3.116 \(\int \frac {\tanh ^2(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=100 \[ \frac {b^2 x}{a \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2}-\frac {a \tanh (x)}{a^2+b^2}+\frac {b \text {sech}(x)}{a^2+b^2}+\frac {2 b^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{3/2}} \]

[Out]

a*x/(a^2+b^2)+b^2*x/a/(a^2+b^2)+2*b^3*arctanh((a-b*tanh(1/2*x))/(a^2+b^2)^(1/2))/a/(a^2+b^2)^(3/2)+b*sech(x)/(
a^2+b^2)-a*tanh(x)/(a^2+b^2)

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Rubi [A]  time = 0.22, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3898, 2902, 2606, 8, 3473, 2735, 2660, 618, 204} \[ \frac {b^2 x}{a \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2}+\frac {2 b^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{3/2}}-\frac {a \tanh (x)}{a^2+b^2}+\frac {b \text {sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Csch[x]),x]

[Out]

(a*x)/(a^2 + b^2) + (b^2*x)/(a*(a^2 + b^2)) + (2*b^3*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a*(a^2 + b^2
)^(3/2)) + (b*Sech[x])/(a^2 + b^2) - (a*Tanh[x])/(a^2 + b^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 2735

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

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\tanh ^2(x)}{a+b \text {csch}(x)} \, dx &=i \int \frac {\sinh (x) \tanh ^2(x)}{i b+i a \sinh (x)} \, dx\\ &=\frac {a \int \tanh ^2(x) \, dx}{a^2+b^2}-\frac {b \int \text {sech}(x) \tanh (x) \, dx}{a^2+b^2}+\frac {\left (i b^2\right ) \int \frac {\sinh (x)}{i b+i a \sinh (x)} \, dx}{a^2+b^2}\\ &=\frac {b^2 x}{a \left (a^2+b^2\right )}-\frac {a \tanh (x)}{a^2+b^2}+\frac {a \int 1 \, dx}{a^2+b^2}+\frac {b \operatorname {Subst}(\int 1 \, dx,x,\text {sech}(x))}{a^2+b^2}-\frac {\left (i b^3\right ) \int \frac {1}{i b+i a \sinh (x)} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac {a x}{a^2+b^2}+\frac {b^2 x}{a \left (a^2+b^2\right )}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {a \tanh (x)}{a^2+b^2}-\frac {\left (2 i b^3\right ) \operatorname {Subst}\left (\int \frac {1}{i b+2 i a x-i b x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right )}\\ &=\frac {a x}{a^2+b^2}+\frac {b^2 x}{a \left (a^2+b^2\right )}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {a \tanh (x)}{a^2+b^2}+\frac {\left (4 i b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,2 i a-2 i b \tanh \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right )}\\ &=\frac {a x}{a^2+b^2}+\frac {b^2 x}{a \left (a^2+b^2\right )}+\frac {2 b^3 \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{3/2}}+\frac {b \text {sech}(x)}{a^2+b^2}-\frac {a \tanh (x)}{a^2+b^2}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 82, normalized size = 0.82 \[ -\frac {a \tanh (x)}{a^2+b^2}+\frac {b \text {sech}(x)}{a^2+b^2}+\frac {\frac {2 b^3 \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+x}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/(a + b*Csch[x]),x]

[Out]

(x + (2*b^3*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(3/2))/a + (b*Sech[x])/(a^2 + b^2) - (a*T
anh[x])/(a^2 + b^2)

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fricas [B]  time = 0.69, size = 349, normalized size = 3.49 \[ \frac {2 \, a^{4} + 2 \, a^{2} b^{2} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)^{2} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x \sinh \relax (x)^{2} + {\left (b^{3} \cosh \relax (x)^{2} + 2 \, b^{3} \cosh \relax (x) \sinh \relax (x) + b^{3} \sinh \relax (x)^{2} + b^{3}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x + 2 \, {\left (a^{3} b + a b^{3}\right )} \cosh \relax (x) + 2 \, {\left (a^{3} b + a b^{3} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x \cosh \relax (x)\right )} \sinh \relax (x)}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*csch(x)),x, algorithm="fricas")

[Out]

(2*a^4 + 2*a^2*b^2 + (a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^2 + (a^4 + 2*a^2*b^2 + b^4)*x*sinh(x)^2 + (b^3*cosh(x)^
2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2 + b^3)*sqrt(a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*co
sh(x) + a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x
)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a)) + (a^4 + 2*a^2*b^2 + b^4)*x + 2*(a^3*b + a*b
^3)*cosh(x) + 2*(a^3*b + a*b^3 + (a^4 + 2*a^2*b^2 + b^4)*x*cosh(x))*sinh(x))/(a^5 + 2*a^3*b^2 + a*b^4 + (a^5 +
2*a^3*b^2 + a*b^4)*cosh(x)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x)*sinh(x) + (a^5 + 2*a^3*b^2 + a*b^4)*sinh(x
)^2)

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giac [A]  time = 0.16, size = 102, normalized size = 1.02 \[ -\frac {b^{3} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{3} + a b^{2}\right )} \sqrt {a^{2} + b^{2}}} + \frac {x}{a} + \frac {2 \, {\left (b e^{x} + a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*csch(x)),x, algorithm="giac")

[Out]

-b^3*log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/((a^3 + a*b^2)*sqrt(a^
2 + b^2)) + x/a + 2*(b*e^x + a)/((a^2 + b^2)*(e^(2*x) + 1))

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maple [A]  time = 0.18, size = 95, normalized size = 0.95 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {2 b^{3} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {2 b -2 a \tanh \left (\frac {x}{2}\right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a+b*csch(x)),x)

[Out]

-1/a*ln(tanh(1/2*x)-1)+1/a*ln(tanh(1/2*x)+1)-2/a*b^3/(a^2+b^2)^(3/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^
2)^(1/2))+2/(a^2+b^2)*(b-a*tanh(1/2*x))/(tanh(1/2*x)^2+1)

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maxima [A]  time = 0.42, size = 108, normalized size = 1.08 \[ -\frac {b^{3} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{3} + a b^{2}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (b e^{\left (-x\right )} - a\right )}}{a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )}} + \frac {x}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(a+b*csch(x)),x, algorithm="maxima")

[Out]

-b^3*log((a*e^(-x) - b - sqrt(a^2 + b^2))/(a*e^(-x) - b + sqrt(a^2 + b^2)))/((a^3 + a*b^2)*sqrt(a^2 + b^2)) +
2*(b*e^(-x) - a)/(a^2 + b^2 + (a^2 + b^2)*e^(-2*x)) + x/a

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mupad [B]  time = 2.47, size = 376, normalized size = 3.76 \[ \frac {x}{a}+\frac {\frac {2\,a}{a^2+b^2}+\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{{\mathrm {e}}^{2\,x}+1}+\frac {2\,\mathrm {atan}\left (\left (\frac {a^4\,\sqrt {-a^8-3\,a^6\,b^2-3\,a^4\,b^4-a^2\,b^6}}{2}+\frac {a^2\,b^2\,\sqrt {-a^8-3\,a^6\,b^2-3\,a^4\,b^4-a^2\,b^6}}{2}\right )\,\left ({\mathrm {e}}^x\,\left (\frac {2\,b^3}{a^3\,\left (a^3+a\,b^2\right )\,\sqrt {b^6}\,\left (a^2+b^2\right )}+\frac {2\,\left (a\,b^3\,\sqrt {b^6}+a^3\,b\,\sqrt {b^6}\right )}{a^2\,b^2\,\sqrt {-a^2\,{\left (a^2+b^2\right )}^3}\,\left (a^3+a\,b^2\right )\,\sqrt {-a^8-3\,a^6\,b^2-3\,a^4\,b^4-a^2\,b^6}}\right )-\frac {2\,\left (a^4\,\sqrt {b^6}+a^2\,b^2\,\sqrt {b^6}\right )}{a^2\,b^2\,\sqrt {-a^2\,{\left (a^2+b^2\right )}^3}\,\left (a^3+a\,b^2\right )\,\sqrt {-a^8-3\,a^6\,b^2-3\,a^4\,b^4-a^2\,b^6}}\right )\right )\,\sqrt {b^6}}{\sqrt {-a^8-3\,a^6\,b^2-3\,a^4\,b^4-a^2\,b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a + b/sinh(x)),x)

[Out]

x/a + ((2*a)/(a^2 + b^2) + (2*b*exp(x))/(a^2 + b^2))/(exp(2*x) + 1) + (2*atan(((a^4*(- a^8 - a^2*b^6 - 3*a^4*b
^4 - 3*a^6*b^2)^(1/2))/2 + (a^2*b^2*(- a^8 - a^2*b^6 - 3*a^4*b^4 - 3*a^6*b^2)^(1/2))/2)*(exp(x)*((2*b^3)/(a^3*
(a*b^2 + a^3)*(b^6)^(1/2)*(a^2 + b^2)) + (2*(a*b^3*(b^6)^(1/2) + a^3*b*(b^6)^(1/2)))/(a^2*b^2*(-a^2*(a^2 + b^2
)^3)^(1/2)*(a*b^2 + a^3)*(- a^8 - a^2*b^6 - 3*a^4*b^4 - 3*a^6*b^2)^(1/2))) - (2*(a^4*(b^6)^(1/2) + a^2*b^2*(b^
6)^(1/2)))/(a^2*b^2*(-a^2*(a^2 + b^2)^3)^(1/2)*(a*b^2 + a^3)*(- a^8 - a^2*b^6 - 3*a^4*b^4 - 3*a^6*b^2)^(1/2)))
)*(b^6)^(1/2))/(- a^8 - a^2*b^6 - 3*a^4*b^4 - 3*a^6*b^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{2}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**2/(a+b*csch(x)),x)

[Out]

Integral(tanh(x)**2/(a + b*csch(x)), x)

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