3.113 \(\int \frac {(a+b \coth ^{-1}(c+d x))^2}{(e+f x)^2} \, dx\)

Optimal. Leaf size=480 \[ \frac {2 a b d \log (e+f x)}{f^2-(d e-c f)^2}-\frac {a b d \log (-c-d x+1)}{f (-c f+d e+f)}+\frac {a b d \log (c+d x+1)}{f (-c f+d e-f)}-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{2 f (-c f+d e+f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{2 f (-c f+d e-f)}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \log \left (\frac {2}{-c-d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e+f)}-\frac {b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e-f)}+\frac {2 b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.74, antiderivative size = 485, normalized size of antiderivative = 1.01, number of steps used = 21, number of rules used = 19, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.950, Rules used = {6110, 1982, 705, 31, 632, 6741, 6122, 706, 633, 6688, 12, 6725, 72, 6742, 5919, 2402, 2315, 5921, 2447} \[ \frac {b^2 d \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{2 f (-c f+d e+f)}+\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{2 f (-c f+d e-f)}-\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)}-\frac {a b d \log (-c-d x+1)}{f (-c f+d e+f)}+\frac {a b d \log (c+d x+1)}{f (-c f+d e-f)}-\frac {2 a b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \log \left (\frac {2}{-c-d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e+f)}-\frac {b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{f (-c f+d e-f)}+\frac {2 b^2 d \log \left (\frac {2}{c+d x+1}\right ) \coth ^{-1}(c+d x)}{(-c f+d e+f) (d e-(c+1) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f) (d e-(c+1) f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcCoth[c + d*x])^2/(f*(e + f*x))) + (b^2*d*ArcCoth[c + d*x]*Log[2/(1 - c - d*x)])/(f*(d*e + f - c*f)
) - (a*b*d*Log[1 - c - d*x])/(f*(d*e + f - c*f)) - (b^2*d*ArcCoth[c + d*x]*Log[2/(1 + c + d*x)])/(f*(d*e - f -
c*f)) + (2*b^2*d*ArcCoth[c + d*x]*Log[2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (a*b*d*Log[1 +
c + d*x])/(f*(d*e - f - c*f)) - (2*a*b*d*Log[e + f*x])/((d*e + f - c*f)*(d*e - (1 + c)*f)) - (2*b^2*d*ArcCoth[
c + d*x]*Log[(2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (b^2*d*Po
lyLog[2, -((1 + c + d*x)/(1 - c - d*x))])/(2*f*(d*e + f - c*f)) + (b^2*d*PolyLog[2, 1 - 2/(1 + c + d*x)])/(2*f
*(d*e - f - c*f)) - (b^2*d*PolyLog[2, 1 - 2/(1 + c + d*x)])/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (b^2*d*PolyL
og[2, 1 - (2*d*(e + f*x))/((d*e + f - c*f)*(1 + c + d*x))])/((d*e + f - c*f)*(d*e - (1 + c)*f))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 1982

Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 5921

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcCoth[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 6110

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcCoth[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcCoth[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]

Rule 6122

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.)*((A_.) + (B_.)*(x_) + (C_.)*(x
_)^2)^(q_.), x_Symbol] :> Dist[1/d, Subst[Int[((d*e - c*f)/d + (f*x)/d)^m*(-(C/d^2) + (C*x^2)/d^2)^q*(a + b*Ar
cCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] && EqQ[B*(1 - c^2) + 2*A*c*
d, 0] && EqQ[2*c*C - B*d, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{(e+f x)^2} \, dx &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x) \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x) \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {a+b \coth ^{-1}(x)}{\left (\frac {d e-c f}{d}+\frac {f x}{d}\right ) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {d \left (a+b \coth ^{-1}(x)\right )}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \operatorname {Subst}\left (\int \frac {a+b \coth ^{-1}(x)}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {(2 b d) \operatorname {Subst}\left (\int \left (-\frac {a}{(-1+x) (1+x) (d e-c f+f x)}-\frac {b \coth ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {(2 a b d) \operatorname {Subst}\left (\int \frac {1}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {(2 a b d) \operatorname {Subst}\left (\int \left (\frac {1}{2 (d e+f-c f) (-1+x)}+\frac {1}{2 (-d e+(1+c) f) (1+x)}+\frac {f^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \left (\frac {\coth ^{-1}(x)}{2 (d e+f-c f) (-1+x)}+\frac {\coth ^{-1}(x)}{2 (-d e+(1+c) f) (1+x)}+\frac {f^2 \coth ^{-1}(x)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{1+x} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{-1+x} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d f\right ) \operatorname {Subst}\left (\int \frac {\coth ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 (d e-c f+f x)}{(d e+f-c f) (1+x)}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {\left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {\left (a+b \coth ^{-1}(c+d x)\right )^2}{f (e+f x)}+\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{f (d e+f-c f)}-\frac {a b d \log (1-c-d x)}{f (d e+f-c f)}-\frac {b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {a b d \log (1+c+d x)}{f (d e-f-c f)}-\frac {2 a b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac {2 b^2 d \coth ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {b^2 d \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ \end {align*}

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Mathematica [C]  time = 8.90, size = 470, normalized size = 0.98 \[ \frac {-\frac {a^2}{f}+\frac {2 a b \left (\coth ^{-1}(c+d x) \left (c^2 (-f)+c d (e-f x)+d^2 e x+f\right )-d (e+f x) \log \left (-\frac {d (e+f x)}{(c+d x) \sqrt {1-\frac {1}{(c+d x)^2}}}\right )\right )}{(-c f+d e+f) (d e-(c+1) f)}+\frac {b^2 d \left (1-(c+d x)^2\right ) (e+f x) \left (\frac {f \left (-\text {Li}_2\left (\exp \left (-2 \left (\coth ^{-1}(c+d x)+\tanh ^{-1}\left (\frac {f}{d e-c f}\right )\right )\right )\right )-2 \tanh ^{-1}\left (\frac {f}{c f-d e}\right ) \log \left (1-\exp \left (-2 \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )\right )+\coth ^{-1}(c+d x) \left (2 \log \left (1-\exp \left (-2 \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )\right )+2 \tanh ^{-1}\left (\frac {f}{d e-c f}\right )+i \pi \right )+2 \tanh ^{-1}\left (\frac {f}{c f-d e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {f}{d e-c f}\right )+\coth ^{-1}(c+d x)\right )\right )+i \pi \log \left (\frac {1}{\sqrt {1-\frac {1}{(c+d x)^2}}}\right )-i \pi \log \left (e^{2 \coth ^{-1}(c+d x)}+1\right )\right )}{\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2}+\frac {\coth ^{-1}(c+d x)^2 e^{\tanh ^{-1}\left (\frac {f}{c f-d e}\right )}}{(c f-d e) \sqrt {1-\frac {f^2}{(d e-c f)^2}}}+\frac {\coth ^{-1}(c+d x)^2}{d e+d f x}\right )}{(c+d x)^2 \left (f-\frac {f}{(c+d x)^2}\right )}}{e+f x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2/f) + (2*a*b*((f - c^2*f + d^2*e*x + c*d*(e - f*x))*ArcCoth[c + d*x] - d*(e + f*x)*Log[-((d*(e + f*x))/(
(c + d*x)*Sqrt[1 - (c + d*x)^(-2)]))]))/((d*e + f - c*f)*(d*e - (1 + c)*f)) + (b^2*d*(e + f*x)*(1 - (c + d*x)^
2)*((E^ArcTanh[f/(-(d*e) + c*f)]*ArcCoth[c + d*x]^2)/((-(d*e) + c*f)*Sqrt[1 - f^2/(d*e - c*f)^2]) + ArcCoth[c
+ d*x]^2/(d*e + d*f*x) + (f*((-I)*Pi*Log[1 + E^(2*ArcCoth[c + d*x])] - 2*ArcTanh[f/(-(d*e) + c*f)]*Log[1 - E^(
-2*(ArcCoth[c + d*x] + ArcTanh[f/(d*e - c*f)]))] + ArcCoth[c + d*x]*(I*Pi + 2*ArcTanh[f/(d*e - c*f)] + 2*Log[1
- E^(-2*(ArcCoth[c + d*x] + ArcTanh[f/(d*e - c*f)]))]) + I*Pi*Log[1/Sqrt[1 - (c + d*x)^(-2)]] + 2*ArcTanh[f/(
-(d*e) + c*f)]*Log[I*Sinh[ArcCoth[c + d*x] + ArcTanh[f/(d*e - c*f)]]] - PolyLog[2, E^(-2*(ArcCoth[c + d*x] + A
rcTanh[f/(d*e - c*f)]))]))/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)))/((c + d*x)^2*(f - f/(c + d*x)^2)))/(e + f*
x)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arcoth}\left (d x + c\right ) + a^{2}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arccoth(d*x + c)^2 + 2*a*b*arccoth(d*x + c) + a^2)/(f^2*x^2 + 2*e*f*x + e^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{2}}{{\left (f x + e\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 783, normalized size = 1.63 \[ -\frac {d \,a^{2}}{\left (d f x +d e \right ) f}-\frac {d \,b^{2} \mathrm {arccoth}\left (d x +c \right )^{2}}{\left (d f x +d e \right ) f}-\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {2 d \,b^{2} \mathrm {arccoth}\left (d x +c \right ) \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \right )}+\frac {d \,b^{2} \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d \,b^{2} \dilog \left (\frac {\left (d x +c \right ) f +f}{c f -d e +f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}-\frac {d \,b^{2} \ln \left (\left (d x +c \right ) f -c f +d e \right ) \ln \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}-\frac {d \,b^{2} \dilog \left (\frac {\left (d x +c \right ) f -f}{c f -d e -f}\right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d \,b^{2} \ln \left (d x +c -1\right )^{2}}{4 f \left (c f -d e -f \right )}-\frac {d \,b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e -f \right )}-\frac {d \,b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e -f \right )}+\frac {d \,b^{2} \ln \left (d x +c +1\right )^{2}}{4 f \left (c f -d e +f \right )}+\frac {d \,b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e +f \right )}-\frac {d \,b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{2 f \left (c f -d e +f \right )}+\frac {d \,b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 f \left (c f -d e +f \right )}-\frac {2 d a b \,\mathrm {arccoth}\left (d x +c \right )}{\left (d f x +d e \right ) f}-\frac {2 d a b \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {2 d a b \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {2 d a b \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*a^2/(d*f*x+d*e)/f-d*b^2/(d*f*x+d*e)/f*arccoth(d*x+c)^2-2*d*b^2*arccoth(d*x+c)/(c*f-d*e-f)/(c*f-d*e+f)*ln((d
*x+c)*f-c*f+d*e)+2*d*b^2/f*arccoth(d*x+c)/(2*c*f-2*d*e-2*f)*ln(d*x+c-1)-2*d*b^2/f*arccoth(d*x+c)/(2*c*f-2*d*e+
2*f)*ln(d*x+c+1)+d*b^2/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f+f)/(c*f-d*e+f))+d*b^2/(c*f-
d*e-f)/(c*f-d*e+f)*dilog(((d*x+c)*f+f)/(c*f-d*e+f))-d*b^2/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)*ln(((d
*x+c)*f-f)/(c*f-d*e-f))-d*b^2/(c*f-d*e-f)/(c*f-d*e+f)*dilog(((d*x+c)*f-f)/(c*f-d*e-f))+1/4*d*b^2/f/(c*f-d*e-f)
*ln(d*x+c-1)^2-1/2*d*b^2/f/(c*f-d*e-f)*dilog(1/2+1/2*d*x+1/2*c)-1/2*d*b^2/f/(c*f-d*e-f)*ln(d*x+c-1)*ln(1/2+1/2
*d*x+1/2*c)+1/4*d*b^2/f/(c*f-d*e+f)*ln(d*x+c+1)^2+1/2*d*b^2/f/(c*f-d*e+f)*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*
x+1/2*c)-1/2*d*b^2/f/(c*f-d*e+f)*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+1/2*d*b^2/f/(c*f-d*e+f)*dilog(1/2+1/2*d*x+
1/2*c)-2*d*a*b/(d*f*x+d*e)/f*arccoth(d*x+c)-2*d*a*b/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)+2*d*a*b/f/(2
*c*f-2*d*e-2*f)*ln(d*x+c-1)-2*d*a*b/f/(2*c*f-2*d*e+2*f)*ln(d*x+c+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d e f - {\left (c + 1\right )} f^{2}} - \frac {\log \left (d x + c - 1\right )}{d e f - {\left (c - 1\right )} f^{2}} - \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} - 1\right )} f^{2}}\right )} - \frac {2 \, \operatorname {arcoth}\left (d x + c\right )}{f^{2} x + e f}\right )} a b - \frac {1}{4} \, b^{2} {\left (\frac {\log \left (d x + c + 1\right )^{2}}{f^{2} x + e f} + \int -\frac {{\left (d f x + c f + f\right )} \log \left (d x + c - 1\right )^{2} + 2 \, {\left (d f x + d e - {\left (d f x + c f + f\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )}{d f^{3} x^{3} + c e^{2} f + e^{2} f + {\left (2 \, d e f^{2} + c f^{3} + f^{3}\right )} x^{2} + {\left (d e^{2} f + 2 \, c e f^{2} + 2 \, e f^{2}\right )} x}\,{d x}\right )} - \frac {a^{2}}{f^{2} x + e f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*(log(d*x + c + 1)/(d*e*f - (c + 1)*f^2) - log(d*x + c - 1)/(d*e*f - (c - 1)*f^2) - 2*log(f*x + e)/(d^2*e^2
- 2*c*d*e*f + (c^2 - 1)*f^2)) - 2*arccoth(d*x + c)/(f^2*x + e*f))*a*b - 1/4*b^2*(log(d*x + c + 1)^2/(f^2*x + e
*f) + integrate(-((d*f*x + c*f + f)*log(d*x + c - 1)^2 + 2*(d*f*x + d*e - (d*f*x + c*f + f)*log(d*x + c - 1))*
log(d*x + c + 1))/(d*f^3*x^3 + c*e^2*f + e^2*f + (2*d*e*f^2 + c*f^3 + f^3)*x^2 + (d*e^2*f + 2*c*e*f^2 + 2*e*f^
2)*x), x)) - a^2/(f^2*x + e*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right )}^2}{{\left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c + d x \right )}\right )^{2}}{\left (e + f x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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