3.1.78 \(\int (a+b \coth (c+d x))^4 \, dx\) [78]

3.1.78.1 Optimal result
3.1.78.2 Mathematica [A] (verified)
3.1.78.3 Rubi [C] (verified)
3.1.78.4 Maple [A] (verified)
3.1.78.5 Fricas [B] (verification not implemented)
3.1.78.6 Sympy [B] (verification not implemented)
3.1.78.7 Maxima [B] (verification not implemented)
3.1.78.8 Giac [A] (verification not implemented)
3.1.78.9 Mupad [B] (verification not implemented)

3.1.78.1 Optimal result

Integrand size = 12, antiderivative size = 101 \[ \int (a+b \coth (c+d x))^4 \, dx=\left (a^4+6 a^2 b^2+b^4\right ) x-\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}-\frac {a b (a+b \coth (c+d x))^2}{d}-\frac {b (a+b \coth (c+d x))^3}{3 d}+\frac {4 a b \left (a^2+b^2\right ) \log (\sinh (c+d x))}{d} \]

output
(a^4+6*a^2*b^2+b^4)*x-b^2*(3*a^2+b^2)*coth(d*x+c)/d-a*b*(a+b*coth(d*x+c))^ 
2/d-1/3*b*(a+b*coth(d*x+c))^3/d+4*a*b*(a^2+b^2)*ln(sinh(d*x+c))/d
 
3.1.78.2 Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08 \[ \int (a+b \coth (c+d x))^4 \, dx=-\frac {6 b^2 \left (6 a^2+b^2\right ) \coth (c+d x)+12 a b^3 \coth ^2(c+d x)+2 b^4 \coth ^3(c+d x)+3 (a+b)^4 \log (1-\tanh (c+d x))-24 a b \left (a^2+b^2\right ) \log (\tanh (c+d x))-3 (a-b)^4 \log (1+\tanh (c+d x))}{6 d} \]

input
Integrate[(a + b*Coth[c + d*x])^4,x]
 
output
-1/6*(6*b^2*(6*a^2 + b^2)*Coth[c + d*x] + 12*a*b^3*Coth[c + d*x]^2 + 2*b^4 
*Coth[c + d*x]^3 + 3*(a + b)^4*Log[1 - Tanh[c + d*x]] - 24*a*b*(a^2 + b^2) 
*Log[Tanh[c + d*x]] - 3*(a - b)^4*Log[1 + Tanh[c + d*x]])/d
 
3.1.78.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 3963, 3042, 4011, 3042, 4008, 26, 3042, 26, 3956}

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.

\(\displaystyle \int (a+b \coth (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3963

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4008

\(\displaystyle -4 i a b \left (a^2+b^2\right ) \int i \coth (c+d x)dx-\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}+x \left (a^4+6 a^2 b^2+b^4\right )-\frac {b (a+b \coth (c+d x))^3}{3 d}-\frac {a b (a+b \coth (c+d x))^2}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle 4 a b \left (a^2+b^2\right ) \int \coth (c+d x)dx-\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}+x \left (a^4+6 a^2 b^2+b^4\right )-\frac {b (a+b \coth (c+d x))^3}{3 d}-\frac {a b (a+b \coth (c+d x))^2}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 a b \left (a^2+b^2\right ) \int -i \tan \left (i c+i d x+\frac {\pi }{2}\right )dx-\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}+x \left (a^4+6 a^2 b^2+b^4\right )-\frac {b (a+b \coth (c+d x))^3}{3 d}-\frac {a b (a+b \coth (c+d x))^2}{d}\)

\(\Big \downarrow \) 26

\(\displaystyle -4 i a b \left (a^2+b^2\right ) \int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )dx-\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}+x \left (a^4+6 a^2 b^2+b^4\right )-\frac {b (a+b \coth (c+d x))^3}{3 d}-\frac {a b (a+b \coth (c+d x))^2}{d}\)

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {b^2 \left (3 a^2+b^2\right ) \coth (c+d x)}{d}+\frac {4 a b \left (a^2+b^2\right ) \log (-i \sinh (c+d x))}{d}+x \left (a^4+6 a^2 b^2+b^4\right )-\frac {b (a+b \coth (c+d x))^3}{3 d}-\frac {a b (a+b \coth (c+d x))^2}{d}\)

input
Int[(a + b*Coth[c + d*x])^4,x]
 
output
(a^4 + 6*a^2*b^2 + b^4)*x - (b^2*(3*a^2 + b^2)*Coth[c + d*x])/d - (a*b*(a 
+ b*Coth[c + d*x])^2)/d - (b*(a + b*Coth[c + d*x])^3)/(3*d) + (4*a*b*(a^2 
+ b^2)*Log[(-I)*Sinh[c + d*x]])/d
 

3.1.78.3.1 Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 3963
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d 
*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 
+ b^2, 0] && GtQ[n, 1]
 

rule 4008
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), 
x] + Simp[(b*c + a*d)   Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
 

rule 4011
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 
3.1.78.4 Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12

method result size
parallelrisch \(\frac {12 \left (-a^{3} b -a \,b^{3}\right ) \ln \left (1-\tanh \left (d x +c \right )\right )+12 \left (a^{3} b +a \,b^{3}\right ) \ln \left (\tanh \left (d x +c \right )\right )-b^{4} \coth \left (d x +c \right )^{3}-6 a \,b^{3} \coth \left (d x +c \right )^{2}+3 \left (-6 a^{2} b^{2}-b^{4}\right ) \coth \left (d x +c \right )+3 d x \left (a -b \right )^{4}}{3 d}\) \(113\)
derivativedivides \(\frac {-\frac {b^{4} \coth \left (d x +c \right )^{3}}{3}-2 a \,b^{3} \coth \left (d x +c \right )^{2}-6 a^{2} b^{2} \coth \left (d x +c \right )-b^{4} \coth \left (d x +c \right )-\frac {\left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \ln \left (\coth \left (d x +c \right )-1\right )}{2}+\frac {\left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) \ln \left (\coth \left (d x +c \right )+1\right )}{2}}{d}\) \(134\)
default \(\frac {-\frac {b^{4} \coth \left (d x +c \right )^{3}}{3}-2 a \,b^{3} \coth \left (d x +c \right )^{2}-6 a^{2} b^{2} \coth \left (d x +c \right )-b^{4} \coth \left (d x +c \right )-\frac {\left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \ln \left (\coth \left (d x +c \right )-1\right )}{2}+\frac {\left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) \ln \left (\coth \left (d x +c \right )+1\right )}{2}}{d}\) \(134\)
parts \(x \,a^{4}+\frac {b^{4} \left (-\frac {\coth \left (d x +c \right )^{3}}{3}-\coth \left (d x +c \right )-\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {4 a^{3} b \ln \left (\sinh \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2} \left (-\coth \left (d x +c \right )-\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {4 a \,b^{3} \left (-\frac {\coth \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2}\right )}{d}\) \(155\)
risch \(x \,a^{4}-4 b \,a^{3} x +6 a^{2} b^{2} x -4 a \,b^{3} x +b^{4} x -\frac {8 a^{3} b c}{d}-\frac {8 a \,b^{3} c}{d}-\frac {4 b^{2} \left (9 a^{2} {\mathrm e}^{4 d x +4 c}+6 a b \,{\mathrm e}^{4 d x +4 c}+3 \,{\mathrm e}^{4 d x +4 c} b^{2}-18 a^{2} {\mathrm e}^{2 d x +2 c}-6 a b \,{\mathrm e}^{2 d x +2 c}-3 \,{\mathrm e}^{2 d x +2 c} b^{2}+9 a^{2}+2 b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}+\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}+\frac {4 b^{3} a \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d}\) \(211\)

input
int((a+b*coth(d*x+c))^4,x,method=_RETURNVERBOSE)
 
output
1/3*(12*(-a^3*b-a*b^3)*ln(1-tanh(d*x+c))+12*(a^3*b+a*b^3)*ln(tanh(d*x+c))- 
b^4*coth(d*x+c)^3-6*a*b^3*coth(d*x+c)^2+3*(-6*a^2*b^2-b^4)*coth(d*x+c)+3*d 
*x*(a-b)^4)/d
 
3.1.78.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1396 vs. \(2 (99) = 198\).

Time = 0.27 (sec) , antiderivative size = 1396, normalized size of antiderivative = 13.82 \[ \int (a+b \coth (c+d x))^4 \, dx=\text {Too large to display} \]

input
integrate((a+b*coth(d*x+c))^4,x, algorithm="fricas")
 
output
1/3*(3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x*cosh(d*x + c)^6 + 1 
8*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x*cosh(d*x + c)*sinh(d*x + 
 c)^5 + 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x*sinh(d*x + c)^6 
- 3*(12*a^2*b^2 + 8*a*b^3 + 4*b^4 + 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 
 + b^4)*d*x)*cosh(d*x + c)^4 + 3*(15*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 
+ b^4)*d*x*cosh(d*x + c)^2 - 12*a^2*b^2 - 8*a*b^3 - 4*b^4 - 3*(a^4 - 4*a^3 
*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x)*sinh(d*x + c)^4 - 36*a^2*b^2 - 8*b^4 
+ 12*(5*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x*cosh(d*x + c)^3 - 
(12*a^2*b^2 + 8*a*b^3 + 4*b^4 + 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b 
^4)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 - 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4 
*a*b^3 + b^4)*d*x + 3*(24*a^2*b^2 + 8*a*b^3 + 4*b^4 + 3*(a^4 - 4*a^3*b + 6 
*a^2*b^2 - 4*a*b^3 + b^4)*d*x)*cosh(d*x + c)^2 + 3*(15*(a^4 - 4*a^3*b + 6* 
a^2*b^2 - 4*a*b^3 + b^4)*d*x*cosh(d*x + c)^4 + 24*a^2*b^2 + 8*a*b^3 + 4*b^ 
4 + 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x - 6*(12*a^2*b^2 + 8* 
a*b^3 + 4*b^4 + 3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d*x)*cosh(d* 
x + c)^2)*sinh(d*x + c)^2 + 12*((a^3*b + a*b^3)*cosh(d*x + c)^6 + 6*(a^3*b 
 + a*b^3)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3*b + a*b^3)*sinh(d*x + c)^6 
- 3*(a^3*b + a*b^3)*cosh(d*x + c)^4 - 3*(a^3*b + a*b^3 - 5*(a^3*b + a*b^3) 
*cosh(d*x + c)^2)*sinh(d*x + c)^4 - a^3*b - a*b^3 + 4*(5*(a^3*b + a*b^3)*c 
osh(d*x + c)^3 - 3*(a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(...
 
3.1.78.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (92) = 184\).

Time = 1.20 (sec) , antiderivative size = 444, normalized size of antiderivative = 4.40 \[ \int (a+b \coth (c+d x))^4 \, dx=\begin {cases} x \left (a + b \coth {\left (c \right )}\right )^{4} & \text {for}\: d = 0 \\- \frac {a^{4} \log {\left (- e^{- d x} \right )}}{d} - \frac {4 a^{3} b \log {\left (- e^{- d x} \right )} \coth {\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} - \frac {6 a^{2} b^{2} \log {\left (- e^{- d x} \right )} \coth ^{2}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} - \frac {4 a b^{3} \log {\left (- e^{- d x} \right )} \coth ^{3}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} - \frac {b^{4} \log {\left (- e^{- d x} \right )} \coth ^{4}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} & \text {for}\: c = \log {\left (- e^{- d x} \right )} \\a^{4} x + 4 a^{3} b x \coth {\left (d x + \log {\left (e^{- d x} \right )} \right )} + 6 a^{2} b^{2} x \coth ^{2}{\left (d x + \log {\left (e^{- d x} \right )} \right )} + 4 a b^{3} x \coth ^{3}{\left (d x + \log {\left (e^{- d x} \right )} \right )} + b^{4} x \coth ^{4}{\left (d x + \log {\left (e^{- d x} \right )} \right )} & \text {for}\: c = \log {\left (e^{- d x} \right )} \\a^{4} x + 4 a^{3} b x - \frac {4 a^{3} b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 a^{3} b \log {\left (\tanh {\left (c + d x \right )} \right )}}{d} + 6 a^{2} b^{2} x - \frac {6 a^{2} b^{2}}{d \tanh {\left (c + d x \right )}} + 4 a b^{3} x - \frac {4 a b^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 a b^{3} \log {\left (\tanh {\left (c + d x \right )} \right )}}{d} - \frac {2 a b^{3}}{d \tanh ^{2}{\left (c + d x \right )}} + b^{4} x - \frac {b^{4}}{d \tanh {\left (c + d x \right )}} - \frac {b^{4}}{3 d \tanh ^{3}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

input
integrate((a+b*coth(d*x+c))**4,x)
 
output
Piecewise((x*(a + b*coth(c))**4, Eq(d, 0)), (-a**4*log(-exp(-d*x))/d - 4*a 
**3*b*log(-exp(-d*x))*coth(d*x + log(-exp(-d*x)))/d - 6*a**2*b**2*log(-exp 
(-d*x))*coth(d*x + log(-exp(-d*x)))**2/d - 4*a*b**3*log(-exp(-d*x))*coth(d 
*x + log(-exp(-d*x)))**3/d - b**4*log(-exp(-d*x))*coth(d*x + log(-exp(-d*x 
)))**4/d, Eq(c, log(-exp(-d*x)))), (a**4*x + 4*a**3*b*x*coth(d*x + log(exp 
(-d*x))) + 6*a**2*b**2*x*coth(d*x + log(exp(-d*x)))**2 + 4*a*b**3*x*coth(d 
*x + log(exp(-d*x)))**3 + b**4*x*coth(d*x + log(exp(-d*x)))**4, Eq(c, log( 
exp(-d*x)))), (a**4*x + 4*a**3*b*x - 4*a**3*b*log(tanh(c + d*x) + 1)/d + 4 
*a**3*b*log(tanh(c + d*x))/d + 6*a**2*b**2*x - 6*a**2*b**2/(d*tanh(c + d*x 
)) + 4*a*b**3*x - 4*a*b**3*log(tanh(c + d*x) + 1)/d + 4*a*b**3*log(tanh(c 
+ d*x))/d - 2*a*b**3/(d*tanh(c + d*x)**2) + b**4*x - b**4/(d*tanh(c + d*x) 
) - b**4/(3*d*tanh(c + d*x)**3), True))
 
3.1.78.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (99) = 198\).

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.17 \[ \int (a+b \coth (c+d x))^4 \, dx=\frac {1}{3} \, b^{4} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 4 \, a b^{3} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + 6 \, a^{2} b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + a^{4} x + \frac {4 \, a^{3} b \log \left (\sinh \left (d x + c\right )\right )}{d} \]

input
integrate((a+b*coth(d*x+c))^4,x, algorithm="maxima")
 
output
1/3*b^4*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) - 2)/(d* 
(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 4*a*b 
^3*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d + 2*e^(-2* 
d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) - 1))) + 6*a^2*b^2*(x 
 + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) + a^4*x + 4*a^3*b*log(sinh(d*x + c) 
)/d
 
3.1.78.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.51 \[ \int (a+b \coth (c+d x))^4 \, dx=\frac {3 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (a^{3} b + a b^{3}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) - \frac {4 \, {\left (9 \, a^{2} b^{2} + 2 \, b^{4} + 3 \, {\left (3 \, a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} e^{\left (4 \, d x + 4 \, c\right )} - 3 \, {\left (6 \, a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

input
integrate((a+b*coth(d*x+c))^4,x, algorithm="giac")
 
output
1/3*(3*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(d*x + c) + 12*(a^3*b + 
 a*b^3)*log(abs(e^(2*d*x + 2*c) - 1)) - 4*(9*a^2*b^2 + 2*b^4 + 3*(3*a^2*b^ 
2 + 2*a*b^3 + b^4)*e^(4*d*x + 4*c) - 3*(6*a^2*b^2 + 2*a*b^3 + b^4)*e^(2*d* 
x + 2*c))/(e^(2*d*x + 2*c) - 1)^3)/d
 
3.1.78.9 Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int (a+b \coth (c+d x))^4 \, dx=x\,{\left (a-b\right )}^4-\frac {4\,\left (3\,a^2\,b^2+2\,a\,b^3+b^4\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {4\,\left (b^4+2\,a\,b^3\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )\,\left (4\,a^3\,b+4\,a\,b^3\right )}{d}-\frac {8\,b^4}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]

input
int((a + b*coth(c + d*x))^4,x)
 
output
x*(a - b)^4 - (4*(2*a*b^3 + b^4 + 3*a^2*b^2))/(d*(exp(2*c + 2*d*x) - 1)) - 
 (4*(2*a*b^3 + b^4))/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) + (lo 
g(exp(2*c)*exp(2*d*x) - 1)*(4*a*b^3 + 4*a^3*b))/d - (8*b^4)/(3*d*(3*exp(2* 
c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1))