3.35 \(\int \frac {1}{\sqrt {a+b \coth ^2(x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \coth (x)}{\sqrt {a+b \coth ^2(x)}}\right )}{\sqrt {a+b}} \]

[Out]

arctanh(coth(x)*(a+b)^(1/2)/(a+b*coth(x)^2)^(1/2))/(a+b)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3661, 377, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \coth (x)}{\sqrt {a+b \coth ^2(x)}}\right )}{\sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Coth[x]^2],x]

[Out]

ArcTanh[(Sqrt[a + b]*Coth[x])/Sqrt[a + b*Coth[x]^2]]/Sqrt[a + b]

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])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \coth ^2(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\coth (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\coth (x)}{\sqrt {a+b \coth ^2(x)}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \coth (x)}{\sqrt {a+b \coth ^2(x)}}\right )}{\sqrt {a+b}}\\ \end {align*}

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Mathematica [B]  time = 0.12, size = 77, normalized size = 2.48 \[ \frac {\coth (x) \sqrt {\frac {b \coth ^2(x)}{a}+1} \tanh ^{-1}\left (\frac {\sqrt {\frac {(a+b) \coth ^2(x)}{a}}}{\sqrt {\frac {b \coth ^2(x)}{a}+1}}\right )}{\sqrt {\frac {(a+b) \coth ^2(x)}{a}} \sqrt {a+b \coth ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Coth[x]^2],x]

[Out]

(ArcTanh[Sqrt[((a + b)*Coth[x]^2)/a]/Sqrt[1 + (b*Coth[x]^2)/a]]*Coth[x]*Sqrt[1 + (b*Coth[x]^2)/a])/(Sqrt[((a +
b)*Coth[x]^2)/a]*Sqrt[a + b*Coth[x]^2])

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fricas [B]  time = 0.50, size = 1357, normalized size = 43.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(x)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(a + b)*log(((a*b^2 + b^3)*cosh(x)^8 + 8*(a*b^2 + b^3)*cosh(x)*sinh(x)^7 + (a*b^2 + b^3)*sinh(x)^8 +
2*(a*b^2 + 2*b^3)*cosh(x)^6 + 2*(a*b^2 + 2*b^3 + 14*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^6 + 4*(14*(a*b^2 + b^3)*
cosh(x)^3 + 3*(a*b^2 + 2*b^3)*cosh(x))*sinh(x)^5 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^4 + (70*(a*b^2 + b^
3)*cosh(x)^4 + a^3 - a^2*b + 4*a*b^2 + 6*b^3 + 30*(a*b^2 + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a*b^2 + b^3)*c
osh(x)^5 + 10*(a*b^2 + 2*b^3)*cosh(x)^3 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x))*sinh(x)^3 + a^3 + 3*a^2*b +
3*a*b^2 + b^3 - 2*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2 + 2*(14*(a*b^2 + b^3)*cosh(x)^6 + 15*(a*b^2 + 2*b^3)*cosh
(x)^4 - a^3 + 3*a*b^2 + 2*b^3 + 3*(a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh(x)^
6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 + 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 + b^2)*sinh(x)^4 + 4*(5*b^2
*cosh(x)^3 + 3*b^2*cosh(x))*sinh(x)^3 - (a^2 - 2*a*b - 3*b^2)*cosh(x)^2 + (15*b^2*cosh(x)^4 + 18*b^2*cosh(x)^2
- a^2 + 2*a*b + 3*b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*(3*b^2*cosh(x)^5 + 6*b^2*cosh(x)^3 - (a^2 - 2*a*b -
3*b^2)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(
x)*sinh(x) + sinh(x)^2)) + 4*(2*(a*b^2 + b^3)*cosh(x)^7 + 3*(a*b^2 + 2*b^3)*cosh(x)^5 + (a^3 - a^2*b + 4*a*b^2
+ 6*b^3)*cosh(x)^3 - (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^
4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + sqrt(a + b
)*log(-((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 - 2*a*cosh(x)^2 + 2*(3*(a + b)*cos
h(x)^2 - a)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(a + b)*sqrt(((a + b)*cosh
(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*((a + b)*cosh(x)^3 - a*cos
h(x))*sinh(x) + a + b)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)))/(a + b), -1/2*(sqrt(-a - b)*arctan(sqrt(2
)*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + a + b)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sin
h(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a*b + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*si
nh(x)^3 + (a*b + b^2)*sinh(x)^4 - (a^2 - a*b - 2*b^2)*cosh(x)^2 + (6*(a*b + b^2)*cosh(x)^2 - a^2 + a*b + 2*b^2
)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*(2*(a*b + b^2)*cosh(x)^3 - (a^2 - a*b - 2*b^2)*cosh(x))*sinh(x))) + sqrt(-
a - b)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (
a + b)*sinh(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*
sinh(x)^3 + (a + b)*sinh(x)^4 - 2*(a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 - a + b)*sinh(x)^2 + 4*((a + b)*c
osh(x)^3 - (a - b)*cosh(x))*sinh(x) + a + b)))/(a + b)]

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giac [B]  time = 0.35, size = 196, normalized size = 6.32 \[ -\frac {\frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} + \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{\sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{\sqrt {a + b}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{\sqrt {a + b}}}{2 \, \mathrm {sgn}\left (e^{\left (2 \, x\right )} - 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/2*(log(abs(-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) - 2*a*e^(2*x) + 2*b*e^(2*x) + a + b))*(a + b)
+ sqrt(a + b)*(a - b)))/sqrt(a + b) - log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) - 2*a*e^(2*x)
+ 2*b*e^(2*x) + a + b) + sqrt(a + b)))/sqrt(a + b) + log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x
) - 2*a*e^(2*x) + 2*b*e^(2*x) + a + b) - sqrt(a + b)))/sqrt(a + b))/sgn(e^(2*x) - 1)

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maple [B]  time = 0.12, size = 114, normalized size = 3.68 \[ \frac {\ln \left (\frac {2 a +2 b +2 \left (\coth \relax (x )-1\right ) b +2 \sqrt {a +b}\, \sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}}{\coth \relax (x )-1}\right )}{2 \sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\coth \relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}}{1+\coth \relax (x )}\right )}{2 \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*coth(x)^2)^(1/2),x)

[Out]

1/2/(a+b)^(1/2)*ln((2*a+2*b+2*(coth(x)-1)*b+2*(a+b)^(1/2)*((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(1/2))/(coth(x
)-1))-1/2/(a+b)^(1/2)*ln((2*a+2*b-2*(1+coth(x))*b+2*(a+b)^(1/2)*((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2))/(
1+coth(x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \coth \relax (x)^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(x)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*coth(x)^2 + a), x)

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mupad [B]  time = 1.48, size = 25, normalized size = 0.81 \[ \frac {\mathrm {atanh}\left (\frac {\mathrm {coth}\relax (x)\,\sqrt {a+b}}{\sqrt {b\,{\mathrm {coth}\relax (x)}^2+a}}\right )}{\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*coth(x)^2)^(1/2),x)

[Out]

atanh((coth(x)*(a + b)^(1/2))/(a + b*coth(x)^2)^(1/2))/(a + b)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b \coth ^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*coth(x)**2), x)

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