\(\int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx\) [53]

Optimal result

Integrand size = 20, antiderivative size = 156 \[ \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx=\frac {(c+d x)^3}{3 (a+b) d}-\frac {b (c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3} \] Output:

1/3*(d*x+c)^3/(a+b)/d-b*(d*x+c)^2*ln(1-(a-b)/(a+b)/exp(2*f*x+2*e))/(a^2-b^ 
2)/f+b*d*(d*x+c)*polylog(2,(a-b)/(a+b)/exp(2*f*x+2*e))/(a^2-b^2)/f^2+1/2*b 
*d^2*polylog(3,(a-b)/(a+b)/exp(2*f*x+2*e))/(a^2-b^2)/f^3
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx=\frac {1}{6} \left (\frac {4 b (c+d x)^3}{(a+b) d \left (a \left (-1+e^{2 e}\right )+b \left (1+e^{2 e}\right )\right )}-\frac {6 b (c+d x)^2 \log \left (1+\frac {(-a+b) e^{-2 (e+f x)}}{a+b}\right )}{(a-b) (a+b) f}+\frac {3 b d \left (2 f (c+d x) \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )+d \operatorname {PolyLog}\left (3,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )\right )}{(a-b) (a+b) f^3}+\frac {2 x \left (3 c^2+3 c d x+d^2 x^2\right ) \sinh (e)}{b \cosh (e)+a \sinh (e)}\right ) \] Input:

Integrate[(c + d*x)^2/(a + b*Coth[e + f*x]),x]
 

Output:

((4*b*(c + d*x)^3)/((a + b)*d*(a*(-1 + E^(2*e)) + b*(1 + E^(2*e)))) - (6*b 
*(c + d*x)^2*Log[1 + (-a + b)/((a + b)*E^(2*(e + f*x)))])/((a - b)*(a + b) 
*f) + (3*b*d*(2*f*(c + d*x)*PolyLog[2, (a - b)/((a + b)*E^(2*(e + f*x)))] 
+ d*PolyLog[3, (a - b)/((a + b)*E^(2*(e + f*x)))]))/((a - b)*(a + b)*f^3) 
+ (2*x*(3*c^2 + 3*c*d*x + d^2*x^2)*Sinh[e])/(b*Cosh[e] + a*Sinh[e]))/6
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4214, 25, 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.

\(\displaystyle \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4214

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2620

\(\displaystyle \frac {(c+d x)^3}{3 d (a+b)}-2 b \left (\frac {(c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f \left (a^2-b^2\right )}-\frac {d \int (c+d x) \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )dx}{f \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {(c+d x)^3}{3 d (a+b)}-2 b \left (\frac {(c+d x)^2 \log \left (1-\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f \left (a^2-b^2\right )}-\frac {d \left (\frac {(c+d x) \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f}-\frac {d \int \operatorname {PolyLog}\left (2,\frac {(a-b) e^{-2 (e+f x)}}{a+b}\right )dx}{2 f}\right )}{f \left (a^2-b^2\right )}\right )\)

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[(c + d*x)^2/(a + b*Coth[e + f*x]),x]
 

Output:

(c + d*x)^3/(3*(a + b)*d) - 2*b*(((c + d*x)^2*Log[1 - (a - b)/((a + b)*E^( 
2*(e + f*x)))])/(2*(a^2 - b^2)*f) - (d*(((c + d*x)*PolyLog[2, (a - b)/((a 
+ b)*E^(2*(e + f*x)))])/(2*f) + (d*PolyLog[3, (a - b)/((a + b)*E^(2*(e + f 
*x)))])/(4*f^2)))/((a^2 - b^2)*f))
 

Defintions of rubi rules used

rule 25

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

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 4214

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + Pi*(k_.) + (f_.)*( 
x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp 
[2*I*b   Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^ 
2 + (a^2 + b^2)*E^(2*I*k*Pi)*E^Simp[2*I*(e + f*x), x])), x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && IntegerQ[4*k] && NeQ[a^2 + b^2, 0] && IGtQ[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. \(734\) vs. \(2(158)=316\).

Time = 0.16 (sec) , antiderivative size = 735, normalized size of antiderivative = 4.71

method	result	size
risch	\(\frac {d^{2} x^{3}}{3 a +3 b}+\frac {d c \,x^{2}}{a +b}+\frac {c^{2} x}{a +b}+\frac {c^{3}}{3 \left (a +b \right ) d}-\frac {b \,d^{2} e^{2} \ln \left ({\mathrm e}^{2 f x +2 e} a +{\mathrm e}^{2 f x +2 e} b -a +b \right )}{f^{3} \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {2 b \,d^{2} e^{2} x}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e^{2}}{f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} \operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f \left (a +b \right ) \left (a -b \right )}-\frac {b \,c^{2} \ln \left ({\mathrm e}^{2 f x +2 e} a +{\mathrm e}^{2 f x +2 e} b -a +b \right )}{f \left (a +b \right ) \left (a -b \right )}+\frac {2 b c d e \ln \left ({\mathrm e}^{2 f x +2 e} a +{\mathrm e}^{2 f x +2 e} b -a +b \right )}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {4 b c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {4 b d c e x}{f \left (a +b \right ) \left (a -b \right )}-\frac {2 b d c \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) e}{f^{2} \left (a +b \right ) \left (a -b \right )}+\frac {2 b \,d^{2} x^{3}}{3 \left (a +b \right ) \left (a -b \right )}-\frac {4 b \,d^{2} e^{3}}{3 f^{3} \left (a +b \right ) \left (a -b \right )}-\frac {b \,d^{2} \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x^{2}}{f \left (a +b \right ) \left (a -b \right )}+\frac {b \,d^{2} \operatorname {polylog}\left (3, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{2 f^{3} \left (a +b \right ) \left (a -b \right )}+\frac {2 b d c \,x^{2}}{\left (a +b \right ) \left (a -b \right )}+\frac {2 b d c \,e^{2}}{f^{2} \left (a +b \right ) \left (a -b \right )}-\frac {2 b d c \ln \left (1-\frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right ) x}{f \left (a +b \right ) \left (a -b \right )}-\frac {b d c \operatorname {polylog}\left (2, \frac {\left (a +b \right ) {\mathrm e}^{2 f x +2 e}}{a -b}\right )}{f^{2} \left (a +b \right ) \left (a -b \right )}\)	\(735\)

Input:

int((d*x+c)^2/(a+b*coth(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

1/3/(a+b)*d^2*x^3+1/(a+b)*d*c*x^2+1/(a+b)*c^2*x+1/3/(a+b)/d*c^3-1/f^3*b/(a 
+b)*d^2*e^2/(a-b)*ln(exp(2*f*x+2*e)*a+exp(2*f*x+2*e)*b-a+b)+2/f^3*b/(a+b)* 
d^2*e^2/(a-b)*ln(exp(f*x+e))-2/f^2*b/(a+b)/(a-b)*d^2*e^2*x+1/f^3*b/(a+b)/( 
a-b)*d^2*ln(1-(a+b)*exp(2*f*x+2*e)/(a-b))*e^2-1/f^2*b/(a+b)/(a-b)*d^2*poly 
log(2,(a+b)*exp(2*f*x+2*e)/(a-b))*x+2/f*b/(a+b)*c^2/(a-b)*ln(exp(f*x+e))-1 
/f*b/(a+b)*c^2/(a-b)*ln(exp(2*f*x+2*e)*a+exp(2*f*x+2*e)*b-a+b)+2/f^2*b/(a+ 
b)*c*d*e/(a-b)*ln(exp(2*f*x+2*e)*a+exp(2*f*x+2*e)*b-a+b)-4/f^2*b/(a+b)*c*d 
*e/(a-b)*ln(exp(f*x+e))+4/f*b/(a+b)/(a-b)*d*c*e*x-2/f^2*b/(a+b)/(a-b)*d*c* 
ln(1-(a+b)*exp(2*f*x+2*e)/(a-b))*e+2/3*b/(a+b)/(a-b)*d^2*x^3-4/3/f^3*b/(a+ 
b)/(a-b)*d^2*e^3-1/f*b/(a+b)/(a-b)*d^2*ln(1-(a+b)*exp(2*f*x+2*e)/(a-b))*x^ 
2+1/2/f^3*b/(a+b)/(a-b)*d^2*polylog(3,(a+b)*exp(2*f*x+2*e)/(a-b))+2*b/(a+b 
)/(a-b)*d*c*x^2+2/f^2*b/(a+b)/(a-b)*d*c*e^2-2/f*b/(a+b)/(a-b)*d*c*ln(1-(a+ 
b)*exp(2*f*x+2*e)/(a-b))*x-1/f^2*b/(a+b)/(a-b)*d*c*polylog(2,(a+b)*exp(2*f 
*x+2*e)/(a-b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 492 vs. \(2 (151) = 302\).

Time = 0.10 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.15 \[ \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx=\frac {{\left (a + b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a + b\right )} c d f^{3} x^{2} + 3 \, {\left (a + b\right )} c^{2} f^{3} x + 6 \, b d^{2} {\rm polylog}\left (3, \sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) + 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (2 \, {\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \, {\left (a + b\right )} \sinh \left (f x + e\right ) - 2 \, {\left (a - b\right )} \sqrt {\frac {a + b}{a - b}}\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\sqrt {\frac {a + b}{a - b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right )}{3 \, {\left (a^{2} - b^{2}\right )} f^{3}} \] Input:

integrate((d*x+c)^2/(a+b*coth(f*x+e)),x, algorithm="fricas")
 

Output:

1/3*((a + b)*d^2*f^3*x^3 + 3*(a + b)*c*d*f^3*x^2 + 3*(a + b)*c^2*f^3*x + 6 
*b*d^2*polylog(3, sqrt((a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e))) + 
 6*b*d^2*polylog(3, -sqrt((a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) 
) - 6*(b*d^2*f*x + b*c*d*f)*dilog(sqrt((a + b)/(a - b))*(cosh(f*x + e) + s 
inh(f*x + e))) - 6*(b*d^2*f*x + b*c*d*f)*dilog(-sqrt((a + b)/(a - b))*(cos 
h(f*x + e) + sinh(f*x + e))) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log 
(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) + 2*(a - b)*sqrt((a + b 
)/(a - b))) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(2*(a + b)*cosh(f 
*x + e) + 2*(a + b)*sinh(f*x + e) - 2*(a - b)*sqrt((a + b)/(a - b))) - 3*( 
b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(sqrt((a + b)/ 
(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1) - 3*(b*d^2*f^2*x^2 + 2*b*c*d 
*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-sqrt((a + b)/(a - b))*(cosh(f*x + e 
) + sinh(f*x + e)) + 1))/((a^2 - b^2)*f^3)
 

Sympy [F]

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

integrate((d*x+c)**2/(a+b*coth(f*x+e)),x)
 

Output:

Integral((c + d*x)**2/(a + b*coth(e + f*x)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (151) = 302\).

Time = 0.38 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.13 \[ \int \frac {(c+d x)^2}{a+b \coth (e+f x)} \, dx=-\frac {{\left (2 \, f x \log \left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + {\rm Li}_2\left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right )\right )} b c d}{a^{2} f^{2} - b^{2} f^{2}} - \frac {{\left (2 \, f^{2} x^{2} \log \left (-\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + 2 \, f x {\rm Li}_2\left (\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right ) - {\rm Li}_{3}(\frac {{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b})\right )} b d^{2}}{2 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} - c^{2} {\left (\frac {b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + a + b\right )}{{\left (a^{2} - b^{2}\right )} f} - \frac {f x + e}{{\left (a + b\right )} f}\right )} + \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, {\left (a^{2} f^{3} - b^{2} f^{3}\right )}} + \frac {d^{2} x^{3} + 3 \, c d x^{2}}{3 \, {\left (a + b\right )}} \] Input:

integrate((d*x+c)^2/(a+b*coth(f*x+e)),x, algorithm="maxima")
 

Output:

-(2*f*x*log(-(a*e^(2*e) + b*e^(2*e))*e^(2*f*x)/(a - b) + 1) + dilog((a*e^( 
2*e) + b*e^(2*e))*e^(2*f*x)/(a - b)))*b*c*d/(a^2*f^2 - b^2*f^2) - 1/2*(2*f 
^2*x^2*log(-(a*e^(2*e) + b*e^(2*e))*e^(2*f*x)/(a - b) + 1) + 2*f*x*dilog(( 
a*e^(2*e) + b*e^(2*e))*e^(2*f*x)/(a - b)) - polylog(3, (a*e^(2*e) + b*e^(2 
*e))*e^(2*f*x)/(a - b)))*b*d^2/(a^2*f^3 - b^2*f^3) - c^2*(b*log(-(a - b)*e 
^(-2*f*x - 2*e) + a + b)/((a^2 - b^2)*f) - (f*x + e)/((a + b)*f)) + 2/3*(b 
*d^2*f^3*x^3 + 3*b*c*d*f^3*x^2)/(a^2*f^3 - b^2*f^3) + 1/3*(d^2*x^3 + 3*c*d 
*x^2)/(a + b)
 

Giac [F]

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

integrate((d*x+c)^2/(a+b*coth(f*x+e)),x, algorithm="giac")
 

Output:

integrate((d*x + c)^2/(b*coth(f*x + e) + a), x)
 

Mupad [F(-1)]

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

int((c + d*x)^2/(a + b*coth(e + f*x)),x)
 

Output:

int((c + d*x)^2/(a + b*coth(e + f*x)), x)
 

Reduce [F]

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

int((d*x+c)^2/(a+b*coth(f*x+e)),x)
 

Output:

( - 6*e**(2*e)*int((e**(2*f*x)*x**2)/(e**(2*e + 2*f*x)*a**2 - e**(2*e + 2* 
f*x)*b**2 - a**2 + 2*a*b - b**2),x)*a**2*b*d**2*f + 6*e**(2*e)*int((e**(2* 
f*x)*x**2)/(e**(2*e + 2*f*x)*a**2 - e**(2*e + 2*f*x)*b**2 - a**2 + 2*a*b - 
 b**2),x)*b**3*d**2*f - 12*e**(2*e)*int((e**(2*f*x)*x)/(e**(2*e + 2*f*x)*a 
**2 - e**(2*e + 2*f*x)*b**2 - a**2 + 2*a*b - b**2),x)*a**2*b*c*d*f + 12*e* 
*(2*e)*int((e**(2*f*x)*x)/(e**(2*e + 2*f*x)*a**2 - e**(2*e + 2*f*x)*b**2 - 
 a**2 + 2*a*b - b**2),x)*b**3*c*d*f - 3*log(e**(2*e + 2*f*x)*a + e**(2*e + 
 2*f*x)*b - a + b)*b*c**2 + 3*a*c**2*f*x + 3*a*c*d*f*x**2 + a*d**2*f*x**3 
+ 3*b*c**2*f*x + 3*b*c*d*f*x**2 + b*d**2*f*x**3)/(3*f*(a**2 - b**2))