\(\int x^3 \coth ^2(a+b x) \, dx\) [185]

Optimal result

Integrand size = 12, antiderivative size = 87 \[ \int x^3 \coth ^2(a+b x) \, dx=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac {3 \operatorname {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4} \] Output:

-x^3/b+1/4*x^4-x^3*coth(b*x+a)/b+3*x^2*ln(1-exp(2*b*x+2*a))/b^2+3*x*polylo 
g(2,exp(2*b*x+2*a))/b^3-3/2*polylog(3,exp(2*b*x+2*a))/b^4
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(222\) vs. \(2(87)=174\).

Time = 0.64 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.55 \[ \int x^3 \coth ^2(a+b x) \, dx=\frac {x^4}{4}-\frac {e^{2 a} \left (2 b^3 e^{-2 a} x^3-3 b^2 \left (1-e^{-2 a}\right ) x^2 \log \left (1-e^{-a-b x}\right )-3 b^2 \left (1-e^{-2 a}\right ) x^2 \log \left (1+e^{-a-b x}\right )+6 b \left (1-e^{-2 a}\right ) x \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )+6 b \left (1-e^{-2 a}\right ) x \operatorname {PolyLog}\left (2,e^{-a-b x}\right )+6 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (3,-e^{-a-b x}\right )+6 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (3,e^{-a-b x}\right )\right )}{b^4 \left (-1+e^{2 a}\right )}+\frac {x^3 \text {csch}(a) \text {csch}(a+b x) \sinh (b x)}{b} \] Input:

Integrate[x^3*Coth[a + b*x]^2,x]
 

Output:

x^4/4 - (E^(2*a)*((2*b^3*x^3)/E^(2*a) - 3*b^2*(1 - E^(-2*a))*x^2*Log[1 - E 
^(-a - b*x)] - 3*b^2*(1 - E^(-2*a))*x^2*Log[1 + E^(-a - b*x)] + 6*b*(1 - E 
^(-2*a))*x*PolyLog[2, -E^(-a - b*x)] + 6*b*(1 - E^(-2*a))*x*PolyLog[2, E^( 
-a - b*x)] + 6*(1 - E^(-2*a))*PolyLog[3, -E^(-a - b*x)] + 6*(1 - E^(-2*a)) 
*PolyLog[3, E^(-a - b*x)]))/(b^4*(-1 + E^(2*a))) + (x^3*Csch[a]*Csch[a + b 
*x]*Sinh[b*x])/b
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.51, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 25, 4203, 15, 26, 3042, 26, 4201, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int x^3 \coth ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4203

\(\displaystyle -\frac {3 i \int i x^2 \coth (a+b x)dx}{b}+\int x^3dx-\frac {x^3 \coth (a+b x)}{b}\)

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 26

\(\displaystyle \frac {3 \int x^2 \coth (a+b x)dx}{b}-\frac {x^3 \coth (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4201

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \coth (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\int e^{-2 a-2 b x+i \pi } \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )de^{2 a+2 b x-i \pi }}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \coth (a+b x)}{b}+\frac {x^4}{4}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {3 i \left (2 i \left (\frac {x^2 \log \left (1+e^{2 a+2 b x-i \pi }\right )}{2 b}-\frac {\frac {\operatorname {PolyLog}\left (3,-e^{2 a+2 b x-i \pi }\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-e^{2 a+2 b x-i \pi }\right )}{2 b}}{b}\right )-\frac {i x^3}{3}\right )}{b}-\frac {x^3 \coth (a+b x)}{b}+\frac {x^4}{4}\)

Input:

Int[x^3*Coth[a + b*x]^2,x]
 

Output:

x^4/4 - (x^3*Coth[a + b*x])/b - ((3*I)*((-1/3*I)*x^3 + (2*I)*((x^2*Log[1 + 
 E^(2*a - I*Pi + 2*b*x)])/(2*b) - (-1/2*(x*PolyLog[2, -E^(2*a - I*Pi + 2*b 
*x)])/b + PolyLog[3, -E^(2*a - I*Pi + 2*b*x)]/(4*b^2))/b)))/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

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

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 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(197\) vs. \(2(83)=166\).

Time = 0.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.28

method	result	size
risch	\(\frac {x^{4}}{4}-\frac {2 x^{3}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {6 \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {6 \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 a^{2} x}{b^{3}}+\frac {4 a^{3}}{b^{4}}-\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}+\frac {6 x \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {6 x \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right ) x^{2}}{b^{2}}\)	\(198\)

Input:

int(x^3*coth(b*x+a)^2,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4*x^4-2/b*x^3/(exp(2*b*x+2*a)-1)-6*polylog(3,-exp(b*x+a))/b^4-6*polylog( 
3,exp(b*x+a))/b^4-2/b*x^3+6/b^3*a^2*x+4/b^4*a^3-3/b^4*ln(1-exp(b*x+a))*a^2 
+6*x*polylog(2,-exp(b*x+a))/b^3+3/b^2*ln(1-exp(b*x+a))*x^2+6*x*polylog(2,e 
xp(b*x+a))/b^3+3/b^4*a^2*ln(exp(b*x+a)-1)-6/b^4*a^2*ln(exp(b*x+a))+3/b^2*l 
n(exp(b*x+a)+1)*x^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (82) = 164\).

Time = 0.10 (sec) , antiderivative size = 632, normalized size of antiderivative = 7.26 \[ \int x^3 \coth ^2(a+b x) \, dx =\text {Too large to display} \] Input:

integrate(x^3*coth(b*x+a)^2,x, algorithm="fricas")
 

Output:

-1/4*(b^4*x^4 - 8*a^3 - (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)^2 - 2* 
(b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)*sinh(b*x + a) - (b^4*x^4 - 8*b 
^3*x^3 - 8*a^3)*sinh(b*x + a)^2 - 24*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x 
 + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 - b*x)*dilog(cosh(b*x + a) + sin 
h(b*x + a)) - 24*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) 
+ b*x*sinh(b*x + a)^2 - b*x)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 12*(b 
^2*x^2*cosh(b*x + a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*s 
inh(b*x + a)^2 - b^2*x^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 12*(a^2 
*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2 
 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 12*(b^2*x^2 - (b^2*x^2 - 
a^2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a) - (b^ 
2*x^2 - a^2)*sinh(b*x + a)^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1 
) + 24*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 
- 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 24*(cosh(b*x + a)^2 + 2*c 
osh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, -cosh(b*x + a 
) - sinh(b*x + a)))/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + 
a) + b^4*sinh(b*x + a)^2 - b^4)
 

Sympy [F]

\[ \int x^3 \coth ^2(a+b x) \, dx=\int x^{3} \coth ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate(x**3*coth(b*x+a)**2,x)
 

Output:

Integral(x**3*coth(a + b*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.68 \[ \int x^3 \coth ^2(a+b x) \, dx=-\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} - b x^{4} - 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} - b\right )}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} \] Input:

integrate(x^3*coth(b*x+a)^2,x, algorithm="maxima")
 

Output:

-2*x^3/b + 1/4*(b*x^4*e^(2*b*x + 2*a) - b*x^4 - 8*x^3)/(b*e^(2*b*x + 2*a) 
- b) + 3*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*pol 
ylog(3, -e^(b*x + a)))/b^4 + 3*(b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilo 
g(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b^4
 

Giac [F]

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

integrate(x^3*coth(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate(x^3*coth(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \coth ^2(a+b x) \, dx=\int x^3\,{\mathrm {coth}\left (a+b\,x\right )}^2 \,d x \] Input:

int(x^3*coth(a + b*x)^2,x)
 

Output:

int(x^3*coth(a + b*x)^2, x)
 

Reduce [F]

\[ \int x^3 \coth ^2(a+b x) \, dx=\frac {-24 e^{2 b x +2 a} \left (\int \frac {x^{2}}{e^{4 b x +4 a}-2 e^{2 b x +2 a}+1}d x \right ) b^{3}-24 e^{2 b x +2 a} \left (\int \frac {x}{e^{4 b x +4 a}-2 e^{2 b x +2 a}+1}d x \right ) b^{2}+6 e^{2 b x +2 a} \mathrm {log}\left (e^{b x +a}-1\right )+6 e^{2 b x +2 a} \mathrm {log}\left (e^{b x +a}+1\right )+e^{2 b x +2 a} b^{4} x^{4}-12 e^{2 b x +2 a} b x +24 \left (\int \frac {x^{2}}{e^{4 b x +4 a}-2 e^{2 b x +2 a}+1}d x \right ) b^{3}+24 \left (\int \frac {x}{e^{4 b x +4 a}-2 e^{2 b x +2 a}+1}d x \right ) b^{2}-6 \,\mathrm {log}\left (e^{b x +a}-1\right )-6 \,\mathrm {log}\left (e^{b x +a}+1\right )-b^{4} x^{4}-8 b^{3} x^{3}-12 b^{2} x^{2}}{4 b^{4} \left (e^{2 b x +2 a}-1\right )} \] Input:

int(x^3*coth(b*x+a)^2,x)
 

Output:

( - 24*e**(2*a + 2*b*x)*int(x**2/(e**(4*a + 4*b*x) - 2*e**(2*a + 2*b*x) + 
1),x)*b**3 - 24*e**(2*a + 2*b*x)*int(x/(e**(4*a + 4*b*x) - 2*e**(2*a + 2*b 
*x) + 1),x)*b**2 + 6*e**(2*a + 2*b*x)*log(e**(a + b*x) - 1) + 6*e**(2*a + 
2*b*x)*log(e**(a + b*x) + 1) + e**(2*a + 2*b*x)*b**4*x**4 - 12*e**(2*a + 2 
*b*x)*b*x + 24*int(x**2/(e**(4*a + 4*b*x) - 2*e**(2*a + 2*b*x) + 1),x)*b** 
3 + 24*int(x/(e**(4*a + 4*b*x) - 2*e**(2*a + 2*b*x) + 1),x)*b**2 - 6*log(e 
**(a + b*x) - 1) - 6*log(e**(a + b*x) + 1) - b**4*x**4 - 8*b**3*x**3 - 12* 
b**2*x**2)/(4*b**4*(e**(2*a + 2*b*x) - 1))