\(\int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx\) [204]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a}}+\frac {b \text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3781, 1265, 974, 744, 738, 212} \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {b \text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 a^{3/2}}-\frac {\text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a}}+\frac {\text {arctanh}\left (\frac {2 a+(b+2 c) \tanh ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a} \]

[In]

Int[Coth[x]^3/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]

[Out]

-1/2*ArcTanh[(2*a + b*Tanh[x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])]/Sqrt[a] + (b*ArcTanh[(2*a +
b*Tanh[x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])])/(4*a^(3/2)) + ArcTanh[(2*a + b + (b + 2*c)*Tanh
[x]^2)/(2*Sqrt[a + b + c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])]/(2*Sqrt[a + b + c]) - (Coth[x]^2*Sqrt[a + b*Ta
nh[x]^2 + c*Tanh[x]^4])/(2*a)

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 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 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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] && EqQ[m + 2*p + 3, 0]

Rule 974

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 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 3781

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(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*Tan[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 {1}{x^3 \left (1+x^2\right ) \sqrt {a-b x^2+c x^4}} \, dx,x,i \tanh (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 (1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x^2 \sqrt {a-b x+c x^2}}-\frac {1}{x \sqrt {a-b x+c x^2}}+\frac {1}{(1+x) \sqrt {a-b x+c x^2}}\right ) \, dx,x,-\tanh ^2(x)\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right ) \\ & = -\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a}-\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )}{4 a}-\text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )+\text {Subst}\left (\int \frac {1}{4 a+4 b+4 c-x^2} \, dx,x,\frac {2 a+b+(b+2 c) \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right ) \\ & = -\frac {\text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a}+\frac {b \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 a} \\ & = -\frac {\text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a}}+\frac {b \text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=-\frac {(2 a-b) \text {arctanh}\left (\frac {2 a+b \tanh ^2(x)}{2 \sqrt {a} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 a^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\coth ^2(x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 a} \]

[In]

Integrate[Coth[x]^3/Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4],x]

[Out]

-1/4*((2*a - b)*ArcTanh[(2*a + b*Tanh[x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])])/a^(3/2) + ArcTan
h[(2*a + b + (b + 2*c)*Tanh[x]^2)/(2*Sqrt[a + b + c]*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])]/(2*Sqrt[a + b + c])
- (Coth[x]^2*Sqrt[a + b*Tanh[x]^2 + c*Tanh[x]^4])/(2*a)

Maple [F]

\[\int \frac {\coth \left (x \right )^{3}}{\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}d x\]

[In]

int(coth(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x)

[Out]

int(coth(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2136 vs. \(2 (149) = 298\).

Time = 1.15 (sec) , antiderivative size = 9168, normalized size of antiderivative = 50.10 \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Too large to display} \]

[In]

integrate(coth(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {\coth ^{3}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )} + c \tanh ^{4}{\left (x \right )}}}\, dx \]

[In]

integrate(coth(x)**3/(a+b*tanh(x)**2+c*tanh(x)**4)**(1/2),x)

[Out]

Integral(coth(x)**3/sqrt(a + b*tanh(x)**2 + c*tanh(x)**4), x)

Maxima [F]

\[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\coth \left (x\right )^{3}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]

[In]

integrate(coth(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(coth(x)^3/sqrt(c*tanh(x)^4 + b*tanh(x)^2 + a), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Timed out} \]

[In]

integrate(coth(x)^3/(a+b*tanh(x)^2+c*tanh(x)^4)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^3(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {{\mathrm {coth}\left (x\right )}^3}{\sqrt {c\,{\mathrm {tanh}\left (x\right )}^4+b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \]

[In]

int(coth(x)^3/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2),x)

[Out]

int(coth(x)^3/(a + b*tanh(x)^2 + c*tanh(x)^4)^(1/2), x)