\(\int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx\) [31]

Optimal result

Integrand size = 14, antiderivative size = 105 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=-\frac {2 \coth (c+d x)}{d \sqrt {b \coth ^3(c+d x)}}-\frac {\arctan \left (\sqrt {\coth (c+d x)}\right ) \coth ^{\frac {3}{2}}(c+d x)}{d \sqrt {b \coth ^3(c+d x)}}+\frac {\text {arctanh}\left (\sqrt {\coth (c+d x)}\right ) \coth ^{\frac {3}{2}}(c+d x)}{d \sqrt {b \coth ^3(c+d x)}} \] Output:

-2*coth(d*x+c)/d/(b*coth(d*x+c)^3)^(1/2)-arctan(coth(d*x+c)^(1/2))*coth(d* 
x+c)^(3/2)/d/(b*coth(d*x+c)^3)^(1/2)+arctanh(coth(d*x+c)^(1/2))*coth(d*x+c 
)^(3/2)/d/(b*coth(d*x+c)^3)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=-\frac {\coth (c+d x) \left (2+\arctan \left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}-\text {arctanh}\left (\sqrt [4]{\coth ^2(c+d x)}\right ) \sqrt [4]{\coth ^2(c+d x)}\right )}{d \sqrt {b \coth ^3(c+d x)}} \] Input:

Integrate[1/Sqrt[b*Coth[c + d*x]^3],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 4141, 3042, 3955, 3042, 3957, 25, 266, 827, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {i b \tan \left (i c+i d x+\frac {\pi }{2}\right )^3}}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \int \frac {1}{\coth ^{\frac {3}{2}}(c+d x)}dx}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \int \frac {1}{\left (-i \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\int \sqrt {\coth (c+d x)}dx-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (-\frac {2}{d \sqrt {\coth (c+d x)}}+\int \sqrt {-i \tan \left (i c+i d x+\frac {\pi }{2}\right )}dx\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (-\frac {\int -\frac {\sqrt {\coth (c+d x)}}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\frac {\int \frac {\sqrt {\coth (c+d x)}}{1-\coth ^2(c+d x)}d\coth (c+d x)}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\frac {2 \int \frac {\coth (c+d x)}{1-\coth ^2(c+d x)}d\sqrt {\coth (c+d x)}}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{1-\coth (c+d x)}d\sqrt {\coth (c+d x)}-\frac {1}{2} \int \frac {1}{\coth (c+d x)+1}d\sqrt {\coth (c+d x)}\right )}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{1-\coth (c+d x)}d\sqrt {\coth (c+d x)}-\frac {1}{2} \arctan \left (\sqrt {\coth (c+d x)}\right )\right )}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\coth ^{\frac {3}{2}}(c+d x) \left (\frac {2 \left (\frac {1}{2} \text {arctanh}\left (\sqrt {\coth (c+d x)}\right )-\frac {1}{2} \arctan \left (\sqrt {\coth (c+d x)}\right )\right )}{d}-\frac {2}{d \sqrt {\coth (c+d x)}}\right )}{\sqrt {b \coth ^3(c+d x)}}\)

Input:

Int[1/Sqrt[b*Coth[c + d*x]^3],x]
 

Output:

(((2*(-1/2*ArcTan[Sqrt[Coth[c + d*x]]] + ArcTanh[Sqrt[Coth[c + d*x]]]/2))/ 
d - 2/(d*Sqrt[Coth[c + d*x]]))*Coth[c + d*x]^(3/2))/Sqrt[b*Coth[c + d*x]^3 
]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

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

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3955

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

rule 3957

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

rule 4141

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[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]])
 

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {\coth \left (d x +c \right ) \left (2 b^{\frac {5}{2}}+\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \coth \left (d x +c \right )^{3}}\, b^{\frac {5}{2}}}\)	\(92\)
default	\(-\frac {\coth \left (d x +c \right ) \left (2 b^{\frac {5}{2}}+\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}-\operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right ) b^{2} \sqrt {b \coth \left (d x +c \right )}\right )}{d \sqrt {b \coth \left (d x +c \right )^{3}}\, b^{\frac {5}{2}}}\)	\(92\)

Input:

int(1/(b*coth(d*x+c)^3)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/d*coth(d*x+c)*(2*b^(5/2)+arctan((b*coth(d*x+c))^(1/2)/b^(1/2))*b^2*(b*c 
oth(d*x+c))^(1/2)-arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))*b^2*(b*coth(d*x+c 
))^(1/2))/(b*coth(d*x+c)^3)^(1/2)/b^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (91) = 182\).

Time = 0.12 (sec) , antiderivative size = 899, normalized size of antiderivative = 8.56 \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^ 
2 + 1)*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + 
sinh(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh( 
d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2)) + (cosh 
(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b 
)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d 
*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x 
 + c)^4 - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + 
c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - 2*b)/(cosh(d*x + 
c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 
 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) + 8*(cosh(d*x + c)^ 
2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b*cosh(d*x + 
 c)/sinh(d*x + c)))/(b*d*cosh(d*x + c)^2 + 2*b*d*cosh(d*x + c)*sinh(d*x + 
c) + b*d*sinh(d*x + c)^2 + b*d), 1/4*(2*(cosh(d*x + c)^2 + 2*cosh(d*x + c) 
*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(b)*arctan((cosh(d*x + c)^2 + 2* 
cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b*cosh(d*x + c)/si 
nh(d*x + c))/sqrt(b)) + (cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + 
 sinh(d*x + c)^2 + 1)*sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^ 
3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c) 
*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*...
 

Sympy [F]

\[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\int \frac {1}{\sqrt {b \coth ^{3}{\left (c + d x \right )}}}\, dx \] Input:

integrate(1/(b*coth(d*x+c)**3)**(1/2),x)
 

Output:

Integral(1/sqrt(b*coth(c + d*x)**3), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \coth \left (d x + c\right )^{3}}} \,d x } \] Input:

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

Output:

integrate(1/sqrt(b*coth(d*x + c)^3), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {coth}\left (c+d\,x\right )}^3}} \,d x \] Input:

int(1/(b*coth(c + d*x)^3)^(1/2),x)
 

Output:

int(1/(b*coth(c + d*x)^3)^(1/2), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {b \coth ^3(c+d x)}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\coth \left (d x +c \right )}}{\coth \left (d x +c \right )^{2}}d x \right )}{b} \] Input:

int(1/(b*coth(d*x+c)^3)^(1/2),x)
 

Output:

(sqrt(b)*int(sqrt(coth(c + d*x))/coth(c + d*x)**2,x))/b