\(\int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx\) [210]

Optimal result

Integrand size = 21, antiderivative size = 132 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \coth ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \]

[Out]

-1/4*(b+2*c)*arctanh(1/2*(b+2*c*coth(x)^2)/c^(1/2)/(a+b*coth(x)^2+c*coth(x)^4)^(1/2))/c^(1/2)+1/2*arctanh(1/2*
(2*a+b+(b+2*c)*coth(x)^2)/(a+b+c)^(1/2)/(a+b*coth(x)^2+c*coth(x)^4)^(1/2))*(a+b+c)^(1/2)-1/2*(a+b*coth(x)^2+c*
coth(x)^4)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3782, 1261, 748, 857, 635, 212, 738} \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+(b+2 c) \coth ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \]

[In]

Int[Coth[x]*Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4],x]

[Out]

-1/4*((b + 2*c)*ArcTanh[(b + 2*c*Coth[x]^2)/(2*Sqrt[c]*Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4])])/Sqrt[c] + (Sqrt[
a + b + c]*ArcTanh[(2*a + b + (b + 2*c)*Coth[x]^2)/(2*Sqrt[a + b + c]*Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4])])/2
- Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4]/2

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 3782

Int[cot[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*(cot[(d_.) + (e_.)*(x_)]*(f_.))^(n_.) + (c_.)*(cot[(d_.) + (e
_.)*(x_)]*(f_.))^(n2_.))^(p_), x_Symbol] :> Dist[-f/e, Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2
)), x], x, f*Cot[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x \sqrt {a-b x^2+c x^4}}{1+x^2} \, dx,x,-i \coth (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a-b x+c x^2}}{1+x} \, dx,x,-\coth ^2(x)\right )\right ) \\ & = -\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}+\frac {1}{4} \text {Subst}\left (\int \frac {-2 a-b+(b+2 c) x}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right ) \\ & = -\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}+\frac {1}{2} (-a-b-c) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right )+\frac {1}{4} (b+2 c) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right ) \\ & = -\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}+(a+b+c) \text {Subst}\left (\int \frac {1}{4 a+4 b+4 c-x^2} \, dx,x,\frac {2 a+b+(b+2 c) \coth ^2(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )+\frac {1}{2} (b+2 c) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {-b-2 c \coth ^2(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right ) \\ & = -\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \coth ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.39 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=-\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \tanh ^2(x) \left ((b+2 c) \text {arctanh}\left (\frac {2 c+b \tanh ^2(x)}{2 \sqrt {c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )-2 \sqrt {c} \sqrt {a+b+c} \text {arctanh}\left (\frac {b+2 c+(2 a+b) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )+2 \sqrt {c} \coth ^2(x) \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}\right )}{4 \sqrt {c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}} \]

[In]

Integrate[Coth[x]*Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4],x]

[Out]

-1/4*(Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4]*Tanh[x]^2*((b + 2*c)*ArcTanh[(2*c + b*Tanh[x]^2)/(2*Sqrt[c]*Sqrt[c +
b*Tanh[x]^2 + a*Tanh[x]^4])] - 2*Sqrt[c]*Sqrt[a + b + c]*ArcTanh[(b + 2*c + (2*a + b)*Tanh[x]^2)/(2*Sqrt[a +
b + c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])] + 2*Sqrt[c]*Coth[x]^2*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4]))/(Sqrt
[c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(-\frac {\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\coth \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{\coth \left (x \right )^{2}-1}\right )}{2}\)	\(165\)
default	\(-\frac {\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\coth \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{\coth \left (x \right )^{2}-1}\right )}{2}\)	\(165\)

[In]

int(coth(x)*(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*((coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1)+a+b+c)^(1/2)-1/4*(b+2*c)*ln((1/2*b+c+c*(coth(x)^2-1))/c^(1/2)+((
coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1)+a+b+c)^(1/2))/c^(1/2)+1/2*(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(coth(x
)^2-1)+2*(a+b+c)^(1/2)*((coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1)+a+b+c)^(1/2))/(coth(x)^2-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1840 vs. \(2 (108) = 216\).

Time = 1.44 (sec) , antiderivative size = 7964, normalized size of antiderivative = 60.33 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\text {Too large to display} \]

[In]

integrate(coth(x)*(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int \sqrt {a + b \coth ^{2}{\left (x \right )} + c \coth ^{4}{\left (x \right )}} \coth {\left (x \right )}\, dx \]

[In]

integrate(coth(x)*(a+b*coth(x)**2+c*coth(x)**4)**(1/2),x)

[Out]

Integral(sqrt(a + b*coth(x)**2 + c*coth(x)**4)*coth(x), x)

Maxima [F]

\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int { \sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a} \coth \left (x\right ) \,d x } \]

[In]

integrate(coth(x)*(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*coth(x)^4 + b*coth(x)^2 + a)*coth(x), x)

Giac [F]

\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int { \sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a} \coth \left (x\right ) \,d x } \]

[In]

integrate(coth(x)*(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*coth(x)^4 + b*coth(x)^2 + a)*coth(x), x)

Mupad [F(-1)]

Timed out. \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int \mathrm {coth}\left (x\right )\,\sqrt {c\,{\mathrm {coth}\left (x\right )}^4+b\,{\mathrm {coth}\left (x\right )}^2+a} \,d x \]

[In]

int(coth(x)*(a + b*coth(x)^2 + c*coth(x)^4)^(1/2),x)

[Out]

int(coth(x)*(a + b*coth(x)^2 + c*coth(x)^4)^(1/2), x)