\(\int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx\) [42]

Optimal result

Integrand size = 14, antiderivative size = 50 \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=-\frac {\coth (c+d x)}{d \sqrt {b \coth ^4(c+d x)}}+\frac {x \coth ^2(c+d x)}{\sqrt {b \coth ^4(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\frac {x \coth ^2(c+d x)}{\sqrt {b \coth ^4(c+d x)}}-\frac {\coth (c+d x)}{d \sqrt {b \coth ^4(c+d x)}} \]

[In]

Int[1/Sqrt[b*Coth[c + d*x]^4],x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^2(c+d x) \int \tanh ^2(c+d x) \, dx}{\sqrt {b \coth ^4(c+d x)}} \\ & = -\frac {\coth (c+d x)}{d \sqrt {b \coth ^4(c+d x)}}+\frac {\coth ^2(c+d x) \int 1 \, dx}{\sqrt {b \coth ^4(c+d x)}} \\ & = -\frac {\coth (c+d x)}{d \sqrt {b \coth ^4(c+d x)}}+\frac {x \coth ^2(c+d x)}{\sqrt {b \coth ^4(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\frac {\coth (c+d x) (-1+\text {arctanh}(\tanh (c+d x)) \coth (c+d x))}{d \sqrt {b \coth ^4(c+d x)}} \]

[In]

Integrate[1/Sqrt[b*Coth[c + d*x]^4],x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(-\frac {\coth \left (d x +c \right ) \left (\ln \left (\coth \left (d x +c \right )-1\right ) \coth \left (d x +c \right )-\ln \left (\coth \left (d x +c \right )+1\right ) \coth \left (d x +c \right )+2\right )}{2 d \sqrt {b \coth \left (d x +c \right )^{4}}}\)	\(59\)
default	\(-\frac {\coth \left (d x +c \right ) \left (\ln \left (\coth \left (d x +c \right )-1\right ) \coth \left (d x +c \right )-\ln \left (\coth \left (d x +c \right )+1\right ) \coth \left (d x +c \right )+2\right )}{2 d \sqrt {b \coth \left (d x +c \right )^{4}}}\)	\(59\)
risch	\(\frac {{\mathrm e}^{4 d x +4 c} d x +2 \,{\mathrm e}^{2 d x +2 c} d x +d x +2 \,{\mathrm e}^{2 d x +2 c}+2}{\sqrt {\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}}\, \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} d}\)	\(89\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (46) = 92\).

Time = 0.28 (sec) , antiderivative size = 422, normalized size of antiderivative = 8.44 \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\frac {{\left (d x \cosh \left (d x + c\right )^{2} + {\left (d x e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x e^{\left (2 \, d x + 2 \, c\right )} + d x\right )} \sinh \left (d x + c\right )^{2} + d x + {\left (d x \cosh \left (d x + c\right )^{2} + d x + 2\right )} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, {\left (d x \cosh \left (d x + c\right )^{2} + d x + 2\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (d x \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d x \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + d x \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2\right )} \sqrt {\frac {b e^{\left (8 \, d x + 8 \, c\right )} + 4 \, b e^{\left (6 \, d x + 6 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (8 \, d x + 8 \, c\right )} - 4 \, e^{\left (6 \, d x + 6 \, c\right )} + 6 \, e^{\left (4 \, d x + 4 \, c\right )} - 4 \, e^{\left (2 \, d x + 2 \, c\right )} + 1}}}{b d \cosh \left (d x + c\right )^{2} + {\left (b d e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d e^{\left (2 \, d x + 2 \, c\right )} + b d\right )} \sinh \left (d x + c\right )^{2} + b d + {\left (b d \cosh \left (d x + c\right )^{2} + b d\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (b d \cosh \left (d x + c\right )^{2} + b d\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (b d \cosh \left (d x + c\right ) e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b d \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + b d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \]

[In]

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

[Out]

(d*x*cosh(d*x + c)^2 + (d*x*e^(4*d*x + 4*c) - 2*d*x*e^(2*d*x + 2*c) + d*x)*sinh(d*x + c)^2 + d*x + (d*x*cosh(d
*x + c)^2 + d*x + 2)*e^(4*d*x + 4*c) - 2*(d*x*cosh(d*x + c)^2 + d*x + 2)*e^(2*d*x + 2*c) + 2*(d*x*cosh(d*x + c
)*e^(4*d*x + 4*c) - 2*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + d*x*cosh(d*x + c))*sinh(d*x + c) + 2)*sqrt((b*e^(8*d
*x + 8*c) + 4*b*e^(6*d*x + 6*c) + 6*b*e^(4*d*x + 4*c) + 4*b*e^(2*d*x + 2*c) + b)/(e^(8*d*x + 8*c) - 4*e^(6*d*x
+ 6*c) + 6*e^(4*d*x + 4*c) - 4*e^(2*d*x + 2*c) + 1))/(b*d*cosh(d*x + c)^2 + (b*d*e^(4*d*x + 4*c) + 2*b*d*e^(2
*d*x + 2*c) + b*d)*sinh(d*x + c)^2 + b*d + (b*d*cosh(d*x + c)^2 + b*d)*e^(4*d*x + 4*c) + 2*(b*d*cosh(d*x + c)^
2 + b*d)*e^(2*d*x + 2*c) + 2*(b*d*cosh(d*x + c)*e^(4*d*x + 4*c) + 2*b*d*cosh(d*x + c)*e^(2*d*x + 2*c) + b*d*co
sh(d*x + c))*sinh(d*x + c))

Sympy [F]

\[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {b \coth ^{4}{\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(b*coth(c + d*x)**4), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\frac {d x + c}{\sqrt {b} d} - \frac {2 \, \sqrt {b}}{{\left (b e^{\left (-2 \, d x - 2 \, c\right )} + b\right )} d} \]

[In]

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

[Out]

(d*x + c)/(sqrt(b)*d) - 2*sqrt(b)/((b*e^(-2*d*x - 2*c) + b)*d)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\frac {\frac {d x + c}{\sqrt {b}} + \frac {2}{\sqrt {b} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \]

[In]

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

[Out]

((d*x + c)/sqrt(b) + 2/(sqrt(b)*(e^(2*d*x + 2*c) + 1)))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \coth ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^4}} \,d x \]

[In]

int(1/(b*coth(c + d*x)^4)^(1/2),x)

[Out]

int(1/(b*coth(c + d*x)^4)^(1/2), x)