2 - 2*(a^2*b + 2*b^3)*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x
) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2
+ b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) - 6*(a^4 + a^2*b^2)*cosh(x) + 2*((a^2*b^2 + b^4)
*cosh(x)^4 + (a^2*b^2 + b^4)*sinh(x)^4 + a^2*b^2 + b^4 + 2*(a^3*b + a*b^3)*cosh(x)^3 + 2*(a^3*b + a*b^3 + 2*(a
^2*b^2 + b^4)*cosh(x))*sinh(x)^3 - 2*(a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^2*b^2 + b^4 - 3*(a^2*b^2 + b^4)*cosh(x)^
2 - 3*(a^3*b + a*b^3)*cosh(x))*sinh(x)^2 - 2*(a^3*b + a*b^3)*cosh(x) - 2*(a^3*b + a*b^3 - 2*(a^2*b^2 + b^4)*co
sh(x)^3 - 3*(a^3*b + a*b^3)*cosh(x)^2 + 2*(a^2*b^2 + b^4)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) - 2*((a
^2*b^2 + b^4)*cosh(x)^4 + (a^2*b^2 + b^4)*sinh(x)^4 + a^2*b^2 + b^4 + 2*(a^3*b + a*b^3)*cosh(x)^3 + 2*(a^3*b +
a*b^3 + 2*(a^2*b^2 + b^4)*cosh(x))*sinh(x)^3 - 2*(a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^2*b^2 + b^4 - 3*(a^2*b^2 +
b^4)*cosh(x)^2 - 3*(a^3*b + a*b^3)*cosh(x))*sinh(x)^2 - 2*(a^3*b + a*b^3)*cosh(x) - 2*(a^3*b + a*b^3 - 2*(a^2*
b^2 + b^4)*cosh(x)^3 - 3*(a^3*b + a*b^3)*cosh(x)^2 + 2*(a^2*b^2 + b^4)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x)
- 1) - 2*(3*a^4 + 3*a^2*b^2 - 3*(a^4 + a^2*b^2)*cosh(x)^2 + 4*(a^3*b + a*b^3)*cosh(x))*sinh(x))/(a^5*b + a^3*
b^3 + (a^5*b + a^3*b^3)*cosh(x)^4 + (a^5*b + a^3*b^3)*sinh(x)^4 + 2*(a^6 + a^4*b^2)*cosh(x)^3 + 2*(a^6 + a^4*b
^2 + 2*(a^5*b + a^3*b^3)*cosh(x))*sinh(x)^3 - 2*(a^5*b + a^3*b^3)*cosh(x)^2 - 2*(a^5*b + a^3*b^3 - 3*(a^5*b +
a^3*b^3)*cosh(x)^2 - 3*(a^6 + a^4*b^2)*cosh(x))*sinh(x)^2 - 2*(a^6 + a^4*b^2)*cosh(x) - 2*(a^6 + a^4*b^2 - 2*(
a^5*b + a^3*b^3)*cosh(x)^3 - 3*(a^6 + a^4*b^2)*cosh(x)^2 + 2*(a^5*b + a^3*b^3)*cosh(x))*sinh(x))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{2}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)**2/(a+b*sinh(x))**2,x)
[Out]
Integral(coth(x)**2/(a + b*sinh(x))**2, x)
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Giac [A] time = 1.18626, size = 200, normalized size = 2.5 \begin{align*} \frac{2 \, b \log \left (e^{x} + 1\right )}{a^{3}} - \frac{2 \, b \log \left ({\left | e^{x} - 1 \right |}\right )}{a^{3}} + \frac{{\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3}} + \frac{2 \,{\left (a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - 3 \, a e^{x} + 2 \, b\right )}}{{\left (b e^{\left (4 \, x\right )} + 2 \, a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - 2 \, a e^{x} + b\right )} a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sinh(x))^2,x, algorithm="giac")
[Out]
2*b*log(e^x + 1)/a^3 - 2*b*log(abs(e^x - 1))/a^3 + (a^2 + 2*b^2)*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/ab
s(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^3) + 2*(a*e^(3*x) - 2*b*e^(2*x) - 3*a*e^x + 2*b)/((b*
e^(4*x) + 2*a*e^(3*x) - 2*b*e^(2*x) - 2*a*e^x + b)*a^2)