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

Optimal. Leaf size=206 \[ \frac {1}{6} b c^2 d^3 x^3 \left (a+b \tanh ^{-1}(c x)\right )+b c d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac {4 b d^3 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac {7}{2} a b d^3 x+\frac {11 b^2 d^3 \log \left (1-c^2 x^2\right )}{6 c}-\frac {2 b^2 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c}+\frac {1}{12} b^2 c d^3 x^2-\frac {b^2 d^3 \tanh ^{-1}(c x)}{c}+\frac {7}{2} b^2 d^3 x \tanh ^{-1}(c x)+b^2 d^3 x \]

[Out]

7/2*a*b*d^3*x+b^2*d^3*x+1/12*b^2*c*d^3*x^2-b^2*d^3*arctanh(c*x)/c+7/2*b^2*d^3*x*arctanh(c*x)+b*c*d^3*x^2*(a+b*
arctanh(c*x))+1/6*b*c^2*d^3*x^3*(a+b*arctanh(c*x))+1/4*d^3*(c*x+1)^4*(a+b*arctanh(c*x))^2/c-4*b*d^3*(a+b*arcta
nh(c*x))*ln(2/(-c*x+1))/c+11/6*b^2*d^3*ln(-c^2*x^2+1)/c-2*b^2*d^3*polylog(2,1-2/(-c*x+1))/c

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Rubi [A]  time = 0.21, antiderivative size = 206, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.88, size = 293, normalized size = 1.42 \[ \frac {d^3 \left (3 a^2 c^4 x^4+12 a^2 c^3 x^3+18 a^2 c^2 x^2+12 a^2 c x+2 a b c^3 x^3+12 a b c^2 x^2+12 a b \log \left (1-c^2 x^2\right )+12 a b \log \left (c^2 x^2-1\right )+2 b \tanh ^{-1}(c x) \left (3 a c x \left (c^3 x^3+4 c^2 x^2+6 c x+4\right )+b \left (c^3 x^3+6 c^2 x^2+21 c x-6\right )-24 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+42 a b c x+21 a b \log (1-c x)-21 a b \log (c x+1)+b^2 c^2 x^2+22 b^2 \log \left (1-c^2 x^2\right )+3 b^2 \left (c^4 x^4+4 c^3 x^3+6 c^2 x^2+4 c x-15\right ) \tanh ^{-1}(c x)^2+24 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+12 b^2 c x-b^2\right )}{12 c} \]

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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} c^{3} d^{3} x^{3} + 3 \, a^{2} c^{2} d^{3} x^{2} + 3 \, a^{2} c d^{3} x + a^{2} d^{3} + {\left (b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} + 3 \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} + 3 \, a b c d^{3} x + a b d^{3}\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

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

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)

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maple [B]  time = 0.06, size = 462, normalized size = 2.24 \[ b^{2} d^{3} x +\frac {c^{3} d^{3} a b \arctanh \left (c x \right ) x^{4}}{2}+d^{3} b^{2} \arctanh \left (c x \right )^{2} x +\frac {d^{3} b^{2} \arctanh \left (c x \right )^{2}}{4 c}-\frac {2 d^{3} b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{c}+\frac {7 b^{2} d^{3} x \arctanh \left (c x \right )}{2}+\frac {3 c \,d^{3} a^{2} x^{2}}{2}+c^{2} d^{3} a^{2} x^{3}+\frac {d^{3} b^{2} \ln \left (c x -1\right )^{2}}{c}+\frac {c^{3} d^{3} a^{2} x^{4}}{4}+\frac {7 d^{3} b^{2} \ln \left (c x -1\right )}{3 c}+\frac {4 d^{3} b^{2} \ln \left (c x +1\right )}{3 c}+3 c \,d^{3} a b \arctanh \left (c x \right ) x^{2}+2 c^{2} d^{3} a b \arctanh \left (c x \right ) x^{3}-\frac {13 d^{3} b^{2}}{12 c}+\frac {d^{3} a^{2}}{4 c}+\frac {4 d^{3} a b \ln \left (c x -1\right )}{c}+\frac {4 d^{3} b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{c}+\frac {d^{3} a b \arctanh \left (c x \right )}{2 c}+c^{2} d^{3} b^{2} \arctanh \left (c x \right )^{2} x^{3}+2 d^{3} a b \arctanh \left (c x \right ) x +\frac {c^{2} d^{3} b^{2} \arctanh \left (c x \right ) x^{3}}{6}+c \,d^{3} b^{2} \arctanh \left (c x \right ) x^{2}+\frac {c^{3} d^{3} b^{2} \arctanh \left (c x \right )^{2} x^{4}}{4}+\frac {3 c \,d^{3} b^{2} \arctanh \left (c x \right )^{2} x^{2}}{2}+a^{2} x \,d^{3}+\frac {c^{2} d^{3} a b \,x^{3}}{6}+c \,d^{3} a b \,x^{2}-\frac {2 d^{3} b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{c}+\frac {7 a b \,d^{3} x}{2}+\frac {b^{2} c \,d^{3} x^{2}}{12} \]

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

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maxima [B]  time = 0.56, size = 627, normalized size = 3.04 \[ \frac {1}{4} \, a^{2} c^{3} d^{3} x^{4} + a^{2} c^{2} d^{3} x^{3} + \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 c^{3} d^{3} + {\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 c^{2} d^{3} + \frac {3}{2} \, a^{2} c d^{3} 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 c d^{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 {2 \, {\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} d^{3}}{c} + \frac {4 \, b^{2} d^{3} \log \left (c x + 1\right )}{3 \, c} + \frac {7 \, b^{2} d^{3} \log \left (c x - 1\right )}{3 \, c} + \frac {4 \, b^{2} c^{2} d^{3} x^{2} + 48 \, b^{2} c d^{3} x + 3 \, {\left (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}\right )} \log \left (c x + 1\right )^{2} + 3 \, {\left (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}\right )} \log \left (-c x + 1\right )^{2} + 4 \, {\left (b^{2} c^{3} d^{3} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 21 \, b^{2} c d^{3} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (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 \, {\left (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}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{48 \, c} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{3} \left (\int a^{2}\, dx + \int b^{2} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int 2 a b \operatorname {atanh}{\left (c x \right )}\, dx + \int 3 a^{2} c x\, dx + \int 3 a^{2} c^{2} x^{2}\, dx + \int a^{2} c^{3} x^{3}\, dx + \int 3 b^{2} c x \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int 3 b^{2} c^{2} x^{2} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int b^{2} c^{3} x^{3} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int 6 a b c x \operatorname {atanh}{\left (c x \right )}\, dx + \int 6 a b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}\, dx + \int 2 a b c^{3} x^{3} \operatorname {atanh}{\left (c x \right )}\, dx\right ) \]

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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