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

Optimal. Leaf size=240 \[ -\frac {4 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac {6 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+a b^2 x+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {(c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {4 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {2 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{c}+b^3 x \tanh ^{-1}(c x) \]

[Out]

a*b^2*x+b^3*x*arctanh(c*x)+5/2*b*(a+b*arctanh(c*x))^2/c+3*b*x*(a+b*arctanh(c*x))^2+1/2*b*c*x^2*(a+b*arctanh(c*
x))^2+1/3*(c*x+1)^3*(a+b*arctanh(c*x))^3/c-6*b^2*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c-4*b*(a+b*arctanh(c*x))^2*
ln(2/(-c*x+1))/c+1/2*b^3*ln(-c^2*x^2+1)/c-3*b^3*polylog(2,1-2/(-c*x+1))/c-4*b^2*(a+b*arctanh(c*x))*polylog(2,1
-2/(-c*x+1))/c+2*b^3*polylog(3,1-2/(-c*x+1))/c

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Rubi [A]  time = 0.44, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {5928, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 1586, 6058, 6610} \[ -\frac {4 b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c}+\frac {2 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{c}-\frac {6 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+a b^2 x+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {(c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {4 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c}+b^3 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + c*x)^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

a*b^2*x + b^3*x*ArcTanh[c*x] + (5*b*(a + b*ArcTanh[c*x])^2)/(2*c) + 3*b*x*(a + b*ArcTanh[c*x])^2 + (b*c*x^2*(a
+ b*ArcTanh[c*x])^2)/2 + ((1 + c*x)^3*(a + b*ArcTanh[c*x])^3)/(3*c) - (6*b^2*(a + b*ArcTanh[c*x])*Log[2/(1 -
c*x)])/c - (4*b*(a + b*ArcTanh[c*x])^2*Log[2/(1 - c*x)])/c + (b^3*Log[1 - c^2*x^2])/(2*c) - (3*b^3*PolyLog[2,
1 - 2/(1 - c*x)])/c - (4*b^2*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/c + (2*b^3*PolyLog[3, 1 - 2/(1
- c*x)])/c

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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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 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 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 (1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-b \int \left (-3 \left (a+b \tanh ^{-1}(c x)\right )^2-c x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {4 (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx\\ &=\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}+(3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx-(4 b) \int \frac {(1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx+(b c) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-(4 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx-\left (6 b^2 c\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\left (b^2 c^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}+b^2 \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx-b^2 \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (6 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (8 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b^2 x+\frac {5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+b^3 \int \tanh ^{-1}(c x) \, dx+\left (4 b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (6 b^3\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b^2 x+b^3 x \tanh ^{-1}(c x)+\frac {5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}-\frac {4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {2 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{c}-\frac {\left (6 b^3\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c}-\left (b^3 c\right ) \int \frac {x}{1-c^2 x^2} \, dx\\ &=a b^2 x+b^3 x \tanh ^{-1}(c x)+\frac {5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac {6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c}-\frac {4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}-\frac {4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {2 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{c}\\ \end {align*}

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Mathematica [B]  time = 0.99, size = 488, normalized size = 2.03 \[ \frac {2 a^3 c^3 x^3+6 a^3 c^2 x^2+6 a^3 c x+6 a^2 b c^3 x^3 \tanh ^{-1}(c x)+3 a^2 b c^2 x^2+18 a^2 b c^2 x^2 \tanh ^{-1}(c x)+18 a^2 b c x+21 a^2 b \log (1-c x)+3 a^2 b \log (c x+1)+18 a^2 b c x \tanh ^{-1}(c x)+6 a b^2 c^3 x^3 \tanh ^{-1}(c x)^2+18 a b^2 \log \left (1-c^2 x^2\right )+18 a b^2 c^2 x^2 \tanh ^{-1}(c x)^2+6 a b^2 c^2 x^2 \tanh ^{-1}(c x)+6 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (4 a+4 b \tanh ^{-1}(c x)+3 b\right )+6 a b^2 c x-42 a b^2 \tanh ^{-1}(c x)^2+18 a b^2 c x \tanh ^{-1}(c x)^2-6 a b^2 \tanh ^{-1}(c x)+36 a b^2 c x \tanh ^{-1}(c x)-48 a b^2 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+2 b^3 c^3 x^3 \tanh ^{-1}(c x)^3+3 b^3 \log \left (1-c^2 x^2\right )+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^3+3 b^3 c^2 x^2 \tanh ^{-1}(c x)^2+12 b^3 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-14 b^3 \tanh ^{-1}(c x)^3+6 b^3 c x \tanh ^{-1}(c x)^3-21 b^3 \tanh ^{-1}(c x)^2+18 b^3 c x \tanh ^{-1}(c x)^2+6 b^3 c x \tanh ^{-1}(c x)-24 b^3 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-36 b^3 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{6 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + c*x)^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

(6*a^3*c*x + 18*a^2*b*c*x + 6*a*b^2*c*x + 6*a^3*c^2*x^2 + 3*a^2*b*c^2*x^2 + 2*a^3*c^3*x^3 - 6*a*b^2*ArcTanh[c*
x] + 18*a^2*b*c*x*ArcTanh[c*x] + 36*a*b^2*c*x*ArcTanh[c*x] + 6*b^3*c*x*ArcTanh[c*x] + 18*a^2*b*c^2*x^2*ArcTanh
[c*x] + 6*a*b^2*c^2*x^2*ArcTanh[c*x] + 6*a^2*b*c^3*x^3*ArcTanh[c*x] - 42*a*b^2*ArcTanh[c*x]^2 - 21*b^3*ArcTanh
[c*x]^2 + 18*a*b^2*c*x*ArcTanh[c*x]^2 + 18*b^3*c*x*ArcTanh[c*x]^2 + 18*a*b^2*c^2*x^2*ArcTanh[c*x]^2 + 3*b^3*c^
2*x^2*ArcTanh[c*x]^2 + 6*a*b^2*c^3*x^3*ArcTanh[c*x]^2 - 14*b^3*ArcTanh[c*x]^3 + 6*b^3*c*x*ArcTanh[c*x]^3 + 6*b
^3*c^2*x^2*ArcTanh[c*x]^3 + 2*b^3*c^3*x^3*ArcTanh[c*x]^3 - 48*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])]
- 36*b^3*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 24*b^3*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 21*a
^2*b*Log[1 - c*x] + 3*a^2*b*Log[1 + c*x] + 18*a*b^2*Log[1 - c^2*x^2] + 3*b^3*Log[1 - c^2*x^2] + 6*b^2*(4*a + 3
*b + 4*b*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 12*b^3*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(6*c)

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} c^{2} x^{2} + 2 \, a^{3} c x + {\left (b^{3} c^{2} x^{2} + 2 \, b^{3} c x + b^{3}\right )} \operatorname {artanh}\left (c x\right )^{3} + a^{3} + 3 \, {\left (a b^{2} c^{2} x^{2} + 2 \, a b^{2} c x + a b^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 3 \, {\left (a^{2} b c^{2} x^{2} + 2 \, a^{2} b c x + a^{2} b\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(a^3*c^2*x^2 + 2*a^3*c*x + (b^3*c^2*x^2 + 2*b^3*c*x + b^3)*arctanh(c*x)^3 + a^3 + 3*(a*b^2*c^2*x^2 + 2
*a*b^2*c*x + a*b^2)*arctanh(c*x)^2 + 3*(a^2*b*c^2*x^2 + 2*a^2*b*c*x + a^2*b)*arctanh(c*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x + 1\right )}^{2} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((c*x + 1)^2*(b*arctanh(c*x) + a)^3, x)

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maple [C]  time = 0.50, size = 811, normalized size = 3.38 \[ \frac {a^{3}}{3 c}+b^{3} \arctanh \left (c x \right )^{3} x +3 b^{3} \arctanh \left (c x \right )^{2} x +\frac {b^{3} \arctanh \left (c x \right )^{3}}{3 c}+\frac {5 b^{3} \arctanh \left (c x \right )^{2}}{2 c}+\frac {2 b^{3} \polylog \left (3, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{c}+\frac {b^{3} \arctanh \left (c x \right )}{c}+3 x \,a^{2} b +\frac {c^{2} x^{3} a^{3}}{3}+c \,x^{2} a^{3}-\frac {b^{3} \ln \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{c}-\frac {a \,b^{2}}{c}-\frac {4 a \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{c}+\frac {8 a \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{c}+c^{2} a \,b^{2} \arctanh \left (c x \right )^{2} x^{3}+c^{2} a^{2} b \arctanh \left (c x \right ) x^{3}+c a \,b^{2} \arctanh \left (c x \right ) x^{2}+3 c \,a^{2} b \arctanh \left (c x \right ) x^{2}+3 c a \,b^{2} \arctanh \left (c x \right )^{2} x^{2}-\frac {4 i b^{3} \pi \arctanh \left (c x \right )^{2}}{c}+x a \,b^{2}-\frac {4 i b^{3} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}}\right )^{3} \arctanh \left (c x \right )^{2}}{c}+\frac {4 i b^{3} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}}\right )^{2} \arctanh \left (c x \right )^{2}}{c}+b^{3} x \arctanh \left (c x \right )-\frac {6 b^{3} \dilog \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c}-\frac {6 b^{3} \dilog \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c}+3 a \,b^{2} \arctanh \left (c x \right )^{2} x +3 a^{2} b \arctanh \left (c x \right ) x +6 a \,b^{2} \arctanh \left (c x \right ) x -\frac {6 b^{3} \arctanh \left (c x \right ) \ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c}-\frac {6 b^{3} \arctanh \left (c x \right ) \ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c}+\frac {c^{2} b^{3} \arctanh \left (c x \right )^{3} x^{3}}{3}+\frac {c \,b^{3} \arctanh \left (c x \right )^{2} x^{2}}{2}-\frac {4 b^{3} \arctanh \left (c x \right ) \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{c}+\frac {4 b^{3} \arctanh \left (c x \right )^{2} \ln \left (c x -1\right )}{c}+\frac {a^{2} b \arctanh \left (c x \right )}{c}+\frac {a \,b^{2} \arctanh \left (c x \right )^{2}}{c}-\frac {4 a \,b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{c}-\frac {4 b^{3} \arctanh \left (c x \right )^{2} \ln \relax (2)}{c}+c \,b^{3} \arctanh \left (c x \right )^{3} x^{2}+\frac {4 a^{2} b \ln \left (c x -1\right )}{c}+\frac {7 a \,b^{2} \ln \left (c x -1\right )}{2 c}+\frac {5 a \,b^{2} \ln \left (c x +1\right )}{2 c}+\frac {2 a \,b^{2} \ln \left (c x -1\right )^{2}}{c}+\frac {c \,a^{2} b \,x^{2}}{2}+a^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x+1)^2*(a+b*arctanh(c*x))^3,x)

[Out]

1/3/c*a^3+3*x*a^2*b+b^3*arctanh(c*x)^3*x+3*b^3*arctanh(c*x)^2*x+1/3*c^2*x^3*a^3+c*x^2*a^3+1/3/c*b^3*arctanh(c*
x)^3+5/2/c*b^3*arctanh(c*x)^2-6/c*b^3*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-6/c*b^3*dilog(1+I*(c*x+1)/(-c^2*x^
2+1)^(1/2))-1/c*b^3*ln(1+(c*x+1)^2/(-c^2*x^2+1))+2/c*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/c*a*b^2+1/3*c^2*
b^3*arctanh(c*x)^3*x^3+1/2*c*b^3*arctanh(c*x)^2*x^2-6/c*b^3*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-4/
c*b^3*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-6/c*b^3*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+
4/c*b^3*arctanh(c*x)^2*ln(c*x-1)+8/c*a*b^2*arctanh(c*x)*ln(c*x-1)-4/c*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+c^2*a*b^
2*arctanh(c*x)^2*x^3+c^2*a^2*b*arctanh(c*x)*x^3+c*a*b^2*arctanh(c*x)*x^2+3*c*a^2*b*arctanh(c*x)*x^2+b^3*arctan
h(c*x)/c+b^3*x*arctanh(c*x)+x*a*b^2-4*I/c*b^3*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2+4*I/c*b^3
*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2+1/c*a^2*b*arctanh(c*x)+4/c*a^2*b*ln(c*x-1)+1/c*a*b^2*a
rctanh(c*x)^2+7/2/c*a*b^2*ln(c*x-1)+5/2/c*a*b^2*ln(c*x+1)+2/c*a*b^2*ln(c*x-1)^2-4/c*a*b^2*dilog(1/2+1/2*c*x)-4
/c*b^3*arctanh(c*x)^2*ln(2)+c*b^3*arctanh(c*x)^3*x^2+3*a*b^2*arctanh(c*x)^2*x+3*a^2*b*arctanh(c*x)*x+6*a*b^2*a
rctanh(c*x)*x+1/2*c*a^2*b*x^2+3*c*a*b^2*arctanh(c*x)^2*x^2-4*I/c*b^3*Pi*arctanh(c*x)^2+a^3*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a^{3} c^{2} x^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a^{2} b c^{2} + a^{3} c x^{2} + \frac {3}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a^{2} b c + a^{3} x + \frac {3 \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a^{2} b}{2 \, c} - \frac {{\left (b^{3} c^{3} x^{3} + 3 \, b^{3} c^{2} x^{2} + 3 \, b^{3} c x - 7 \, b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \, {\left (2 \, a b^{2} c^{3} x^{3} + {\left (6 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} + 6 \, {\left (a b^{2} c + b^{3} c\right )} x + {\left (b^{3} c^{3} x^{3} + 3 \, b^{3} c^{2} x^{2} + 3 \, b^{3} c x + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{24 \, c} - \int -\frac {{\left (b^{3} c^{3} x^{3} + b^{3} c^{2} x^{2} - b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \, {\left (a b^{2} c^{3} x^{3} + a b^{2} c^{2} x^{2} - a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} - {\left (4 \, a b^{2} c^{3} x^{3} + 2 \, {\left (6 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} + 3 \, {\left (b^{3} c^{3} x^{3} + b^{3} c^{2} x^{2} - b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} + 12 \, {\left (a b^{2} c + b^{3} c\right )} x + 2 \, {\left ({\left (6 \, a b^{2} c^{3} + b^{3} c^{3}\right )} x^{3} - 6 \, a b^{2} + b^{3} + 3 \, {\left (2 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} - 3 \, {\left (2 \, a b^{2} c - b^{3} c\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \, {\left (c x - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/3*a^3*c^2*x^3 + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a^2*b*c^2 + a^3*c*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^2*b*c + a^3*x + 3/2*(2*c*x*arctanh(c*
x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/24*((b^3*c^3*x^3 + 3*b^3*c^2*x^2 + 3*b^3*c*x - 7*b^3)*log(-c*x + 1)^3 - 3*
(2*a*b^2*c^3*x^3 + (6*a*b^2*c^2 + b^3*c^2)*x^2 + 6*(a*b^2*c + b^3*c)*x + (b^3*c^3*x^3 + 3*b^3*c^2*x^2 + 3*b^3*
c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c - integrate(-1/8*((b^3*c^3*x^3 + b^3*c^2*x^2 - b^3*c*x - b^3)*log(
c*x + 1)^3 + 6*(a*b^2*c^3*x^3 + a*b^2*c^2*x^2 - a*b^2*c*x - a*b^2)*log(c*x + 1)^2 - (4*a*b^2*c^3*x^3 + 2*(6*a*
b^2*c^2 + b^3*c^2)*x^2 + 3*(b^3*c^3*x^3 + b^3*c^2*x^2 - b^3*c*x - b^3)*log(c*x + 1)^2 + 12*(a*b^2*c + b^3*c)*x
+ 2*((6*a*b^2*c^3 + b^3*c^3)*x^3 - 6*a*b^2 + b^3 + 3*(2*a*b^2*c^2 + b^3*c^2)*x^2 - 3*(2*a*b^2*c - b^3*c)*x)*l
og(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3*(c*x + 1)^2,x)

[Out]

int((a + b*atanh(c*x))^3*(c*x + 1)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3} \left (c x + 1\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)**2*(a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3*(c*x + 1)**2, x)

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