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

Optimal. Leaf size=206 \[ -\frac{2 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{4 b d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{7}{2} a b d^3 x+\frac{11 b^2 d^3 \log \left (1-c^2 x^2\right )}{6 c}+\frac{1}{12} b^2 c d^3 x^2-\frac{b^2 d^3 \tanh ^{-1}(c x)}{c}+\frac{7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b^2 d^3 x \]

[Out]

(7*a*b*d^3*x)/2 + b^2*d^3*x + (b^2*c*d^3*x^2)/12 - (b^2*d^3*ArcTanh[c*x])/c + (7*b^2*d^3*x*ArcTanh[c*x])/2 + b
*c*d^3*x^2*(a + b*ArcTanh[c*x]) + (b*c^2*d^3*x^3*(a + b*ArcTanh[c*x]))/6 + (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x
])^2)/(4*c) - (4*b*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c + (11*b^2*d^3*Log[1 - c^2*x^2])/(6*c) - (2*b^2
*d^3*PolyLog[2, 1 - 2/(1 - c*x)])/c

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Rubi [A]  time = 0.214082, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.632, Rules used = {5928, 5910, 260, 5916, 321, 206, 266, 43, 1586, 5918, 2402, 2315} \[ -\frac{2 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{4 b d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{7}{2} a b d^3 x+\frac{11 b^2 d^3 \log \left (1-c^2 x^2\right )}{6 c}+\frac{1}{12} b^2 c d^3 x^2-\frac{b^2 d^3 \tanh ^{-1}(c x)}{c}+\frac{7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b^2 d^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a*b*d^3*x)/2 + b^2*d^3*x + (b^2*c*d^3*x^2)/12 - (b^2*d^3*ArcTanh[c*x])/c + (7*b^2*d^3*x*ArcTanh[c*x])/2 + b
*c*d^3*x^2*(a + b*ArcTanh[c*x]) + (b*c^2*d^3*x^3*(a + b*ArcTanh[c*x]))/6 + (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x
])^2)/(4*c) - (4*b*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/c + (11*b^2*d^3*Log[1 - c^2*x^2])/(6*c) - (2*b^2
*d^3*PolyLog[2, 1 - 2/(1 - c*x)])/c

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 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 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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int (d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{b \int \left (-7 d^4 \left (a+b \tanh ^{-1}(c x)\right )-4 c d^4 x \left (a+b \tanh ^{-1}(c x)\right )-c^2 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{8 \left (d^4+c d^4 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2}\right ) \, dx}{2 d}\\ &=\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{(4 b) \int \frac{\left (d^4+c d^4 x\right ) \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{d}+\frac{1}{2} \left (7 b d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\left (2 b c d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx+\frac{1}{2} \left (b c^2 d^3\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=\frac{7}{2} a b d^3 x+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{(4 b) \int \frac{a+b \tanh ^{-1}(c x)}{\frac{1}{d^4}-\frac{c x}{d^4}} \, dx}{d}+\frac{1}{2} \left (7 b^2 d^3\right ) \int \tanh ^{-1}(c x) \, dx-\left (b^2 c^2 d^3\right ) \int \frac{x^2}{1-c^2 x^2} \, dx-\frac{1}{6} \left (b^2 c^3 d^3\right ) \int \frac{x^3}{1-c^2 x^2} \, dx\\ &=\frac{7}{2} a b d^3 x+b^2 d^3 x+\frac{7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{4 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\left (b^2 d^3\right ) \int \frac{1}{1-c^2 x^2} \, dx+\left (4 b^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx-\frac{1}{2} \left (7 b^2 c d^3\right ) \int \frac{x}{1-c^2 x^2} \, dx-\frac{1}{12} \left (b^2 c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )\\ &=\frac{7}{2} a b d^3 x+b^2 d^3 x-\frac{b^2 d^3 \tanh ^{-1}(c x)}{c}+\frac{7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{4 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}+\frac{7 b^2 d^3 \log \left (1-c^2 x^2\right )}{4 c}-\frac{\left (4 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c}-\frac{1}{12} \left (b^2 c^3 d^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{7}{2} a b d^3 x+b^2 d^3 x+\frac{1}{12} b^2 c d^3 x^2-\frac{b^2 d^3 \tanh ^{-1}(c x)}{c}+\frac{7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac{4 b d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}+\frac{11 b^2 d^3 \log \left (1-c^2 x^2\right )}{6 c}-\frac{2 b^2 d^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}\\ \end{align*}

Mathematica [A]  time = 0.862084, size = 293, normalized size = 1.42 \[ \frac{d^3 \left (24 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+3 a^2 c^4 x^4+12 a^2 c^3 x^3+18 a^2 c^2 x^2+12 a^2 c x+2 a b c^3 x^3+12 a b c^2 x^2+12 a b \log \left (1-c^2 x^2\right )+12 a b \log \left (c^2 x^2-1\right )+2 b \tanh ^{-1}(c x) \left (3 a c x \left (c^3 x^3+4 c^2 x^2+6 c x+4\right )+b \left (c^3 x^3+6 c^2 x^2+21 c x-6\right )-24 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+42 a b c x+21 a b \log (1-c x)-21 a b \log (c x+1)+b^2 c^2 x^2+22 b^2 \log \left (1-c^2 x^2\right )+3 b^2 \left (c^4 x^4+4 c^3 x^3+6 c^2 x^2+4 c x-15\right ) \tanh ^{-1}(c x)^2+12 b^2 c x-b^2\right )}{12 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-b^2 + 12*a^2*c*x + 42*a*b*c*x + 12*b^2*c*x + 18*a^2*c^2*x^2 + 12*a*b*c^2*x^2 + b^2*c^2*x^2 + 12*a^2*c^3
*x^3 + 2*a*b*c^3*x^3 + 3*a^2*c^4*x^4 + 3*b^2*(-15 + 4*c*x + 6*c^2*x^2 + 4*c^3*x^3 + c^4*x^4)*ArcTanh[c*x]^2 +
2*b*ArcTanh[c*x]*(3*a*c*x*(4 + 6*c*x + 4*c^2*x^2 + c^3*x^3) + b*(-6 + 21*c*x + 6*c^2*x^2 + c^3*x^3) - 24*b*Log
[1 + E^(-2*ArcTanh[c*x])]) + 21*a*b*Log[1 - c*x] - 21*a*b*Log[1 + c*x] + 12*a*b*Log[1 - c^2*x^2] + 22*b^2*Log[
1 - c^2*x^2] + 12*a*b*Log[-1 + c^2*x^2] + 24*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])]))/(12*c)

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Maple [B]  time = 0.05, size = 462, normalized size = 2.2 \begin{align*}{b}^{2}{d}^{3}x+3\,c{d}^{3}ab{\it Artanh} \left ( cx \right ){x}^{2}+2\,{d}^{3}ab{\it Artanh} \left ( cx \right ) x+{\frac{7\,ab{d}^{3}x}{2}}+{\frac{{b}^{2}c{d}^{3}{x}^{2}}{12}}+{\frac{{c}^{2}{d}^{3}ab{x}^{3}}{6}}+{\frac{{c}^{3}{d}^{3}ab{\it Artanh} \left ( cx \right ){x}^{4}}{2}}+2\,{c}^{2}{d}^{3}ab{\it Artanh} \left ( cx \right ){x}^{3}-{\frac{13\,{d}^{3}{b}^{2}}{12\,c}}+{\frac{{d}^{3}{a}^{2}}{4\,c}}+x{a}^{2}{d}^{3}+{\frac{{d}^{3}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{4\,c}}+{d}^{3}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}x-2\,{\frac{{d}^{3}{b}^{2}{\it dilog} \left ( 1/2+1/2\,cx \right ) }{c}}+{\frac{7\,{d}^{3}{b}^{2}\ln \left ( cx-1 \right ) }{3\,c}}+{\frac{{d}^{3}{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{c}}+{\frac{4\,{d}^{3}{b}^{2}\ln \left ( cx+1 \right ) }{3\,c}}+{\frac{7\,{b}^{2}{d}^{3}x{\it Artanh} \left ( cx \right ) }{2}}+{c}^{2}{x}^{3}{a}^{2}{d}^{3}+{\frac{3\,c{x}^{2}{a}^{2}{d}^{3}}{2}}+{\frac{{c}^{3}{x}^{4}{a}^{2}{d}^{3}}{4}}-2\,{\frac{{d}^{3}{b}^{2}\ln \left ( cx-1 \right ) \ln \left ( 1/2+1/2\,cx \right ) }{c}}+{\frac{3\,c{d}^{3}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}{x}^{2}}{2}}+{\frac{{d}^{3}ab{\it Artanh} \left ( cx \right ) }{2\,c}}+{c}^{2}{d}^{3}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}{x}^{3}+{\frac{{c}^{3}{d}^{3}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}{x}^{4}}{4}}+c{d}^{3}{b}^{2}{\it Artanh} \left ( cx \right ){x}^{2}+{\frac{{c}^{2}{d}^{3}{b}^{2}{\it Artanh} \left ( cx \right ){x}^{3}}{6}}+4\,{\frac{{d}^{3}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{c}}+c{d}^{3}ab{x}^{2}+4\,{\frac{{d}^{3}ab\ln \left ( cx-1 \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*d^3*x+3*c*d^3*a*b*arctanh(c*x)*x^2+2*c^2*d^3*a*b*arctanh(c*x)*x^3+1/2*c^3*d^3*a*b*arctanh(c*x)*x^4+7/2*a*b
*d^3*x+1/12*b^2*c*d^3*x^2+7/2*b^2*d^3*x*arctanh(c*x)+1/6*c^2*d^3*a*b*x^3-13/12/c*d^3*b^2+1/4/c*d^3*a^2+x*a^2*d
^3-2/c*d^3*b^2*dilog(1/2+1/2*c*x)+7/3/c*d^3*b^2*ln(c*x-1)+1/c*d^3*b^2*ln(c*x-1)^2+1/4/c*d^3*b^2*arctanh(c*x)^2
+d^3*b^2*arctanh(c*x)^2*x+4/3/c*d^3*b^2*ln(c*x+1)+c^2*x^3*a^2*d^3+3/2*c*x^2*a^2*d^3+1/4*c^3*x^4*a^2*d^3-2/c*d^
3*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+c*d^3*a*b*x^2+2*d^3*a*b*arctanh(c*x)*x+3/2*c*d^3*b^2*arctanh(c*x)^2*x^2+1/2/c*
d^3*a*b*arctanh(c*x)+4/c*d^3*a*b*ln(c*x-1)+c^2*d^3*b^2*arctanh(c*x)^2*x^3+1/4*c^3*d^3*b^2*arctanh(c*x)^2*x^4+c
*d^3*b^2*arctanh(c*x)*x^2+1/6*c^2*d^3*b^2*arctanh(c*x)*x^3+4/c*d^3*b^2*arctanh(c*x)*ln(c*x-1)

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Maxima [B]  time = 1.76048, size = 846, normalized size = 4.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*c^3*d^3*x^4 + a^2*c^2*d^3*x^3 + 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*c^3*d^3 + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*c^2*d^3
+ 3/2*a^2*c*d^3*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*c*d^3 +
a^2*d^3*x + (2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*d^3/c + 2*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(
1/2*c*x + 1/2))*b^2*d^3/c + 4/3*b^2*d^3*log(c*x + 1)/c + 7/3*b^2*d^3*log(c*x - 1)/c + 1/48*(4*b^2*c^2*d^3*x^2
+ 48*b^2*c*d^3*x + 3*(b^2*c^4*d^3*x^4 + 4*b^2*c^3*d^3*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c*d^3*x + b^2*d^3)*log(c
*x + 1)^2 + 3*(b^2*c^4*d^3*x^4 + 4*b^2*c^3*d^3*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c*d^3*x - 15*b^2*d^3)*log(-c*x
+ 1)^2 + 4*(b^2*c^3*d^3*x^3 + 6*b^2*c^2*d^3*x^2 + 21*b^2*c*d^3*x)*log(c*x + 1) - 2*(2*b^2*c^3*d^3*x^3 + 12*b^2
*c^2*d^3*x^2 + 42*b^2*c*d^3*x + 3*(b^2*c^4*d^3*x^4 + 4*b^2*c^3*d^3*x^3 + 6*b^2*c^2*d^3*x^2 + 4*b^2*c*d^3*x + b
^2*d^3)*log(c*x + 1))*log(-c*x + 1))/c

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{2} c^{3} d^{3} x^{3} + 3 \, a^{2} c^{2} d^{3} x^{2} + 3 \, a^{2} c d^{3} x + a^{2} d^{3} +{\left (b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} + 3 \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \operatorname{artanh}\left (c x\right )^{2} + 2 \,{\left (a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} + 3 \, a b c d^{3} x + a b d^{3}\right )} \operatorname{artanh}\left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int a^{2}\, dx + \int b^{2} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b \operatorname{atanh}{\left (c x \right )}\, dx + \int 3 a^{2} c x\, dx + \int 3 a^{2} c^{2} x^{2}\, dx + \int a^{2} c^{3} x^{3}\, dx + \int 3 b^{2} c x \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 3 b^{2} c^{2} x^{2} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int b^{2} c^{3} x^{3} \operatorname{atanh}^{2}{\left (c x \right )}\, dx + \int 6 a b c x \operatorname{atanh}{\left (c x \right )}\, dx + \int 6 a b c^{2} x^{2} \operatorname{atanh}{\left (c x \right )}\, dx + \int 2 a b c^{3} x^{3} \operatorname{atanh}{\left (c x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(a**2, x) + Integral(b**2*atanh(c*x)**2, x) + Integral(2*a*b*atanh(c*x), x) + Integral(3*a**2*c*
x, x) + Integral(3*a**2*c**2*x**2, x) + Integral(a**2*c**3*x**3, x) + Integral(3*b**2*c*x*atanh(c*x)**2, x) +
Integral(3*b**2*c**2*x**2*atanh(c*x)**2, x) + Integral(b**2*c**3*x**3*atanh(c*x)**2, x) + Integral(6*a*b*c*x*a
tanh(c*x), x) + Integral(6*a*b*c**2*x**2*atanh(c*x), x) + Integral(2*a*b*c**3*x**3*atanh(c*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c d x + d\right )}^{3}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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