\(\int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx\) [14]

Optimal result

Integrand size = 12, antiderivative size = 198 \[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{b \coth (c+d x)}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 b^{4/3} d}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{b}+2 \sqrt [3]{b \coth (c+d x)}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 b^{4/3} d}+\frac {\text {arctanh}\left (\frac {\sqrt [3]{b \coth (c+d x)}}{\sqrt [3]{b}}\right )}{b^{4/3} d}+\frac {\text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [3]{b \coth (c+d x)}}{b^{2/3}+(b \coth (c+d x))^{2/3}}\right )}{2 b^{4/3} d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.23 \[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=-\frac {6+\sqrt [6]{\coth ^2(c+d x)} \log \left (1-\sqrt [6]{\coth ^2(c+d x)}\right )-\sqrt [6]{\coth ^2(c+d x)} \log \left (1+\sqrt [6]{\coth ^2(c+d x)}\right )+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)} \log \left (1-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )-\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)} \log \left (1+\sqrt [3]{-1} \sqrt [6]{\coth ^2(c+d x)}\right )+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)} \log \left (1-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )-(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)} \log \left (1+(-1)^{2/3} \sqrt [6]{\coth ^2(c+d x)}\right )}{2 b d \sqrt [3]{b \coth (c+d x)}} \] Input:

Integrate[(b*Coth[c + d*x])^(-4/3),x]
 

Output:

-1/2*(6 + (Coth[c + d*x]^2)^(1/6)*Log[1 - (Coth[c + d*x]^2)^(1/6)] - (Coth 
[c + d*x]^2)^(1/6)*Log[1 + (Coth[c + d*x]^2)^(1/6)] + (-1)^(1/3)*(Coth[c + 
 d*x]^2)^(1/6)*Log[1 - (-1)^(1/3)*(Coth[c + d*x]^2)^(1/6)] - (-1)^(1/3)*(C 
oth[c + d*x]^2)^(1/6)*Log[1 + (-1)^(1/3)*(Coth[c + d*x]^2)^(1/6)] + (-1)^( 
2/3)*(Coth[c + d*x]^2)^(1/6)*Log[1 - (-1)^(2/3)*(Coth[c + d*x]^2)^(1/6)] - 
 (-1)^(2/3)*(Coth[c + d*x]^2)^(1/6)*Log[1 + (-1)^(2/3)*(Coth[c + d*x]^2)^( 
1/6)])/(b*d*(b*Coth[c + d*x])^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {3042, 3955, 3042, 3957, 25, 266, 825, 27, 219, 1142, 25, 1082, 217, 1103}

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.

\(\displaystyle \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (-i b \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\)

\(\Big \downarrow \) 3955

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {\int -\frac {(b \coth (c+d x))^{2/3}}{b^2-b^2 \coth ^2(c+d x)}d(b \coth (c+d x))}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(b \coth (c+d x))^{2/3}}{b^2-b^2 \coth ^2(c+d x)}d(b \coth (c+d x))}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 \int \frac {b^4 \coth ^4(c+d x)}{b^2-b^6 \coth ^6(c+d x)}d\sqrt [3]{b \coth (c+d x)}}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 825

\(\displaystyle \frac {3 \left (\frac {1}{3} \int \frac {1}{b^{2/3}-b^2 \coth ^2(c+d x)}d\sqrt [3]{b \coth (c+d x)}+\frac {\int -\frac {\sqrt [3]{b}-\sqrt [3]{b \coth (c+d x)}}{2 \left (b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}\right )}d\sqrt [3]{b \coth (c+d x)}}{3 \sqrt [3]{b}}+\frac {\int -\frac {\sqrt [3]{b}+\sqrt [3]{b \coth (c+d x)}}{2 \left (b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}\right )}d\sqrt [3]{b \coth (c+d x)}}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \left (\frac {1}{3} \int \frac {1}{b^{2/3}-b^2 \coth ^2(c+d x)}d\sqrt [3]{b \coth (c+d x)}-\frac {\int \frac {\sqrt [3]{b}-\sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}-\frac {\int \frac {\sqrt [3]{b}+\sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 \left (-\frac {\int \frac {\sqrt [3]{b}-\sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}-\frac {\int \frac {\sqrt [3]{b}+\sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 \left (-\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}+\frac {1}{2} \int -\frac {\sqrt [3]{b}-2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}-\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}-\frac {1}{2} \int \frac {\sqrt [3]{b}+2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 \left (-\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}-\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}-\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}-\frac {1}{2} \int \frac {\sqrt [3]{b}+2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 \left (-\frac {3 \int \frac {1}{-b^2 \coth ^2(c+d x)-3}d\left (1-2 b^{2/3} \coth (c+d x)\right )-\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}-\frac {-3 \int \frac {1}{-b^2 \coth ^2(c+d x)-3}d\left (2 b^{2/3} \coth (c+d x)+1\right )-\frac {1}{2} \int \frac {\sqrt [3]{b}+2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \left (-\frac {-\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)-b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}-\sqrt {3} \arctan \left (\frac {1-2 b^{2/3} \coth (c+d x)}{\sqrt {3}}\right )}{6 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {2 b^{2/3} \coth (c+d x)+1}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {\sqrt [3]{b}+2 \sqrt [3]{b \coth (c+d x)}}{b^2 \coth ^2(c+d x)+b^{4/3} \coth (c+d x)+b^{2/3}}d\sqrt [3]{b \coth (c+d x)}}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 \left (-\frac {\frac {1}{2} \log \left (-b^{4/3} \coth (c+d x)+b^{2/3}+b^2 \coth ^2(c+d x)\right )-\sqrt {3} \arctan \left (\frac {1-2 b^{2/3} \coth (c+d x)}{\sqrt {3}}\right )}{6 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {2 b^{2/3} \coth (c+d x)+1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (b^{4/3} \coth (c+d x)+b^{2/3}+b^2 \coth ^2(c+d x)\right )}{6 \sqrt [3]{b}}+\frac {\text {arctanh}\left (b^{2/3} \coth (c+d x)\right )}{3 \sqrt [3]{b}}\right )}{b d}-\frac {3}{b d \sqrt [3]{b \coth (c+d x)}}\)

Input:

Int[(b*Coth[c + d*x])^(-4/3),x]
 

Output:

-3/(b*d*(b*Coth[c + d*x])^(1/3)) + (3*(ArcTanh[b^(2/3)*Coth[c + d*x]]/(3*b 
^(1/3)) - (-(Sqrt[3]*ArcTan[(1 - 2*b^(2/3)*Coth[c + d*x])/Sqrt[3]]) + Log[ 
b^(2/3) - b^(4/3)*Coth[c + d*x] + b^2*Coth[c + d*x]^2]/2)/(6*b^(1/3)) - (S 
qrt[3]*ArcTan[(1 + 2*b^(2/3)*Coth[c + d*x])/Sqrt[3]] - Log[b^(2/3) + b^(4/ 
3)*Coth[c + d*x] + b^2*Coth[c + d*x]^2]/2)/(6*b^(1/3))))/(b*d)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[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 825

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator 
[Rt[-a/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r*Cos[2*k 
*m*(Pi/n)] - s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[2*k*(Pi/n)]*x + 
s^2*x^2), x] + Int[(r*Cos[2*k*m*(Pi/n)] + s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 
 + 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x]; 2*(r^(m + 2)/(a*n*s^m))   Int[1/ 
(r^2 - s^2*x^2), x] + 2*(r^(m + 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], 
x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1 
] && NegQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

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]
 

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {1}{3}}\right )}{2 d \,b^{\frac {4}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-b^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}}\right )}{4 d \,b^{\frac {4}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}-1\right )}{3}\right )}{2 d \,b^{\frac {4}{3}}}-\frac {3}{b d \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}-b^{\frac {1}{3}}\right )}{2 d \,b^{\frac {4}{3}}}+\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+b^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}}\right )}{4 d \,b^{\frac {4}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d \,b^{\frac {4}{3}}}\)	\(211\)
default	\(\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {1}{3}}\right )}{2 d \,b^{\frac {4}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-b^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}}\right )}{4 d \,b^{\frac {4}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}-1\right )}{3}\right )}{2 d \,b^{\frac {4}{3}}}-\frac {3}{b d \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}-b^{\frac {1}{3}}\right )}{2 d \,b^{\frac {4}{3}}}+\frac {\ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+b^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}}\right )}{4 d \,b^{\frac {4}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d \,b^{\frac {4}{3}}}\)	\(211\)

Input:

int(1/(b*coth(d*x+c))^(4/3),x,method=_RETURNVERBOSE)
 

Output:

1/2/d/b^(4/3)*ln((b*coth(d*x+c))^(1/3)+b^(1/3))-1/4/d/b^(4/3)*ln((b*coth(d 
*x+c))^(2/3)-b^(1/3)*(b*coth(d*x+c))^(1/3)+b^(2/3))-1/2/d/b^(4/3)*3^(1/2)* 
arctan(1/3*3^(1/2)*(2*(b*coth(d*x+c))^(1/3)/b^(1/3)-1))-3/b/d/(b*coth(d*x+ 
c))^(1/3)-1/2/d/b^(4/3)*ln((b*coth(d*x+c))^(1/3)-b^(1/3))+1/4/d/b^(4/3)*ln 
((b*coth(d*x+c))^(2/3)+b^(1/3)*(b*coth(d*x+c))^(1/3)+b^(2/3))-1/2/d/b^(4/3 
)*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b*coth(d*x+c))^(1/3)/b^(1/3)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (154) = 308\).

Time = 0.17 (sec) , antiderivative size = 3348, normalized size of antiderivative = 16.91 \[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\int \frac {1}{\left (b \coth {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(b*coth(d*x+c))**(4/3),x)
 

Output:

Integral((b*coth(c + d*x))**(-4/3), x)
 

Maxima [F]

\[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\int { \frac {1}{\left (b \coth \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

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

Output:

integrate((b*coth(d*x + c))^(-4/3), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Minimal poly. in rootof must be fra 
ction free Error: Bad Argument ValueMinimal poly. in rootof must be fracti 
on free E
 

Mupad [B] (verification not implemented)

Time = 3.00 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=-\frac {3}{b\,d\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{1/3}}-\frac {\mathrm {atan}\left (\frac {{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{1/3}\,1{}\mathrm {i}}{b^{1/3}}\right )\,1{}\mathrm {i}}{b^{4/3}\,d}-\frac {\mathrm {atan}\left (\frac {b^9\,d^4\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{1/3}\,486{}\mathrm {i}}{243\,b^{28/3}\,d^4-\sqrt {3}\,b^{28/3}\,d^4\,243{}\mathrm {i}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b^{4/3}\,d}+\frac {\mathrm {atan}\left (\frac {b^9\,d^4\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{1/3}\,486{}\mathrm {i}}{243\,b^{28/3}\,d^4+\sqrt {3}\,b^{28/3}\,d^4\,243{}\mathrm {i}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b^{4/3}\,d} \] Input:

int(1/(b*coth(c + d*x))^(4/3),x)
 

Output:

(atan((b^9*d^4*(b*coth(c + d*x))^(1/3)*486i)/(243*b^(28/3)*d^4 + 3^(1/2)*b 
^(28/3)*d^4*243i))*(3^(1/2)*1i - 1)*1i)/(2*b^(4/3)*d) - (atan(((b*coth(c + 
 d*x))^(1/3)*1i)/b^(1/3))*1i)/(b^(4/3)*d) - (atan((b^9*d^4*(b*coth(c + d*x 
))^(1/3)*486i)/(243*b^(28/3)*d^4 - 3^(1/2)*b^(28/3)*d^4*243i))*(3^(1/2)*1i 
 + 1)*1i)/(2*b^(4/3)*d) - 3/(b*d*(b*coth(c + d*x))^(1/3))
 

Reduce [F]

\[ \int \frac {1}{(b \coth (c+d x))^{4/3}} \, dx=\frac {\int \frac {1}{\coth \left (d x +c \right )^{\frac {4}{3}}}d x}{b^{\frac {4}{3}}} \] Input:

int(1/(b*coth(d*x+c))^(4/3),x)
 

Output:

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