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

Optimal. Leaf size=236 \[ \frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{12 c^3}-\frac {2 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^3}+\frac {a b d x}{2 c^2}+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{6} b d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^3}-\frac {b^2 d \tanh ^{-1}(c x)}{3 c^3}+\frac {b^2 d x}{3 c^2}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^2}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^3}+\frac {b^2 d x^2}{12 c} \]

[Out]

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

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Rubi [A]  time = 0.53, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 14, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5940, 5916, 5980, 321, 206, 5984, 5918, 2402, 2315, 266, 43, 5910, 260, 5948} \[ -\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{3 c^3}+\frac {a b d x}{2 c^2}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{12 c^3}-\frac {2 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^3}+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{6} b d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {b^2 d \log \left (1-c^2 x^2\right )}{3 c^3}+\frac {b^2 d x}{3 c^2}+\frac {b^2 d x \tanh ^{-1}(c x)}{2 c^2}-\frac {b^2 d \tanh ^{-1}(c x)}{3 c^3}+\frac {b^2 d x^2}{12 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*d*x)/(2*c^2) + (b^2*d*x)/(3*c^2) + (b^2*d*x^2)/(12*c) - (b^2*d*ArcTanh[c*x])/(3*c^3) + (b^2*d*x*ArcTanh[c
*x])/(2*c^2) + (b*d*x^2*(a + b*ArcTanh[c*x]))/(3*c) + (b*d*x^3*(a + b*ArcTanh[c*x]))/6 + (d*(a + b*ArcTanh[c*x
])^2)/(12*c^3) + (d*x^3*(a + b*ArcTanh[c*x])^2)/3 + (c*d*x^4*(a + b*ArcTanh[c*x])^2)/4 - (2*b*d*(a + b*ArcTanh
[c*x])*Log[2/(1 - c*x)])/(3*c^3) + (b^2*d*Log[1 - c^2*x^2])/(3*c^3) - (b^2*d*PolyLog[2, 1 - 2/(1 - c*x)])/(3*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 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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]

Rubi steps

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

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Mathematica [A]  time = 0.57, size = 234, normalized size = 0.99 \[ \frac {d \left (3 a^2 c^4 x^4+4 a^2 c^3 x^3+2 a b c^3 x^3+4 a b c^2 x^2+4 a b \log \left (c^2 x^2-1\right )+2 b \tanh ^{-1}(c x) \left (a c^3 x^3 (3 c x+4)+b \left (c^3 x^3+2 c^2 x^2+3 c x-2\right )-4 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )+6 a b c x+3 a b \log (1-c x)-3 a b \log (c x+1)+b^2 c^2 x^2+4 b^2 \log \left (1-c^2 x^2\right )+b^2 \left (3 c^4 x^4+4 c^3 x^3-7\right ) \tanh ^{-1}(c x)^2+4 b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+4 b^2 c x-b^2\right )}{12 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} c d x^{3} + a^{2} d x^{2} + {\left (b^{2} c d x^{3} + b^{2} d x^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c d x^{3} + a b d x^{2}\right )} \operatorname {artanh}\left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 383, normalized size = 1.62 \[ \frac {c d a b \arctanh \left (c x \right ) x^{4}}{2}-\frac {d \,b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{3 c^{3}}+\frac {d \,b^{2} \arctanh \left (c x \right ) x^{3}}{6}+\frac {d \,b^{2} \arctanh \left (c x \right )^{2} x^{3}}{3}+\frac {d a b \,x^{3}}{6}+\frac {7 d \,b^{2} \ln \left (c x -1\right )^{2}}{48 c^{3}}+\frac {d \,b^{2} \ln \left (c x +1\right )}{6 c^{3}}+\frac {d \,b^{2} \ln \left (c x -1\right )}{2 c^{3}}-\frac {d \,b^{2} \ln \left (c x +1\right )^{2}}{48 c^{3}}+\frac {c \,a^{2} d \,x^{4}}{4}+\frac {a b d x}{2 c^{2}}+\frac {b^{2} d x}{3 c^{2}}+\frac {b^{2} d \,x^{2}}{12 c}-\frac {7 d \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{24 c^{3}}+\frac {b^{2} d x \arctanh \left (c x \right )}{2 c^{2}}+\frac {d \,b^{2} \arctanh \left (c x \right ) x^{2}}{3 c}+\frac {2 d a b \arctanh \left (c x \right ) x^{3}}{3}+\frac {7 d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{12 c^{3}}+\frac {d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{12 c^{3}}+\frac {c d \,b^{2} \arctanh \left (c x \right )^{2} x^{4}}{4}+\frac {d a b \,x^{2}}{3 c}+\frac {7 d a b \ln \left (c x -1\right )}{12 c^{3}}+\frac {d a b \ln \left (c x +1\right )}{12 c^{3}}-\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{24 c^{3}}+\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{24 c^{3}}+\frac {a^{2} d \,x^{3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*d*a*b*arctanh(c*x)*x^4+1/6*d*a*b*x^3+7/48/c^3*d*b^2*ln(c*x-1)^2+1/6/c^3*d*b^2*ln(c*x+1)+1/2/c^3*d*b^2*ln
(c*x-1)-1/3/c^3*d*b^2*dilog(1/2+1/2*c*x)+1/6*d*b^2*arctanh(c*x)*x^3+1/3*d*b^2*arctanh(c*x)^2*x^3-1/48/c^3*d*b^
2*ln(c*x+1)^2+1/4*c*a^2*d*x^4+1/2*a*b*d*x/c^2+1/2*b^2*d*x*arctanh(c*x)/c^2+1/3*b^2*d*x/c^2+1/12*b^2*d*x^2/c-7/
24/c^3*d*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/3/c*d*b^2*arctanh(c*x)*x^2+1/3/c*d*a*b*x^2+2/3*d*a*b*arctanh(c*x)*x^3
+7/12/c^3*d*a*b*ln(c*x-1)+1/12/c^3*d*a*b*ln(c*x+1)-1/24/c^3*d*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+7/12/c^3*d*
b^2*arctanh(c*x)*ln(c*x-1)+1/12/c^3*d*b^2*arctanh(c*x)*ln(c*x+1)+1/24/c^3*d*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/4
*c*d*b^2*arctanh(c*x)^2*x^4+1/3*a^2*d*x^3

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maxima [A]  time = 0.66, size = 402, normalized size = 1.70 \[ \frac {1}{4} \, a^{2} c d x^{4} + \frac {1}{3} \, a^{2} d x^{3} + \frac {1}{12} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a b c d + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a b d + \frac {{\left (\log \left (c x + 1\right ) \log \left (-\frac {1}{2} \, c x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c x + \frac {1}{2}\right )\right )} b^{2} d}{3 \, c^{3}} + \frac {b^{2} d \log \left (c x + 1\right )}{6 \, c^{3}} + \frac {b^{2} d \log \left (c x - 1\right )}{2 \, c^{3}} + \frac {4 \, b^{2} c^{2} d x^{2} + 16 \, b^{2} c d x + {\left (3 \, b^{2} c^{4} d x^{4} + 4 \, b^{2} c^{3} d x^{3} + b^{2} d\right )} \log \left (c x + 1\right )^{2} + {\left (3 \, b^{2} c^{4} d x^{4} + 4 \, b^{2} c^{3} d x^{3} - 7 \, b^{2} d\right )} \log \left (-c x + 1\right )^{2} + 4 \, {\left (b^{2} c^{3} d x^{3} + 2 \, b^{2} c^{2} d x^{2} + 3 \, b^{2} c d x\right )} \log \left (c x + 1\right ) - 2 \, {\left (2 \, b^{2} c^{3} d x^{3} + 4 \, b^{2} c^{2} d x^{2} + 6 \, b^{2} c d x + {\left (3 \, b^{2} c^{4} d x^{4} + 4 \, b^{2} c^{3} d x^{3} + b^{2} d\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{48 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*c*d*x^4 + 1/3*a^2*d*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*d + 1/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b*d + 1/3*(log(c
*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*d/c^3 + 1/6*b^2*d*log(c*x + 1)/c^3 + 1/2*b^2*d*log(c*x
- 1)/c^3 + 1/48*(4*b^2*c^2*d*x^2 + 16*b^2*c*d*x + (3*b^2*c^4*d*x^4 + 4*b^2*c^3*d*x^3 + b^2*d)*log(c*x + 1)^2
+ (3*b^2*c^4*d*x^4 + 4*b^2*c^3*d*x^3 - 7*b^2*d)*log(-c*x + 1)^2 + 4*(b^2*c^3*d*x^3 + 2*b^2*c^2*d*x^2 + 3*b^2*c
*d*x)*log(c*x + 1) - 2*(2*b^2*c^3*d*x^3 + 4*b^2*c^2*d*x^2 + 6*b^2*c*d*x + (3*b^2*c^4*d*x^4 + 4*b^2*c^3*d*x^3 +
b^2*d)*log(c*x + 1))*log(-c*x + 1))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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