\(\int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx\) [205]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \coth ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c} \]

[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^(3/2)+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*c
oth(x)^4)^(1/2)/c

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3782, 1265, 1667, 857, 635, 212, 738} \[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+(b+2 c) \coth ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c} \]

[In]

Int[Coth[x]^5/Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4],x]

[Out]

((b - 2*c)*ArcTanh[(b + 2*c*Coth[x]^2)/(2*Sqrt[c]*Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4])])/(4*c^(3/2)) + 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 + c])
- Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4]/(2*c)

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 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 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rule 1667

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m
+ q + 2*p + 1))), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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^5}{\left (1+x^2\right ) \sqrt {a-b x^2+c x^4}} \, dx,x,-i \coth (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right )\right ) \\ & = -\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c}-\frac {\text {Subst}\left (\int \frac {\frac {b}{2}+\frac {1}{2} (b-2 c) x}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right )}{2 c} \\ & = -\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right )-\frac {(b-2 c) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x+c x^2}} \, dx,x,-\coth ^2(x)\right )}{4 c} \\ & = -\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c}-\frac {(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 )}{2 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 {(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 c^{3/2}}+\frac {\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 )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}{2 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.47 \[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \tanh ^2(x) \left ((b-2 c) (a+b+c) \text {arctanh}\left (\frac {2 c+b \tanh ^2(x)}{2 \sqrt {c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )+2 c^{3/2} \sqrt {a+b+c} \text {arctanh}\left (\frac {b+2 c+(2 a+b) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )-2 \sqrt {c} (a+b+c) \coth ^2(x) \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}\right )}{4 c^{3/2} (a+b+c) \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}} \]

[In]

Integrate[Coth[x]^5/Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4],x]

[Out]

(Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4]*Tanh[x]^2*((b - 2*c)*(a + b + c)*ArcTanh[(2*c + b*Tanh[x]^2)/(2*Sqrt[c]*S
qrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])] + 2*c^(3/2)*Sqrt[a + b + c]*ArcTanh[(b + 2*c + (2*a + b)*Tanh[x]^2)/(2*Sq
rt[a + b + c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])] - 2*Sqrt[c]*(a + b + c)*Coth[x]^2*Sqrt[c + b*Tanh[x]^2 + a
*Tanh[x]^4]))/(4*c^(3/2)*(a + b + c)*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \coth \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \coth \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \coth \left (x \right )^{2}+2 c \coth \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(149\)
default	\(-\frac {\ln \left (\frac {\frac {b}{2}+c \coth \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \coth \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \coth \left (x \right )^{2}+2 c \coth \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \coth \left (x \right )^{2}+c \coth \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\)	\(149\)

[In]

int(coth(x)^5/(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln((1/2*b+c*coth(x)^2)/c^(1/2)+(a+b*coth(x)^2+c*coth(x)^4)^(1/2))/c^(1/2)-1/2*(a+b*coth(x)^2+c*coth(x)^4)
^(1/2)/c+1/4*b/c^(3/2)*ln((1/2*b+c*coth(x)^2)/c^(1/2)+(a+b*coth(x)^2+c*coth(x)^4)^(1/2))+1/2/(a+b+c)^(1/2)*arc
tanh(1/2*(b*coth(x)^2+2*c*coth(x)^2+2*a+b)/(a+b+c)^(1/2)/(a+b*coth(x)^2+c*coth(x)^4)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2086 vs. \(2 (111) = 222\).

Time = 1.11 (sec) , antiderivative size = 8951, normalized size of antiderivative = 66.30 \[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\text {Too large to display} \]

[In]

integrate(coth(x)^5/(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\int \frac {\coth ^{5}{\left (x \right )}}{\sqrt {a + b \coth ^{2}{\left (x \right )} + c \coth ^{4}{\left (x \right )}}}\, dx \]

[In]

integrate(coth(x)**5/(a+b*coth(x)**2+c*coth(x)**4)**(1/2),x)

[Out]

Integral(coth(x)**5/sqrt(a + b*coth(x)**2 + c*coth(x)**4), x)

Maxima [F]

\[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\int { \frac {\coth \left (x\right )^{5}}{\sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a}} \,d x } \]

[In]

integrate(coth(x)^5/(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(coth(x)^5/sqrt(c*coth(x)^4 + b*coth(x)^2 + a), x)

Giac [F]

\[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\int { \frac {\coth \left (x\right )^{5}}{\sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a}} \,d x } \]

[In]

integrate(coth(x)^5/(a+b*coth(x)^2+c*coth(x)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(coth(x)^5/sqrt(c*coth(x)^4 + b*coth(x)^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^5(x)}{\sqrt {a+b \coth ^2(x)+c \coth ^4(x)}} \, dx=\int \frac {{\mathrm {coth}\left (x\right )}^5}{\sqrt {c\,{\mathrm {coth}\left (x\right )}^4+b\,{\mathrm {coth}\left (x\right )}^2+a}} \,d x \]

[In]

int(coth(x)^5/(a + b*coth(x)^2 + c*coth(x)^4)^(1/2),x)

[Out]

int(coth(x)^5/(a + b*coth(x)^2 + c*coth(x)^4)^(1/2), x)