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

Optimal. Leaf size=359 \[ -\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac {a b e x \left (6 c^2 d^2+e^2\right )}{2 c^3}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {2 b d \left (c^2 d^2+e^2\right ) \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 d e^2 x}{c^2}+\frac {b^2 e^3 x^2}{12 c^2}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac {b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}-\frac {b^2 d \left (c^2 d^2+e^2\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3}+\frac {b^2 e x \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}(c x)}{2 c^3} \]

[Out]

b^2*d*e^2*x/c^2+1/2*a*b*e*(6*c^2*d^2+e^2)*x/c^3+1/12*b^2*e^3*x^2/c^2-b^2*d*e^2*arctanh(c*x)/c^3+1/2*b^2*e*(6*c
^2*d^2+e^2)*x*arctanh(c*x)/c^3+b*d*e^2*x^2*(a+b*arctanh(c*x))/c+1/6*b*e^3*x^3*(a+b*arctanh(c*x))/c+d*(c^2*d^2+
e^2)*(a+b*arctanh(c*x))^2/c^3-1/4*(c^4*d^4+6*c^2*d^2*e^2+e^4)*(a+b*arctanh(c*x))^2/c^4/e+1/4*(e*x+d)^4*(a+b*ar
ctanh(c*x))^2/e-2*b*d*(c^2*d^2+e^2)*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^3+1/12*b^2*e^3*ln(-c^2*x^2+1)/c^4+1/4*
b^2*e*(6*c^2*d^2+e^2)*ln(-c^2*x^2+1)/c^4-b^2*d*(c^2*d^2+e^2)*polylog(2,1-2/(-c*x+1))/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 14, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {5928, 5910, 260, 5916, 321, 206, 266, 43, 6048, 5948, 5984, 5918, 2402, 2315} \[ -\frac {b^2 d \left (c^2 d^2+e^2\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^3}+\frac {a b e x \left (6 c^2 d^2+e^2\right )}{2 c^3}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {\left (6 c^2 d^2 e^2+c^4 d^4+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}-\frac {2 b d \left (c^2 d^2+e^2\right ) \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac {b^2 e x \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}(c x)}{2 c^3}+\frac {b^2 d e^2 x}{c^2}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 e^3 x^2}{12 c^2}+\frac {b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*d*e^2*x)/c^2 + (a*b*e*(6*c^2*d^2 + e^2)*x)/(2*c^3) + (b^2*e^3*x^2)/(12*c^2) - (b^2*d*e^2*ArcTanh[c*x])/c^
3 + (b^2*e*(6*c^2*d^2 + e^2)*x*ArcTanh[c*x])/(2*c^3) + (b*d*e^2*x^2*(a + b*ArcTanh[c*x]))/c + (b*e^3*x^3*(a +
b*ArcTanh[c*x]))/(6*c) + (d*(c^2*d^2 + e^2)*(a + b*ArcTanh[c*x])^2)/c^3 - ((c^4*d^4 + 6*c^2*d^2*e^2 + e^4)*(a
+ b*ArcTanh[c*x])^2)/(4*c^4*e) + ((d + e*x)^4*(a + b*ArcTanh[c*x])^2)/(4*e) - (2*b*d*(c^2*d^2 + e^2)*(a + b*Ar
cTanh[c*x])*Log[2/(1 - c*x)])/c^3 + (b^2*e^3*Log[1 - c^2*x^2])/(12*c^4) + (b^2*e*(6*c^2*d^2 + e^2)*Log[1 - c^2
*x^2])/(4*c^4) - (b^2*d*(c^2*d^2 + e^2)*PolyLog[2, 1 - 2/(1 - c*x)])/c^3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 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 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]

Rubi steps

\begin {align*} \int (d+e x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {(b c) \int \left (-\frac {e^2 \left (6 c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4}-\frac {4 d e^3 x \left (a+b \tanh ^{-1}(c x)\right )}{c^2}-\frac {e^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c^2}+\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {b \int \frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4+4 c^2 d e \left (c^2 d^2+e^2\right ) x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{2 c^3 e}+\frac {\left (2 b d e^2\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c}+\frac {\left (b e^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c}+\frac {\left (b e \left (6 c^2 d^2+e^2\right )\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{2 c^3}\\ &=\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {b \int \left (\frac {c^4 d^4 \left (1+\frac {6 c^2 d^2 e^2+e^4}{c^4 d^4}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}+\frac {4 c^2 d e \left (c^2 d^2+e^2\right ) x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}\right ) \, dx}{2 c^3 e}-\left (b^2 d e^2\right ) \int \frac {x^2}{1-c^2 x^2} \, dx-\frac {1}{6} \left (b^2 e^3\right ) \int \frac {x^3}{1-c^2 x^2} \, dx+\frac {\left (b^2 e \left (6 c^2 d^2+e^2\right )\right ) \int \tanh ^{-1}(c x) \, dx}{2 c^3}\\ &=\frac {b^2 d e^2 x}{c^2}+\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {\left (b^2 d e^2\right ) \int \frac {1}{1-c^2 x^2} \, dx}{c^2}-\frac {1}{12} \left (b^2 e^3\right ) \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )-\frac {\left (2 b d \left (c^2 d^2+e^2\right )\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c}-\frac {\left (b^2 e \left (6 c^2 d^2+e^2\right )\right ) \int \frac {x}{1-c^2 x^2} \, dx}{2 c^2}-\frac {\left (b \left (c^4 d^4+6 c^2 d^2 e^2+e^4\right )\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{2 c^3 e}\\ &=\frac {b^2 d e^2 x}{c^2}+\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac {1}{12} \left (b^2 e^3\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac {\left (2 b d \left (c^2 d^2+e^2\right )\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^2}\\ &=\frac {b^2 d e^2 x}{c^2}+\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac {b^2 e^3 x^2}{12 c^2}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac {\left (2 b^2 d \left (c^2 d^2+e^2\right )\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^2}\\ &=\frac {b^2 d e^2 x}{c^2}+\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac {b^2 e^3 x^2}{12 c^2}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac {\left (2 b^2 d \left (c^2 d^2+e^2\right )\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^3}\\ &=\frac {b^2 d e^2 x}{c^2}+\frac {a b e \left (6 c^2 d^2+e^2\right ) x}{2 c^3}+\frac {b^2 e^3 x^2}{12 c^2}-\frac {b^2 d e^2 \tanh ^{-1}(c x)}{c^3}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) x \tanh ^{-1}(c x)}{2 c^3}+\frac {b d e^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {b e^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 c}+\frac {d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}-\frac {\left (c^4 d^4+6 c^2 d^2 e^2+e^4\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^4 e}+\frac {(d+e x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 e}-\frac {2 b d \left (c^2 d^2+e^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^2 e^3 \log \left (1-c^2 x^2\right )}{12 c^4}+\frac {b^2 e \left (6 c^2 d^2+e^2\right ) \log \left (1-c^2 x^2\right )}{4 c^4}-\frac {b^2 d \left (c^2 d^2+e^2\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 506, normalized size = 1.41 \[ \frac {12 a^2 c^4 d^3 x+18 a^2 c^4 d^2 e x^2+12 a^2 c^4 d e^2 x^3+3 a^2 c^4 e^3 x^4+36 a b c^3 d^2 e x+12 a b c^3 d e^2 x^2+2 a b c^3 e^3 x^3+18 a b c^2 d^2 e \log (1-c x)-18 a b c^2 d^2 e \log (c x+1)+12 a b c d e^2 \log \left (c^2 x^2-1\right )+12 a b c^3 d^3 \log \left (1-c^2 x^2\right )+2 b c \tanh ^{-1}(c x) \left (3 a c^3 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+b e \left (18 c^2 d^2 x+6 d e \left (c^2 x^2-1\right )+e^2 x \left (c^2 x^2+3\right )\right )-12 b d \left (c^2 d^2+e^2\right ) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+6 a b c e^3 x+3 a b e^3 \log (1-c x)-3 a b e^3 \log (c x+1)+12 b^2 c d \left (c^2 d^2+e^2\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+18 b^2 c^2 d^2 e \log \left (1-c^2 x^2\right )+12 b^2 c^2 d e^2 x+b^2 c^2 e^3 x^2+4 b^2 e^3 \log \left (1-c^2 x^2\right )+3 b^2 \tanh ^{-1}(c x)^2 \left (c^4 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )-4 c^3 d^3-6 c^2 d^2 e-4 c d e^2-e^3\right )-b^2 e^3}{12 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b^2*e^3) + 12*a^2*c^4*d^3*x + 36*a*b*c^3*d^2*e*x + 12*b^2*c^2*d*e^2*x + 6*a*b*c*e^3*x + 18*a^2*c^4*d^2*e*x^
2 + 12*a*b*c^3*d*e^2*x^2 + b^2*c^2*e^3*x^2 + 12*a^2*c^4*d*e^2*x^3 + 2*a*b*c^3*e^3*x^3 + 3*a^2*c^4*e^3*x^4 + 3*
b^2*(-4*c^3*d^3 - 6*c^2*d^2*e - 4*c*d*e^2 - e^3 + c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3))*ArcTanh[c
*x]^2 + 2*b*c*ArcTanh[c*x]*(3*a*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b*e*(18*c^2*d^2*x + 6*d*e*
(-1 + c^2*x^2) + e^2*x*(3 + c^2*x^2)) - 12*b*d*(c^2*d^2 + e^2)*Log[1 + E^(-2*ArcTanh[c*x])]) + 18*a*b*c^2*d^2*
e*Log[1 - c*x] + 3*a*b*e^3*Log[1 - c*x] - 18*a*b*c^2*d^2*e*Log[1 + c*x] - 3*a*b*e^3*Log[1 + c*x] + 12*a*b*c^3*
d^3*Log[1 - c^2*x^2] + 18*b^2*c^2*d^2*e*Log[1 - c^2*x^2] + 4*b^2*e^3*Log[1 - c^2*x^2] + 12*a*b*c*d*e^2*Log[-1
+ c^2*x^2] + 12*b^2*c*d*(c^2*d^2 + e^2)*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(12*c^4)

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} e^{3} x^{3} + 3 \, a^{2} d e^{2} x^{2} + 3 \, a^{2} d^{2} e x + a^{2} d^{3} + {\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b e^{3} x^{3} + 3 \, a b d e^{2} x^{2} + 3 \, a b d^{2} e x + a b d^{3}\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [B]  time = 0.07, size = 1430, normalized size = 3.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2/e*d^4+a^2*e^2*x^3*d+3/2*a^2*e*x^2*d^2+b^2*arctanh(c*x)^2*x*d^3+1/4*b^2/e*arctanh(c*x)^2*d^4+1/16*b^2/e
*ln(c*x-1)^2*d^4+1/16*b^2/e*ln(c*x+1)^2*d^4+1/4*b^2*e^3*arctanh(c*x)^2*x^4+1/16/c^4*b^2*e^3*ln(c*x+1)^2+1/3/c^
4*b^2*e^3*ln(c*x-1)+1/3/c^4*b^2*e^3*ln(c*x+1)+1/16/c^4*b^2*e^3*ln(c*x-1)^2+1/4/c*b^2*ln(c*x-1)^2*d^3-1/c*b^2*d
ilog(1/2+1/2*c*x)*d^3-1/4/c*b^2*ln(c*x+1)^2*d^3-1/2/c*b^2*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)*d^3-1/2/c*b^2*ln(c*
x-1)*ln(1/2+1/2*c*x)*d^3+1/2*a*b/e*arctanh(c*x)*d^4+1/4*a*b/e*ln(c*x-1)*d^4-1/4*a*b/e*ln(c*x+1)*d^4-1/8*b^2/e*
ln(c*x+1)*ln(-1/2*c*x+1/2)*d^4+1/8*b^2/e*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)*d^4+1/c*b^2*arctanh(c*x)*ln(c*x-1)*d
^3+1/2*a*b*e^3*arctanh(c*x)*x^4+1/2/c*b^2*ln(c*x+1)*ln(-1/2*c*x+1/2)*d^3+1/6/c*b^2*e^3*arctanh(c*x)*x^3+1/c*b^
2*arctanh(c*x)*ln(c*x+1)*d^3+1/2*a*b/c^3*x*e^3+1/c*a*b*ln(c*x+1)*d^3+1/c*a*b*ln(c*x-1)*d^3+1/4/c^3*b^2*e^2*ln(
c*x-1)^2*d-1/c^3*b^2*e^2*dilog(1/2+1/2*c*x)*d+1/2/c^3*b^2*e^3*arctanh(c*x)*x-1/4/c^3*b^2*e^2*ln(c*x+1)^2*d+3/8
/c^2*b^2*e*ln(c*x+1)^2*d^2-1/4/c^4*a*b*e^3*ln(c*x+1)+1/4/c^4*a*b*e^3*ln(c*x-1)+3/2/c^2*b^2*e*ln(c*x-1)*d^2+3/2
/c^2*b^2*e*ln(c*x+1)*d^2+1/2/c^3*b^2*e^2*ln(c*x-1)*d-1/2/c^3*b^2*e^2*ln(c*x+1)*d+b^2*e^2*arctanh(c*x)^2*x^3*d+
3/2*b^2*e*arctanh(c*x)^2*x^2*d^2+1/12*b^2*e^3*x^2/c^2-1/8/c^4*b^2*e^3*ln(c*x+1)*ln(-1/2*c*x+1/2)+1/8/c^4*b^2*e
^3*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)+1/4/c^4*b^2*e^3*arctanh(c*x)*ln(c*x-1)-1/4/c^4*b^2*e^3*arctanh(c*x)*ln(c*x
+1)-1/8/c^4*b^2*e^3*ln(c*x-1)*ln(1/2+1/2*c*x)+1/6/c*a*b*x^3*e^3+3/8/c^2*b^2*e*ln(c*x-1)^2*d^2+a^2*x*d^3+1/4*a^
2*e^3*x^4+1/4*b^2/e*arctanh(c*x)*ln(c*x-1)*d^4-1/8*b^2/e*ln(c*x-1)*ln(1/2+1/2*c*x)*d^4-1/4*b^2/e*arctanh(c*x)*
ln(c*x+1)*d^4+2*a*b*arctanh(c*x)*x*d^3+b^2*d*e^2*x/c^2+1/c^3*b^2*e^2*arctanh(c*x)*ln(c*x+1)*d+3/2/c^2*b^2*e*ar
ctanh(c*x)*ln(c*x-1)*d^2+3/4/c^2*b^2*e*ln(1/2+1/2*c*x)*ln(-1/2*c*x+1/2)*d^2-1/2/c^3*b^2*e^2*ln(1/2+1/2*c*x)*ln
(-1/2*c*x+1/2)*d-3/4/c^2*b^2*e*ln(c*x-1)*ln(1/2+1/2*c*x)*d^2-1/2/c^3*b^2*e^2*ln(c*x-1)*ln(1/2+1/2*c*x)*d+3/c*b
^2*e*arctanh(c*x)*x*d^2+1/c*b^2*e^2*arctanh(c*x)*x^2*d+3*a*b/c*e*x*d^2+1/c*a*b*e^2*x^2*d+2*a*b*e^2*arctanh(c*x
)*x^3*d+3*a*b*e*arctanh(c*x)*x^2*d^2+3/2/c^2*a*b*e*ln(c*x-1)*d^2+1/c^3*a*b*e^2*ln(c*x-1)*d-3/2/c^2*a*b*e*ln(c*
x+1)*d^2+1/c^3*a*b*e^2*ln(c*x+1)*d-3/4/c^2*b^2*e*ln(c*x+1)*ln(-1/2*c*x+1/2)*d^2+1/2/c^3*b^2*e^2*ln(c*x+1)*ln(-
1/2*c*x+1/2)*d+1/c^3*b^2*e^2*arctanh(c*x)*ln(c*x-1)*d-3/2/c^2*b^2*e*arctanh(c*x)*ln(c*x+1)*d^2

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maxima [B]  time = 0.56, size = 782, normalized size = 2.18 \[ \frac {1}{4} \, a^{2} e^{3} x^{4} + a^{2} d e^{2} x^{3} + \frac {3}{2} \, a^{2} d^{2} e x^{2} + \frac {3}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a b d^{2} e + {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a b d e^{2} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a b e^{3} + a^{2} d^{3} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a b d^{3}}{c} + \frac {{\left (c^{2} d^{3} + d e^{2}\right )} {\left (\log \left (c x + 1\right ) \log \left (-\frac {1}{2} \, c x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c x + \frac {1}{2}\right )\right )} b^{2}}{c^{3}} + \frac {{\left (9 \, c^{2} d^{2} e - 3 \, c d e^{2} + 2 \, e^{3}\right )} b^{2} \log \left (c x + 1\right )}{6 \, c^{4}} + \frac {{\left (9 \, c^{2} d^{2} e + 3 \, c d e^{2} + 2 \, e^{3}\right )} b^{2} \log \left (c x - 1\right )}{6 \, c^{4}} + \frac {4 \, b^{2} c^{2} e^{3} x^{2} + 48 \, b^{2} c^{2} d e^{2} x + 3 \, {\left (b^{2} c^{4} e^{3} x^{4} + 4 \, b^{2} c^{4} d e^{2} x^{3} + 6 \, b^{2} c^{4} d^{2} e x^{2} + 4 \, b^{2} c^{4} d^{3} x + {\left (4 \, c^{3} d^{3} - 6 \, c^{2} d^{2} e + 4 \, c d e^{2} - e^{3}\right )} b^{2}\right )} \log \left (c x + 1\right )^{2} + 3 \, {\left (b^{2} c^{4} e^{3} x^{4} + 4 \, b^{2} c^{4} d e^{2} x^{3} + 6 \, b^{2} c^{4} d^{2} e x^{2} + 4 \, b^{2} c^{4} d^{3} x - {\left (4 \, c^{3} d^{3} + 6 \, c^{2} d^{2} e + 4 \, c d e^{2} + e^{3}\right )} b^{2}\right )} \log \left (-c x + 1\right )^{2} + 4 \, {\left (b^{2} c^{3} e^{3} x^{3} + 6 \, b^{2} c^{3} d e^{2} x^{2} + 3 \, {\left (6 \, c^{3} d^{2} e + c e^{3}\right )} b^{2} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (2 \, b^{2} c^{3} e^{3} x^{3} + 12 \, b^{2} c^{3} d e^{2} x^{2} + 6 \, {\left (6 \, c^{3} d^{2} e + c e^{3}\right )} b^{2} x + 3 \, {\left (b^{2} c^{4} e^{3} x^{4} + 4 \, b^{2} c^{4} d e^{2} x^{3} + 6 \, b^{2} c^{4} d^{2} e x^{2} + 4 \, b^{2} c^{4} d^{3} x + {\left (4 \, c^{3} d^{3} - 6 \, c^{2} d^{2} e + 4 \, c d e^{2} - e^{3}\right )} b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{48 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*e^3*x^4 + a^2*d*e^2*x^3 + 3/2*a^2*d^2*e*x^2 + 3/2*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3
+ log(c*x - 1)/c^3))*a*b*d^2*e + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*d*e^2 + 1/12*(6
*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*a*b*e^3 + a^2*d^3*x +
(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*d^3/c + (c^2*d^3 + d*e^2)*(log(c*x + 1)*log(-1/2*c*x + 1/2) + di
log(1/2*c*x + 1/2))*b^2/c^3 + 1/6*(9*c^2*d^2*e - 3*c*d*e^2 + 2*e^3)*b^2*log(c*x + 1)/c^4 + 1/6*(9*c^2*d^2*e +
3*c*d*e^2 + 2*e^3)*b^2*log(c*x - 1)/c^4 + 1/48*(4*b^2*c^2*e^3*x^2 + 48*b^2*c^2*d*e^2*x + 3*(b^2*c^4*e^3*x^4 +
4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x + (4*c^3*d^3 - 6*c^2*d^2*e + 4*c*d*e^2 - e^3)*b^2)
*log(c*x + 1)^2 + 3*(b^2*c^4*e^3*x^4 + 4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x - (4*c^3*d^
3 + 6*c^2*d^2*e + 4*c*d*e^2 + e^3)*b^2)*log(-c*x + 1)^2 + 4*(b^2*c^3*e^3*x^3 + 6*b^2*c^3*d*e^2*x^2 + 3*(6*c^3*
d^2*e + c*e^3)*b^2*x)*log(c*x + 1) - 2*(2*b^2*c^3*e^3*x^3 + 12*b^2*c^3*d*e^2*x^2 + 6*(6*c^3*d^2*e + c*e^3)*b^2
*x + 3*(b^2*c^4*e^3*x^4 + 4*b^2*c^4*d*e^2*x^3 + 6*b^2*c^4*d^2*e*x^2 + 4*b^2*c^4*d^3*x + (4*c^3*d^3 - 6*c^2*d^2
*e + 4*c*d*e^2 - e^3)*b^2)*log(c*x + 1))*log(-c*x + 1))/c^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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