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

Optimal. Leaf size=546 \[ -\frac {b^2 \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^3}-\frac {6 b^2 f (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^3}+\frac {a b^2 f^2 x}{d^2}-\frac {(d e-c f) \left (\left (c^2+3\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}-\frac {b \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (c+d x) (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}+\frac {b^3 \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right )}{2 d^3}-\frac {3 b^3 f (d e-c f) \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{d^3}+\frac {b^3 f^2 \log \left (1-(c+d x)^2\right )}{2 d^3}+\frac {b^3 f^2 (c+d x) \tanh ^{-1}(c+d x)}{d^3} \]

[Out]

a*b^2*f^2*x/d^2+b^3*f^2*(d*x+c)*arctanh(d*x+c)/d^3-1/2*b*f^2*(a+b*arctanh(d*x+c))^2/d^3+3*b*f*(-c*f+d*e)*(a+b*
arctanh(d*x+c))^2/d^3+3*b*f*(-c*f+d*e)*(d*x+c)*(a+b*arctanh(d*x+c))^2/d^3+1/2*b*f^2*(d*x+c)^2*(a+b*arctanh(d*x
+c))^2/d^3-1/3*(-c*f+d*e)*(d^2*e^2-2*c*d*e*f+(c^2+3)*f^2)*(a+b*arctanh(d*x+c))^3/d^3/f+1/3*(3*d^2*e^2-6*c*d*e*
f+(3*c^2+1)*f^2)*(a+b*arctanh(d*x+c))^3/d^3+1/3*(f*x+e)^3*(a+b*arctanh(d*x+c))^3/f-6*b^2*f*(-c*f+d*e)*(a+b*arc
tanh(d*x+c))*ln(2/(-d*x-c+1))/d^3-b*(3*d^2*e^2-6*c*d*e*f+(3*c^2+1)*f^2)*(a+b*arctanh(d*x+c))^2*ln(2/(-d*x-c+1)
)/d^3+1/2*b^3*f^2*ln(1-(d*x+c)^2)/d^3-3*b^3*f*(-c*f+d*e)*polylog(2,(-d*x-c-1)/(-d*x-c+1))/d^3-b^2*(3*d^2*e^2-6
*c*d*e*f+(3*c^2+1)*f^2)*(a+b*arctanh(d*x+c))*polylog(2,1-2/(-d*x-c+1))/d^3+1/2*b^3*(3*d^2*e^2-6*c*d*e*f+(3*c^2
+1)*f^2)*polylog(3,1-2/(-d*x-c+1))/d^3

________________________________________________________________________________________

Rubi [A]  time = 1.05, antiderivative size = 546, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6111, 5928, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 6048, 6058, 6610} \[ -\frac {b^2 \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^3}+\frac {b^3 \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \text {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )}{2 d^3}-\frac {3 b^3 f (d e-c f) \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{d^3}-\frac {6 b^2 f (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d^3}+\frac {a b^2 f^2 x}{d^2}-\frac {(d e-c f) \left (\left (c^2+3\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}-\frac {b \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (c+d x) (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}+\frac {b^3 f^2 \log \left (1-(c+d x)^2\right )}{2 d^3}+\frac {b^3 f^2 (c+d x) \tanh ^{-1}(c+d x)}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*f^2*x)/d^2 + (b^3*f^2*(c + d*x)*ArcTanh[c + d*x])/d^3 - (b*f^2*(a + b*ArcTanh[c + d*x])^2)/(2*d^3) + (3
*b*f*(d*e - c*f)*(a + b*ArcTanh[c + d*x])^2)/d^3 + (3*b*f*(d*e - c*f)*(c + d*x)*(a + b*ArcTanh[c + d*x])^2)/d^
3 + (b*f^2*(c + d*x)^2*(a + b*ArcTanh[c + d*x])^2)/(2*d^3) - ((d*e - c*f)*(d^2*e^2 - 2*c*d*e*f + (3 + c^2)*f^2
)*(a + b*ArcTanh[c + d*x])^3)/(3*d^3*f) + ((3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*(a + b*ArcTanh[c + d*x])^
3)/(3*d^3) + ((e + f*x)^3*(a + b*ArcTanh[c + d*x])^3)/(3*f) - (6*b^2*f*(d*e - c*f)*(a + b*ArcTanh[c + d*x])*Lo
g[2/(1 - c - d*x)])/d^3 - (b*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*(a + b*ArcTanh[c + d*x])^2*Log[2/(1 - c
- d*x)])/d^3 + (b^3*f^2*Log[1 - (c + d*x)^2])/(2*d^3) - (3*b^3*f*(d*e - c*f)*PolyLog[2, -((1 + c + d*x)/(1 -
c - d*x))])/d^3 - (b^2*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - 2/(1
- c - d*x)])/d^3 + (b^3*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*PolyLog[3, 1 - 2/(1 - c - d*x)])/(2*d^3)

Rule 260

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

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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

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 5928

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

Rule 5948

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

Rule 5980

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

Rule 5984

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

Rule 6048

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

Rule 6058

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

Rule 6111

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int (e+f x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^2 \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {3 f^2 (d e-c f) \left (a+b \tanh ^{-1}(x)\right )^2}{d^3}-\frac {f^3 x \left (a+b \tanh ^{-1}(x)\right )^2}{d^3}+\frac {\left ((d e-c f) \left (d^2 e^2-2 c d e f+3 f^2+c^2 f^2\right )+f \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) x\right ) \left (a+b \tanh ^{-1}(x)\right )^2}{d^3 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{f}\\ &=\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {b \operatorname {Subst}\left (\int \frac {\left ((d e-c f) \left (d^2 e^2-2 c d e f+3 f^2+c^2 f^2\right )+f \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) x\right ) \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d^3 f}+\frac {\left (b f^2\right ) \operatorname {Subst}\left (\int x \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d^3}+\frac {(3 b f (d e-c f)) \operatorname {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d^3}\\ &=\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {b \operatorname {Subst}\left (\int \left (\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2}+\frac {f \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2}\right ) \, dx,x,c+d x\right )}{d^3 f}-\frac {\left (b^2 f^2\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^3}-\frac {\left (6 b^2 f (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}+\frac {\left (b^2 f^2\right ) \operatorname {Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d^3}-\frac {\left (b^2 f^2\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{d^3}-\frac {\left (6 b^2 f (d e-c f)\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d^3}-\frac {\left (b (d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d^3 f}-\frac {\left (b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {a b^2 f^2 x}{d^2}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}-\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {6 b^2 f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^3}+\frac {\left (b^3 f^2\right ) \operatorname {Subst}\left (\int \tanh ^{-1}(x) \, dx,x,c+d x\right )}{d^3}+\frac {\left (6 b^3 f (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^3}-\frac {\left (b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {a b^2 f^2 x}{d^2}+\frac {b^3 f^2 (c+d x) \tanh ^{-1}(c+d x)}{d^3}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}-\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {6 b^2 f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^3}-\frac {b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^3}-\frac {\left (b^3 f^2\right ) \operatorname {Subst}\left (\int \frac {x}{1-x^2} \, dx,x,c+d x\right )}{d^3}-\frac {\left (6 b^3 f (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^3}+\frac {\left (2 b^2 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {a b^2 f^2 x}{d^2}+\frac {b^3 f^2 (c+d x) \tanh ^{-1}(c+d x)}{d^3}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}-\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {6 b^2 f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^3}-\frac {b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^3}+\frac {b^3 f^2 \log \left (1-(c+d x)^2\right )}{2 d^3}-\frac {3 b^3 f (d e-c f) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^3}-\frac {b^2 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^3}+\frac {\left (b^3 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^3}\\ &=\frac {a b^2 f^2 x}{d^2}+\frac {b^3 f^2 (c+d x) \tanh ^{-1}(c+d x)}{d^3}-\frac {b f^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}+\frac {3 b f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d^3}+\frac {b f^2 (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d^3}-\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3 f}+\frac {\left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d^3}+\frac {(e+f x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 f}-\frac {6 b^2 f (d e-c f) \left (a+b \tanh ^{-1}(c+d x)\right ) \log \left (\frac {2}{1-c-d x}\right )}{d^3}-\frac {b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d^3}+\frac {b^3 f^2 \log \left (1-(c+d x)^2\right )}{2 d^3}-\frac {3 b^3 f (d e-c f) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^3}-\frac {b^2 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d^3}+\frac {b^3 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \text {Li}_3\left (1-\frac {2}{1-c-d x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [B]  time = 10.16, size = 1868, normalized size = 3.42 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(a*d^2*e^2 + 3*b*d*e*f - 2*b*c*f^2)*x)/d^2 + (a^2*f*(2*a*d*e + b*f)*x^2)/(2*d) + (a^3*f^2*x^3)/3 + a^2*b*
x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcTanh[c + d*x] + ((3*a^2*b*d^2*e^2 - 3*a^2*b*c*d^2*e^2 + 3*a^2*b*d*e*f - 6*a^2
*b*c*d*e*f + 3*a^2*b*c^2*d*e*f + a^2*b*f^2 - 3*a^2*b*c*f^2 + 3*a^2*b*c^2*f^2 - a^2*b*c^3*f^2)*Log[1 - c - d*x]
)/(2*d^3) + ((3*a^2*b*d^2*e^2 + 3*a^2*b*c*d^2*e^2 - 3*a^2*b*d*e*f - 6*a^2*b*c*d*e*f - 3*a^2*b*c^2*d*e*f + a^2*
b*f^2 + 3*a^2*b*c*f^2 + 3*a^2*b*c^2*f^2 + a^2*b*c^3*f^2)*Log[1 + c + d*x])/(2*d^3) + (3*a*b^2*e^2*(ArcTanh[c +
d*x]*(-ArcTanh[c + d*x] + (c + d*x)*ArcTanh[c + d*x] - 2*Log[1 + E^(-2*ArcTanh[c + d*x])]) + PolyLog[2, -E^(-
2*ArcTanh[c + d*x])]))/d - (3*a*b^2*e*f*((1 - (c + d*x)^2)*ArcTanh[c + d*x]^2 + 2*(-((c + d*x)*ArcTanh[c + d*x
]) - c*ArcTanh[c + d*x]^2 + c*(c + d*x)*ArcTanh[c + d*x]^2 - 2*c*ArcTanh[c + d*x]*Log[1 + E^(-2*ArcTanh[c + d*
x])] + Log[1/Sqrt[1 - (c + d*x)^2]]) + 2*c*PolyLog[2, -E^(-2*ArcTanh[c + d*x])]))/d^2 + (b^3*e^2*(ArcTanh[c +
d*x]^2*(-ArcTanh[c + d*x] + (c + d*x)*ArcTanh[c + d*x] - 3*Log[1 + E^(-2*ArcTanh[c + d*x])]) + 3*ArcTanh[c + d
*x]*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] + (3*PolyLog[3, -E^(-2*ArcTanh[c + d*x])])/2))/d + (b^3*e*f*(-(ArcTan
h[c + d*x]*(3*ArcTanh[c + d*x] - 2*c*ArcTanh[c + d*x]^2 + (1 - (c + d*x)^2)*ArcTanh[c + d*x]^2 + (c + d*x)*Arc
Tanh[c + d*x]*(-3 + 2*c*ArcTanh[c + d*x]) + 6*Log[1 + E^(-2*ArcTanh[c + d*x])] - 6*c*ArcTanh[c + d*x]*Log[1 +
E^(-2*ArcTanh[c + d*x])])) + (3 - 6*c*ArcTanh[c + d*x])*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] - 3*c*PolyLog[3,
-E^(-2*ArcTanh[c + d*x])]))/d^2 - (a*b^2*f^2*(1 - (c + d*x)^2)^(3/2)*(-((c + d*x)/Sqrt[1 - (c + d*x)^2]) + (6*
c*(c + d*x)*ArcTanh[c + d*x])/Sqrt[1 - (c + d*x)^2] + (3*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2] -
(3*c^2*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2] + ArcTanh[c + d*x]^2*Cosh[3*ArcTanh[c + d*x]] + 3*
c^2*ArcTanh[c + d*x]^2*Cosh[3*ArcTanh[c + d*x]] + 2*ArcTanh[c + d*x]*Cosh[3*ArcTanh[c + d*x]]*Log[1 + E^(-2*Ar
cTanh[c + d*x])] + 6*c^2*ArcTanh[c + d*x]*Cosh[3*ArcTanh[c + d*x]]*Log[1 + E^(-2*ArcTanh[c + d*x])] - 6*c*Cosh
[3*ArcTanh[c + d*x]]*Log[1/Sqrt[1 - (c + d*x)^2]] + (ArcTanh[c + d*x]*(4 + 3*(1 - 4*c + 3*c^2)*ArcTanh[c + d*x
]) + 6*(ArcTanh[c + d*x] + 3*c^2*ArcTanh[c + d*x])*Log[1 + E^(-2*ArcTanh[c + d*x])] - 18*c*Log[1/Sqrt[1 - (c +
d*x)^2]])/Sqrt[1 - (c + d*x)^2] - (4*(1 + 3*c^2)*PolyLog[2, -E^(-2*ArcTanh[c + d*x])])/(1 - (c + d*x)^2)^(3/2
) - Sinh[3*ArcTanh[c + d*x]] + 6*c*ArcTanh[c + d*x]*Sinh[3*ArcTanh[c + d*x]] - ArcTanh[c + d*x]^2*Sinh[3*ArcTa
nh[c + d*x]] - 3*c^2*ArcTanh[c + d*x]^2*Sinh[3*ArcTanh[c + d*x]]))/(4*d^3) + (b^3*f^2*((-3*c + ArcTanh[c + d*x
] + 3*c^2*ArcTanh[c + d*x])*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] - ((1 - (c + d*x)^2)^(3/2)*((-3*(c + d*x)*Arc
Tanh[c + d*x])/Sqrt[1 - (c + d*x)^2] + (9*c*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2] + (3*(c + d*x)
*ArcTanh[c + d*x]^3)/Sqrt[1 - (c + d*x)^2] - (3*c^2*(c + d*x)*ArcTanh[c + d*x]^3)/Sqrt[1 - (c + d*x)^2] - 9*c*
ArcTanh[c + d*x]^2*Cosh[3*ArcTanh[c + d*x]] + ArcTanh[c + d*x]^3*Cosh[3*ArcTanh[c + d*x]] + 3*c^2*ArcTanh[c +
d*x]^3*Cosh[3*ArcTanh[c + d*x]] - 18*c*ArcTanh[c + d*x]*Cosh[3*ArcTanh[c + d*x]]*Log[1 + E^(-2*ArcTanh[c + d*x
])] + 3*ArcTanh[c + d*x]^2*Cosh[3*ArcTanh[c + d*x]]*Log[1 + E^(-2*ArcTanh[c + d*x])] + 9*c^2*ArcTanh[c + d*x]^
2*Cosh[3*ArcTanh[c + d*x]]*Log[1 + E^(-2*ArcTanh[c + d*x])] + 3*Cosh[3*ArcTanh[c + d*x]]*Log[1/Sqrt[1 - (c + d
*x)^2]] + (3*(ArcTanh[c + d*x]^2*(2 - 9*c + ArcTanh[c + d*x] - 4*c*ArcTanh[c + d*x] + 3*c^2*ArcTanh[c + d*x])
+ 3*ArcTanh[c + d*x]*(-6*c + ArcTanh[c + d*x] + 3*c^2*ArcTanh[c + d*x])*Log[1 + E^(-2*ArcTanh[c + d*x])] + 3*L
og[1/Sqrt[1 - (c + d*x)^2]]))/Sqrt[1 - (c + d*x)^2] - (6*(1 + 3*c^2)*PolyLog[3, -E^(-2*ArcTanh[c + d*x])])/(1
- (c + d*x)^2)^(3/2) - 3*ArcTanh[c + d*x]*Sinh[3*ArcTanh[c + d*x]] + 9*c*ArcTanh[c + d*x]^2*Sinh[3*ArcTanh[c +
d*x]] - ArcTanh[c + d*x]^3*Sinh[3*ArcTanh[c + d*x]] - 3*c^2*ArcTanh[c + d*x]^3*Sinh[3*ArcTanh[c + d*x]]))/12)
)/d^3

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} f^{2} x^{2} + 2 \, a^{3} e f x + a^{3} e^{2} + {\left (b^{3} f^{2} x^{2} + 2 \, b^{3} e f x + b^{3} e^{2}\right )} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, {\left (a b^{2} f^{2} x^{2} + 2 \, a b^{2} e f x + a b^{2} e^{2}\right )} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, {\left (a^{2} b f^{2} x^{2} + 2 \, a^{2} b e f x + a^{2} b e^{2}\right )} \operatorname {artanh}\left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 22.27, size = 12111, normalized size = 22.18 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^3*f^2*x^3 + a^3*e*f*x^2 + 3/2*(2*x^2*arctanh(d*x + c) + d*(2*x/d^2 - (c^2 + 2*c + 1)*log(d*x + c + 1)/d^
3 + (c^2 - 2*c + 1)*log(d*x + c - 1)/d^3))*a^2*b*e*f + 1/2*(2*x^3*arctanh(d*x + c) + d*((d*x^2 - 4*c*x)/d^3 +
(c^3 + 3*c^2 + 3*c + 1)*log(d*x + c + 1)/d^4 - (c^3 - 3*c^2 + 3*c - 1)*log(d*x + c - 1)/d^4))*a^2*b*f^2 + a^3*
e^2*x + 3/2*(2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*a^2*b*e^2/d - 1/24*((b^3*d^3*f^2*x^3 + 3*b^
3*d^3*e*f*x^2 + 3*b^3*d^3*e^2*x + (c^3*f^2 - 3*d^2*e^2 - 3*(d*e*f + f^2)*c^2 - 3*d*e*f + 3*(d^2*e^2 + 2*d*e*f
+ f^2)*c - f^2)*b^3)*log(-d*x - c + 1)^3 - 3*(2*a*b^2*d^3*f^2*x^3 + (6*a*b^2*d^3*e*f + b^3*d^2*f^2)*x^2 + 2*(3
*a*b^2*d^3*e^2 + (3*d^2*e*f - 2*c*d*f^2)*b^3)*x + (b^3*d^3*f^2*x^3 + 3*b^3*d^3*e*f*x^2 + 3*b^3*d^3*e^2*x + (c^
3*f^2 + 3*d^2*e^2 - 3*(d*e*f - f^2)*c^2 - 3*d*e*f + 3*(d^2*e^2 - 2*d*e*f + f^2)*c + f^2)*b^3)*log(d*x + c + 1)
)*log(-d*x - c + 1)^2)/d^3 - integrate(-1/8*((b^3*d^3*f^2*x^3 + (2*d^3*e*f + c*d^2*f^2 - d^2*f^2)*b^3*x^2 + (d
^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f)*b^3*x + (c*d^2*e^2 - d^2*e^2)*b^3)*log(d*x + c + 1)^3 + 6*(a*b^2*d^3*f^2*x^3
+ (2*d^3*e*f + c*d^2*f^2 - d^2*f^2)*a*b^2*x^2 + (d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f)*a*b^2*x + (c*d^2*e^2 - d^
2*e^2)*a*b^2)*log(d*x + c + 1)^2 - (4*a*b^2*d^3*f^2*x^3 + 2*(6*a*b^2*d^3*e*f + b^3*d^2*f^2)*x^2 + 3*(b^3*d^3*f
^2*x^3 + (2*d^3*e*f + c*d^2*f^2 - d^2*f^2)*b^3*x^2 + (d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f)*b^3*x + (c*d^2*e^2 -
d^2*e^2)*b^3)*log(d*x + c + 1)^2 + 4*(3*a*b^2*d^3*e^2 + (3*d^2*e*f - 2*c*d*f^2)*b^3)*x + 2*(6*(c*d^2*e^2 - d^2
*e^2)*a*b^2 + (c^3*f^2 + 3*d^2*e^2 - 3*(d*e*f - f^2)*c^2 - 3*d*e*f + 3*(d^2*e^2 - 2*d*e*f + f^2)*c + f^2)*b^3
+ (6*a*b^2*d^3*f^2 + b^3*d^3*f^2)*x^3 + 3*(b^3*d^3*e*f + 2*(2*d^3*e*f + c*d^2*f^2 - d^2*f^2)*a*b^2)*x^2 + 3*(b
^3*d^3*e^2 + 2*(d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f)*a*b^2)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^3*x + c*d
^2 - d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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