3.52 \(\int \frac {\coth (x)}{\sqrt {a+b \coth ^4(x)}} \, dx\)
Optimal. Leaf size=40 \[ \frac {\tanh ^{-1}\left (\frac {a+b \coth ^2(x)}{\sqrt {a+b} \sqrt {a+b \coth ^4(x)}}\right )}{2 \sqrt {a+b}} \]
[Out]
1/2*arctanh((a+b*coth(x)^2)/(a+b)^(1/2)/(a+b*coth(x)^4)^(1/2))/(a+b)^(1/2)
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Rubi [A] time = 0.08, antiderivative size = 40, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used =
{3670, 1248, 725, 206} \[ \frac {\tanh ^{-1}\left (\frac {a+b \coth ^2(x)}{\sqrt {a+b} \sqrt {a+b \coth ^4(x)}}\right )}{2 \sqrt {a+b}} \]
Antiderivative was successfully verified.
[In]
Int[Coth[x]/Sqrt[a + b*Coth[x]^4],x]
[Out]
ArcTanh[(a + b*Coth[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Coth[x]^4])]/(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 725
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]
Rule 1248
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]
Rule 3670
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(((d*ff*x)/c)^m*(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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))
Rubi steps
\begin {align*} \int \frac {\coth (x)}{\sqrt {a+b \coth ^4(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {x}{\left (1-x^2\right ) \sqrt {a+b x^4}} \, dx,x,\coth (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x^2}} \, dx,x,\coth ^2(x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {-a-b \coth ^2(x)}{\sqrt {a+b \coth ^4(x)}}\right )\right )\\ &=\frac {\tanh ^{-1}\left (\frac {a+b \coth ^2(x)}{\sqrt {a+b} \sqrt {a+b \coth ^4(x)}}\right )}{2 \sqrt {a+b}}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 40, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {a+b \coth ^2(x)}{\sqrt {a+b} \sqrt {a+b \coth ^4(x)}}\right )}{2 \sqrt {a+b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]/Sqrt[a + b*Coth[x]^4],x]
[Out]
ArcTanh[(a + b*Coth[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Coth[x]^4])]/(2*Sqrt[a + b])
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fricas [B] time = 0.68, size = 1290, normalized size = 32.25 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)/(a+b*coth(x)^4)^(1/2),x, algorithm="fricas")
[Out]
[1/4*log(((a^2 + 2*a*b + b^2)*cosh(x)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^7 + (a^2 + 2*a*b + b^2)*sinh(x
)^8 - 4*(a^2 - b^2)*cosh(x)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(x)^2 - a^2 + b^2)*sinh(x)^6 + 8*(7*(a^2 + 2*a*b
+ b^2)*cosh(x)^3 - 3*(a^2 - b^2)*cosh(x))*sinh(x)^5 + 2*(3*a^2 + 2*a*b + 3*b^2)*cosh(x)^4 + 2*(35*(a^2 + 2*a*b
+ b^2)*cosh(x)^4 - 30*(a^2 - b^2)*cosh(x)^2 + 3*a^2 + 2*a*b + 3*b^2)*sinh(x)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cos
h(x)^5 - 10*(a^2 - b^2)*cosh(x)^3 + (3*a^2 + 2*a*b + 3*b^2)*cosh(x))*sinh(x)^3 - 4*(a^2 - b^2)*cosh(x)^2 + 4*(
7*(a^2 + 2*a*b + b^2)*cosh(x)^6 - 15*(a^2 - b^2)*cosh(x)^4 + 3*(3*a^2 + 2*a*b + 3*b^2)*cosh(x)^2 - a^2 + b^2)*
sinh(x)^2 + sqrt(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)*cosh(x)^3 - (a - b)*cosh(x))*sinh(x) + a + b)*sqrt(a
+ b)*sqrt(((a + b)*cosh(x)^4 + (a + b)*sinh(x)^4 - 4*(a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 - 2*a + 2*b)*
sinh(x)^2 + 3*a + 3*b)/(cosh(x)^4 - 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 - 4*cosh(x)*sinh(x)^3 + sinh(x
)^4)) + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(x)^7 - 3*(a^2 - b^2)*cosh(x)^5 + (3*a^2 + 2*a*b + 3*b^
2)*cosh(x)^3 - (a^2 - b^2)*cosh(x))*sinh(x))/(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh
(x)*sinh(x)^3 + sinh(x)^4))/sqrt(a + b), -1/2*sqrt(-a - b)*arctan(sqrt(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
)*cosh(x)^3 - (a - b)*cosh(x))*sinh(x) + a + b)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^4 + (a + b)*sinh(x)^4 - 4*(
a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 - 2*a + 2*b)*sinh(x)^2 + 3*a + 3*b)/(cosh(x)^4 - 4*cosh(x)^3*sinh(x)
+ 6*cosh(x)^2*sinh(x)^2 - 4*cosh(x)*sinh(x)^3 + sinh(x)^4))/((a^2 + 2*a*b + b^2)*cosh(x)^8 + 8*(a^2 + 2*a*b +
b^2)*cosh(x)*sinh(x)^7 + (a^2 + 2*a*b + b^2)*sinh(x)^8 - 4*(a^2 - b^2)*cosh(x)^6 + 4*(7*(a^2 + 2*a*b + b^2)*c
osh(x)^2 - a^2 + b^2)*sinh(x)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(x)^3 - 3*(a^2 - b^2)*cosh(x))*sinh(x)^5 + 6*(a
^2 + 2*a*b + b^2)*cosh(x)^4 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(x)^4 - 30*(a^2 - b^2)*cosh(x)^2 + 3*a^2 + 6*a*b +
3*b^2)*sinh(x)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(x)^5 - 10*(a^2 - b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*cosh
(x))*sinh(x)^3 - 4*(a^2 - b^2)*cosh(x)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(x)^6 - 15*(a^2 - b^2)*cosh(x)^4 + 9*(
a^2 + 2*a*b + b^2)*cosh(x)^2 - a^2 + b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(x)^7 - 3
*(a^2 - b^2)*cosh(x)^5 + 3*(a^2 + 2*a*b + b^2)*cosh(x)^3 - (a^2 - b^2)*cosh(x))*sinh(x)))/(a + b)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \relax (x)}{\sqrt {b \coth \relax (x)^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)/(a+b*coth(x)^4)^(1/2),x, algorithm="giac")
[Out]
integrate(coth(x)/sqrt(b*coth(x)^4 + a), x)
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maple [A] time = 0.14, size = 37, normalized size = 0.92 \[ \frac {\arctanh \left (\frac {2 b \left (\coth ^{2}\relax (x )\right )+2 a}{2 \sqrt {a +b}\, \sqrt {a +b \left (\coth ^{4}\relax (x )\right )}}\right )}{2 \sqrt {a +b}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)/(a+b*coth(x)^4)^(1/2),x)
[Out]
1/2/(a+b)^(1/2)*arctanh(1/2*(2*b*coth(x)^2+2*a)/(a+b)^(1/2)/(a+b*coth(x)^4)^(1/2))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \relax (x)}{\sqrt {b \coth \relax (x)^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)/(a+b*coth(x)^4)^(1/2),x, algorithm="maxima")
[Out]
integrate(coth(x)/sqrt(b*coth(x)^4 + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {coth}\relax (x)}{\sqrt {b\,{\mathrm {coth}\relax (x)}^4+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)/(a + b*coth(x)^4)^(1/2),x)
[Out]
int(coth(x)/(a + b*coth(x)^4)^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth {\relax (x )}}{\sqrt {a + b \coth ^{4}{\relax (x )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)/(a+b*coth(x)**4)**(1/2),x)
[Out]
Integral(coth(x)/sqrt(a + b*coth(x)**4), x)
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