3.44 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^2}{(e+f x)^3} \, dx\)

Optimal. Leaf size=750 \[ -\frac {a b d^2 \log (-c-d x+1)}{2 f (-c f+d e+f)^2}+\frac {a b d^2 \log (c+d x+1)}{2 f (-c f+d e-f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}-\frac {a b d}{(e+f x) \left (f^2-(d e-c f)^2\right )}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{4 f (-c f+d e+f)^2}+\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{4 f (-c f+d e-f)^2}-\frac {b^2 d^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d^2 (d e-c f) \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d^2 \log (-c-d x+1)}{2 (-c f+d e+f)^2 (d e-(c+1) f)}-\frac {b^2 d^2 \log (c+d x+1)}{2 (-c f+d e+f) (d e-(c+1) f)^2}+\frac {b^2 d^2 f \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d^2 \log \left (\frac {2}{-c-d x+1}\right ) \tanh ^{-1}(c+d x)}{2 f (-c f+d e+f)^2}-\frac {b^2 d^2 \log \left (\frac {2}{c+d x+1}\right ) \tanh ^{-1}(c+d x)}{2 f (-c f+d e-f)^2}+\frac {2 b^2 d^2 (d e-c f) \log \left (\frac {2}{c+d x+1}\right ) \tanh ^{-1}(c+d x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(e+f x) (-c f+d e+f) (d e-(c+1) f)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.13, antiderivative size = 750, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 18, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6109, 1982, 709, 800, 6741, 6121, 710, 801, 6725, 5918, 2402, 2315, 5926, 706, 31, 633, 5920, 2447} \[ \frac {b^2 d^2 \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{4 f (-c f+d e+f)^2}+\frac {b^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{4 f (-c f+d e-f)^2}-\frac {b^2 d^2 (d e-c f) \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d^2 (d e-c f) \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}-\frac {a b d^2 \log (-c-d x+1)}{2 f (-c f+d e+f)^2}+\frac {a b d^2 \log (c+d x+1)}{2 f (-c f+d e-f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}-\frac {a b d}{(e+f x) \left (f^2-(d e-c f)^2\right )}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \log (-c-d x+1)}{2 (-c f+d e+f)^2 (d e-(c+1) f)}-\frac {b^2 d^2 \log (c+d x+1)}{2 (-c f+d e+f) (d e-(c+1) f)^2}+\frac {b^2 d^2 f \log (e+f x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d^2 \log \left (\frac {2}{-c-d x+1}\right ) \tanh ^{-1}(c+d x)}{2 f (-c f+d e+f)^2}-\frac {b^2 d^2 \log \left (\frac {2}{c+d x+1}\right ) \tanh ^{-1}(c+d x)}{2 f (-c f+d e-f)^2}+\frac {2 b^2 d^2 (d e-c f) \log \left (\frac {2}{c+d x+1}\right ) \tanh ^{-1}(c+d x)}{(-c f+d e+f)^2 (d e-(c+1) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{(-c f+d e+f)^2 (d e-(c+1) f)^2}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(e+f x) (-c f+d e+f) (d e-(c+1) f)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^2/(e + f*x)^3,x]

[Out]

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

Rule 31

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

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 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 709

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

Rule 710

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[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 5920

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[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*ArcTanh[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 5926

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

Rule 6109

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcTanh[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcTanh[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 6121

Int[((a_.) + ArcTanh[(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
cTanh[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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(e+f x)^3} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {(b d) \int \frac {a+b \tanh ^{-1}(c+d x)}{(e+f x)^2 \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {(b d) \int \frac {a+b \tanh ^{-1}(c+d x)}{(e+f x)^2 \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{f}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b \operatorname {Subst}\left (\int \left (-\frac {a d^2}{(d e-c f+f x)^2 \left (-1+x^2\right )}-\frac {b d^2 \tanh ^{-1}(x)}{(d e-c f+f x)^2 \left (-1+x^2\right )}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}-\frac {\left (a b d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(d e-c f+f x)^2 \left (-1+x^2\right )} \, dx,x,c+d x\right )}{f}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{(d e-c f+f x)^2 \left (-1+x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \left (\frac {\tanh ^{-1}(x)}{2 (d e-(1+c) f)^2 (-1-x)}+\frac {\tanh ^{-1}(x)}{2 (d e+f-c f)^2 (-1+x)}+\frac {f^2 \tanh ^{-1}(x)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)^2}+\frac {2 f^2 (d e-c f) \tanh ^{-1}(x)}{(d e+(1-c) f)^2 (d e-f-c f)^2 (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}+\frac {\left (a b d^2\right ) \operatorname {Subst}\left (\int \frac {d e-c f-f x}{(d e-c f+f x) \left (-1+x^2\right )} \, dx,x,c+d x\right )}{f \left (f^2-(d e-c f)^2\right )}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{-1-x} \, dx,x,c+d x\right )}{2 f (d e-f-c f)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{-1+x} \, dx,x,c+d x\right )}{2 f (d e+f-c f)^2}-\frac {\left (2 b^2 d^2 f (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {\left (b^2 d^2 f\right ) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}(x)}{(d e-c f+f x)^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac {\left (a b d^2\right ) \operatorname {Subst}\left (\int \left (\frac {-d e+(1+c) f}{2 (d e+f-c f) (1-x)}+\frac {-d e-(1-c) f}{2 (d e-(1+c) f) (1+x)}+\frac {2 f^2 (d e-c f)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f \left (f^2-(d e-c f)^2\right )}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)^2}-\frac {a b d^2 \log (1-c-d x)}{2 f (d e+f-c f)^2}-\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)^2}+\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {a b d^2 \log (1+c+d x)}{2 f (d e-f-c f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-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)^2 (d e-(1+c) f)^2}+\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e-f-c f)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e+f-c f)^2}-\frac {\left (2 b^2 d^2 (d e-c f)\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)^2 (d e-(1+c) f)^2}+\frac {\left (2 b^2 d^2 (d e-c f)\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)^2 (d e-(1+c) f)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(d e-c f+f x) \left (1-x^2\right )} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)^2}-\frac {a b d^2 \log (1-c-d x)}{2 f (d e+f-c f)^2}-\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)^2}+\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {a b d^2 \log (1+c+d x)}{2 f (d e-f-c f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-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)^2 (d e-(1+c) f)^2}+\frac {b^2 d^2 (d e-c f) \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)^2 (d e-(1+c) f)^2}+\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c+d x}\right )}{2 f (d e-f-c f)^2}+\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c-d x}\right )}{2 f (d e+f-c f)^2}+\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {-d e+c f+f x}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {\left (b^2 d^2 f^2\right ) \operatorname {Subst}\left (\int \frac {1}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {\left (2 b^2 d^2 (d e-c f)\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)^2 (d e-(1+c) f)^2}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)^2}-\frac {a b d^2 \log (1-c-d x)}{2 f (d e+f-c f)^2}-\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)^2}+\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {a b d^2 \log (1+c+d x)}{2 f (d e-f-c f)^2}+\frac {b^2 d^2 f \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-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)^2 (d e-(1+c) f)^2}+\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{4 f (d e+f-c f)^2}+\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{4 f (d e-f-c f)^2}-\frac {b^2 d^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {b^2 d^2 (d e-c f) \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)^2 (d e-(1+c) f)^2}+\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,c+d x\right )}{2 (d e+f-c f) (d e-(1+c) f)^2}-\frac {\left (b^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,c+d x\right )}{2 (d e+f-c f)^2 (d e-(1+c) f)}\\ &=-\frac {a b d}{\left (f^2-(d e-c f)^2\right ) (e+f x)}+\frac {b^2 d \tanh ^{-1}(c+d x)}{(d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f (e+f x)^2}+\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1-c-d x}\right )}{2 f (d e+f-c f)^2}-\frac {a b d^2 \log (1-c-d x)}{2 f (d e+f-c f)^2}+\frac {b^2 d^2 \log (1-c-d x)}{2 (d e+f-c f)^2 (d e-(1+c) f)}-\frac {b^2 d^2 \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{2 f (d e-f-c f)^2}+\frac {2 b^2 d^2 (d e-c f) \tanh ^{-1}(c+d x) \log \left (\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {a b d^2 \log (1+c+d x)}{2 f (d e-f-c f)^2}-\frac {b^2 d^2 \log (1+c+d x)}{2 (d e+f-c f) (d e-(1+c) f)^2}+\frac {b^2 d^2 f \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 a b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2}-\frac {2 b^2 d^2 (d e-c f) \tanh ^{-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)^2 (d e-(1+c) f)^2}+\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{4 f (d e+f-c f)^2}+\frac {b^2 d^2 \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{4 f (d e-f-c f)^2}-\frac {b^2 d^2 (d e-c f) \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{(d e+f-c f)^2 (d e-(1+c) f)^2}+\frac {b^2 d^2 (d e-c f) \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)^2 (d e-(1+c) f)^2}\\ \end {align*}

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Mathematica [C]  time = 14.79, size = 1968, normalized size = 2.62 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^2/(e + f*x)^3,x]

[Out]

-1/2*a^2/(f*(e + f*x)^2) + (a*b*(d*e - c*f + f*(c + d*x))^3*((f*(2 + ((d*e + f - c*f)*(d*e - (1 + c)*f))/((d*e
- c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])^2)*ArcTanh[c + d*x])/((d*e + f - c*f)^2*(
-(d*e) + f + c*f)^2) - ((c + d*x)*(f - 2*d*e*ArcTanh[c + d*x] + 2*c*f*ArcTanh[c + d*x]))/((d*e - c*f)*(d*e + f
- c*f)*(d*e - (1 + c)*f)*Sqrt[1 - (c + d*x)^2]*((d*e - c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c
+ d*x)^2])) - (2*(d*e - c*f)*Log[(d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sq
rt[1 - (c + d*x)^2]])/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)^2))/(d*(e + f*x)^3) + (b^2*(d*e - c*f + f*(c + d*
x))^3*((f*(1 - (c + d*x)^2)^(3/2)*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/S
qrt[1 - (c + d*x)^2])^3*ArcTanh[c + d*x]^2)/(2*(d*e - f - c*f)*(d*e + f - c*f)*(d*e - c*f + f*(c + d*x))^3*(-(
(d*e)/Sqrt[1 - (c + d*x)^2]) + (c*f)/Sqrt[1 - (c + d*x)^2] - (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])^2) + ((1 - (
c + d*x)^2)^(3/2)*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x
)^2])^3*((f*(c + d*x)*ArcTanh[c + d*x])/Sqrt[1 - (c + d*x)^2] - (d*e*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c
+ d*x)^2] + (c*f*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2]))/((d*e - c*f)*(d*e - f - c*f)*(d*e + f
- c*f)*(d*e - c*f + f*(c + d*x))^3*(-((d*e)/Sqrt[1 - (c + d*x)^2]) + (c*f)/Sqrt[1 - (c + d*x)^2] - (f*(c + d*x
))/Sqrt[1 - (c + d*x)^2])) + (f*(1 - (c + d*x)^2)^(3/2)*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x
)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])^3*(-(f*ArcTanh[c + d*x]) + (d*e - c*f)*Log[(d*e - c*f)/Sqrt[1 - (c
+ d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2]]))/((d*e - c*f)*(d*e - f - c*f)*(d*e + f - c*f)*(-f^2 + (d*e
- c*f)^2)*(d*e - c*f + f*(c + d*x))^3) + (c*(1 - (c + d*x)^2)^(3/2)*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[
1 - (c + d*x)^2] + (f*(c + d*x))/Sqrt[1 - (c + d*x)^2])^3*(ArcTanh[c + d*x]^2/E^ArcTanh[(d*e - c*f)/f] - (I*(d
*e - c*f)*(-((-Pi + (2*I)*ArcTanh[(d*e - c*f)/f])*ArcTanh[c + d*x]) - 2*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[
c + d*x])*Log[1 - E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))] - Pi*Log[1 + E^(2*ArcTanh[c + d*x
])] + Pi*Log[1/Sqrt[1 - (c + d*x)^2]] + (2*I)*ArcTanh[(d*e - c*f)/f]*Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTa
nh[c + d*x]]] + I*PolyLog[2, E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))]))/(f*Sqrt[1 - (d*e - c
*f)^2/f^2])))/((d*e - c*f)*(d*e - f - c*f)*(d*e + f - c*f)*Sqrt[(f^2 - (d*e - c*f)^2)/f^2]*(d*e - c*f + f*(c +
d*x))^3) - (d*e*(1 - (c + d*x)^2)^(3/2)*((d*e)/Sqrt[1 - (c + d*x)^2] - (c*f)/Sqrt[1 - (c + d*x)^2] + (f*(c +
d*x))/Sqrt[1 - (c + d*x)^2])^3*(ArcTanh[c + d*x]^2/E^ArcTanh[(d*e - c*f)/f] - (I*(d*e - c*f)*(-((-Pi + (2*I)*A
rcTanh[(d*e - c*f)/f])*ArcTanh[c + d*x]) - 2*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x])*Log[1 - E^((2*I)*
(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))] - Pi*Log[1 + E^(2*ArcTanh[c + d*x])] + Pi*Log[1/Sqrt[1 - (c
+ d*x)^2]] + (2*I)*ArcTanh[(d*e - c*f)/f]*Log[I*Sinh[ArcTanh[(d*e - c*f)/f] + ArcTanh[c + d*x]]] + I*PolyLog[2
, E^((2*I)*(I*ArcTanh[(d*e - c*f)/f] + I*ArcTanh[c + d*x]))]))/(f*Sqrt[1 - (d*e - c*f)^2/f^2])))/(f*(d*e - c*f
)*(d*e - f - c*f)*(d*e + f - c*f)*Sqrt[(f^2 - (d*e - c*f)^2)/f^2]*(d*e - c*f + f*(c + d*x))^3)))/(d*(e + f*x)^
3)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {artanh}\left (d x + c\right ) + a^{2}}{f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(f*x+e)^3,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(d*x + c)^2 + 2*a*b*arctanh(d*x + c) + a^2)/(f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{2}}{{\left (f x + e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(d*x + c) + a)^2/(f*x + e)^3, x)

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maple [A]  time = 0.09, size = 1428, normalized size = 1.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^2/(f*x+e)^3,x)

[Out]

1/2*d^2*a*b/f/(c*f-d*e+f)^2*ln(d*x+c+1)+d^2*b^2*f/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)-d^2*b^2/(c
*f-d*e-f)/(c*f-d*e+f)/(2*c*f-2*d*e-2*f)*ln(d*x+c-1)+d^2*b^2/(c*f-d*e-f)/(c*f-d*e+f)/(2*c*f-2*d*e+2*f)*ln(d*x+c
+1)-1/2*d^2*a^2/(d*f*x+d*e)^2/f+d^2*b^2*f/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f-f)/(
c*f-d*e-f))*c-d^2*b^2*f/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f+f)/(c*f-d*e+f))*c+2*d^
2*a*b*f/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*c+2*d^2*b^2*f*arctanh(d*x+c)/(c*f-d*e-f)^2/(c*f-d*e+
f)^2*ln((d*x+c)*f-c*f+d*e)*c+1/4*d^2*b^2/f/(c*f-d*e-f)^2*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)-1/4*d^2*b^2/f/(c*f-
d*e+f)^2*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)+1/4*d^2*b^2/f/(c*f-d*e+f)^2*ln(-1/2*d*x-1/2*c+1/2)*ln(d*
x+c+1)-1/2*d^2*a*b/f/(c*f-d*e-f)^2*ln(d*x+c-1)+d^2*a*b/(c*f-d*e-f)/(c*f-d*e+f)/(d*f*x+d*e)+d^3*b^2/(c*f-d*e-f)
^2/(c*f-d*e+f)^2*dilog(((d*x+c)*f+f)/(c*f-d*e+f))*e+1/4*d^2*b^2/f/(c*f-d*e-f)^2*dilog(1/2+1/2*d*x+1/2*c)-1/8*d
^2*b^2/f/(c*f-d*e+f)^2*ln(d*x+c+1)^2-1/4*d^2*b^2/f/(c*f-d*e+f)^2*dilog(1/2+1/2*d*x+1/2*c)-1/8*d^2*b^2/f/(c*f-d
*e-f)^2*ln(d*x+c-1)^2-d^3*b^2/(c*f-d*e-f)^2/(c*f-d*e+f)^2*dilog(((d*x+c)*f-f)/(c*f-d*e-f))*e+1/2*d^2*b^2/f*arc
tanh(d*x+c)/(c*f-d*e+f)^2*ln(d*x+c+1)+d^2*b^2*arctanh(d*x+c)/(c*f-d*e-f)/(c*f-d*e+f)/(d*f*x+d*e)-d^2*a*b/(d*f*
x+d*e)^2/f*arctanh(d*x+c)-1/2*d^2*b^2/f*arctanh(d*x+c)/(c*f-d*e-f)^2*ln(d*x+c-1)-1/2*d^2*b^2/(d*f*x+d*e)^2/f*a
rctanh(d*x+c)^2-d^2*b^2*f/(c*f-d*e-f)^2/(c*f-d*e+f)^2*dilog(((d*x+c)*f+f)/(c*f-d*e+f))*c-2*d^3*a*b/(c*f-d*e-f)
^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*e-d^3*b^2/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)
*f-f)/(c*f-d*e-f))*e+d^3*b^2/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f+f)/(c*f-d*e+f))*e
-2*d^3*b^2*arctanh(d*x+c)/(c*f-d*e-f)^2/(c*f-d*e+f)^2*ln((d*x+c)*f-c*f+d*e)*e+d^2*b^2*f/(c*f-d*e-f)^2/(c*f-d*e
+f)^2*dilog(((d*x+c)*f-f)/(c*f-d*e-f))*c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (d {\left (\frac {d \log \left (d x + c + 1\right )}{d^{2} e^{2} f - 2 \, {\left (c + 1\right )} d e f^{2} + {\left (c^{2} + 2 \, c + 1\right )} f^{3}} - \frac {d \log \left (d x + c - 1\right )}{d^{2} e^{2} f - 2 \, {\left (c - 1\right )} d e f^{2} + {\left (c^{2} - 2 \, c + 1\right )} f^{3}} - \frac {4 \, {\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} - c\right )} d e f^{3} + {\left (c^{4} - 2 \, c^{2} + 1\right )} f^{4}} + \frac {2}{d^{2} e^{3} - 2 \, c d e^{2} f + {\left (c^{2} - 1\right )} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} - 1\right )} f^{3}\right )} x}\right )} - \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} a b - \frac {1}{8} \, b^{2} {\left (\frac {\log \left (-d x - c + 1\right )^{2}}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f} + 2 \, \int -\frac {{\left (d f x + c f - f\right )} \log \left (d x + c + 1\right )^{2} + {\left (d f x + d e - 2 \, {\left (d f x + c f - f\right )} \log \left (d x + c + 1\right )\right )} \log \left (-d x - c + 1\right )}{d f^{4} x^{4} + c e^{3} f - e^{3} f + {\left (3 \, d e f^{3} + c f^{4} - f^{4}\right )} x^{3} + 3 \, {\left (d e^{2} f^{2} + c e f^{3} - e f^{3}\right )} x^{2} + {\left (d e^{3} f + 3 \, c e^{2} f^{2} - 3 \, e^{2} f^{2}\right )} x}\,{d x}\right )} - \frac {a^{2}}{2 \, {\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(f*x+e)^3,x, algorithm="maxima")

[Out]

1/2*(d*(d*log(d*x + c + 1)/(d^2*e^2*f - 2*(c + 1)*d*e*f^2 + (c^2 + 2*c + 1)*f^3) - d*log(d*x + c - 1)/(d^2*e^2
*f - 2*(c - 1)*d*e*f^2 + (c^2 - 2*c + 1)*f^3) - 4*(d^2*e - c*d*f)*log(f*x + e)/(d^4*e^4 - 4*c*d^3*e^3*f + 2*(3
*c^2 - 1)*d^2*e^2*f^2 - 4*(c^3 - c)*d*e*f^3 + (c^4 - 2*c^2 + 1)*f^4) + 2/(d^2*e^3 - 2*c*d*e^2*f + (c^2 - 1)*e*
f^2 + (d^2*e^2*f - 2*c*d*e*f^2 + (c^2 - 1)*f^3)*x)) - 2*arctanh(d*x + c)/(f^3*x^2 + 2*e*f^2*x + e^2*f))*a*b -
1/8*b^2*(log(-d*x - c + 1)^2/(f^3*x^2 + 2*e*f^2*x + e^2*f) + 2*integrate(-((d*f*x + c*f - f)*log(d*x + c + 1)^
2 + (d*f*x + d*e - 2*(d*f*x + c*f - f)*log(d*x + c + 1))*log(-d*x - c + 1))/(d*f^4*x^4 + c*e^3*f - e^3*f + (3*
d*e*f^3 + c*f^4 - f^4)*x^3 + 3*(d*e^2*f^2 + c*e*f^3 - e*f^3)*x^2 + (d*e^3*f + 3*c*e^2*f^2 - 3*e^2*f^2)*x), x))
- 1/2*a^2/(f^3*x^2 + 2*e*f^2*x + e^2*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^2}{{\left (e+f\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))^2/(e + f*x)^3,x)

[Out]

int((a + b*atanh(c + d*x))^2/(e + f*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{2}}{\left (e + f x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**2/(f*x+e)**3,x)

[Out]

Integral((a + b*atanh(c + d*x))**2/(e + f*x)**3, x)

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