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

Optimal. Leaf size=385 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}+\frac {b d^2 \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{e^3}-\frac {d^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}+\frac {2 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac {a b x}{c e}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^2 e}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e^3}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c e^2}+\frac {b^2 x \tanh ^{-1}(c x)}{c e} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 5922} \[ \frac {b d^2 \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}+\frac {b^2 d^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 e^3}-\frac {b^2 d^2 \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^3}+\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c e^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac {d^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^3}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}+\frac {2 b d \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac {a b x}{c e}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^2 e}+\frac {b^2 x \tanh ^{-1}(c x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*x)/(c*e) + (b^2*x*ArcTanh[c*x])/(c*e) - (d*(a + b*ArcTanh[c*x])^2)/(c*e^2) - (a + b*ArcTanh[c*x])^2/(2*c^
2*e) - (d*x*(a + b*ArcTanh[c*x])^2)/e^2 + (x^2*(a + b*ArcTanh[c*x])^2)/(2*e) + (2*b*d*(a + b*ArcTanh[c*x])*Log
[2/(1 - c*x)])/(c*e^2) - (d^2*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/e^3 + (d^2*(a + b*ArcTanh[c*x])^2*Log[(
2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e^3 + (b^2*Log[1 - c^2*x^2])/(2*c^2*e) + (b^2*d*PolyLog[2, 1 - 2/(1 - c
*x)])/(c*e^2) + (b*d^2*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e^3 - (b*d^2*(a + b*ArcTanh[c*x])*Pol
yLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e^3 + (b^2*d^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e^3) - (b^2
*d^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e^3)

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 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 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[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 \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx &=\int \left (-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2 (d+e x)}\right ) \, dx\\ &=-\frac {d \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{e^2}+\frac {d^2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx}{e^2}+\frac {\int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{e}\\ &=-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^3}+\frac {(2 b c d) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{e^2}-\frac {(b c) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{e}\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^3}+\frac {(2 b d) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{e^2}+\frac {b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c e}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c e}\\ &=\frac {a b x}{c e}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e^2}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^3}-\frac {\left (2 b^2 d\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{e^2}+\frac {b^2 \int \tanh ^{-1}(c x) \, dx}{c e}\\ &=\frac {a b x}{c e}+\frac {b^2 x \tanh ^{-1}(c x)}{c e}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e^2}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^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 )}{c e^2}-\frac {b^2 \int \frac {x}{1-c^2 x^2} \, dx}{e}\\ &=\frac {a b x}{c e}+\frac {b^2 x \tanh ^{-1}(c x)}{c e}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{c e^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 e}-\frac {d x \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}+\frac {2 b d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e^2}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^3}+\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^2 e}+\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c e^2}+\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^3}-\frac {b d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^3}+\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^3}-\frac {b^2 d^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^3}\\ \end {align*}

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Mathematica [C]  time = 15.76, size = 1072, normalized size = 2.78 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*a^2*d*e*x + 3*a^2*e^2*x^2 + 6*a^2*d^2*Log[d + e*x] + (6*a*b*(c*e^2*x + I*c^2*d^2*Pi*ArcTanh[c*x] - 2*c^2*d
*e*x*ArcTanh[c*x] + e^2*(-1 + c^2*x^2)*ArcTanh[c*x] + 2*c^2*d^2*ArcTanh[(c*d)/e]*ArcTanh[c*x] - c^2*d^2*ArcTan
h[c*x]^2 + c*d*e*ArcTanh[c*x]^2 - (c*d*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*d)/e] - 2*c^2*d^
2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - I*c^2*d^2*Pi*Log[1 + E^(2*ArcTanh[c*x])] + 2*c^2*d^2*ArcTanh[(c*
d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 2*c^2*d^2*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/
e] + ArcTanh[c*x]))] - c*d*e*Log[1 - c^2*x^2] - (I/2)*c^2*d^2*Pi*Log[1 - c^2*x^2] - 2*c^2*d^2*ArcTanh[(c*d)/e]
*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + c^2*d^2*PolyLog[2, -E^(-2*ArcTanh[c*x])] - c^2*d^2*PolyLog[2,
E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/c^2 + (b^2*(6*c*e^2*x*ArcTanh[c*x] + 6*c*d*e*ArcTanh[c*x]^2 - 6*c^
2*d*e*x*ArcTanh[c*x]^2 + 3*e^2*(-1 + c^2*x^2)*ArcTanh[c*x]^2 - 2*c^2*d^2*ArcTanh[c*x]^3 + 2*c*d*e*ArcTanh[c*x]
^3 + 12*c*d*e*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 6*c^2*d^2*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])
] + 3*e^2*Log[1 - c^2*x^2] + 6*c*d*(-e + c*d*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 3*c^2*d^2*PolyLo
g[3, -E^(-2*ArcTanh[c*x])] - (6*c*d*(-(c*d) + e)*(c*d + e)*(-3*c*d*ArcTanh[c*x]^3 + e*ArcTanh[c*x]^3 - (2*Sqrt
[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - (3*I)*c*d*Pi*ArcTanh[c*x]*Log[(E^(-ArcTanh[c*x]) +
E^ArcTanh[c*x])/2] + 3*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 3*c*d*ArcTanh[c*x]^2*
Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)
/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])))] + 3*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*A
rcTanh[c*x])) + c*d*(1 + E^(2*ArcTanh[c*x])))/(2*E^ArcTanh[c*x])] - 3*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/Sqr
t[1 - c^2*x^2]] - ((3*I)/2)*c*d*Pi*ArcTanh[c*x]*Log[1 - c^2*x^2] - 6*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*S
inh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + 6*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] +
6*c*d*ArcTanh[c*x]*PolyLog[2, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + A
rcTanh[c*x])] - 6*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])]))/(3*c^2*d^2 - 3*e^2)))/c^2)/(6*e^3)

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {artanh}\left (c x\right ) + a^{2} x^{2}}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 4.58, size = 1656, normalized size = 4.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*x/c/e+b^2*x*arctanh(c*x)/c/e-1/2*I*b^2/e^3*d^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(((c*x+1)^2/(-
c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2-1/2*I*b^2/e^3*d^2
*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1))))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e
+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2+1/2*I*b^2/e^3*d^2*Pi*csgn(I/(1+(
c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1))))*csgn(I*(((c*x+1)
^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))*arctanh(c*x)^2+1/c^2*b^2*arct
anh(c*x)/e-1/c^2*b^2/e*ln(1+(c*x+1)^2/(-c^2*x^2+1))-1/2/c^2*b^2*arctanh(c*x)^2/e+1/2*b^2*arctanh(c*x)^2*x^2/e+
1/2*b^2*d^2/e^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+a^2*d^2/e^3*ln(c*e*x+c*d)-a^2*d/e^2*x+1/2*a^2*x^2/e+2/c*b^2
/e^2*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))*d+2/c*b^2/e^2*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))*d-1/c*b^2/e^2*a
rctanh(c*x)^2*d-1/2/c^2*a*b/e*ln(c*e*x+e)+1/2/c^2*a*b/e*ln(c*e*x-e)+a*b*arctanh(c*x)*x^2/e-a*b/e^3*d^2*dilog((
c*e*x+e)/(-c*d+e))+a*b/e^3*d^2*dilog((c*e*x-e)/(-c*d-e))-1/2*b^2*d^2/e^2/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(
-c^2*x^2+1)/(-c*d+e))-b^2*d^2/e^3*arctanh(c*x)^2*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)
))+b^2*arctanh(c*x)^2*d^2/e^3*ln(c*e*x+c*d)+1/c*a*b*d/e^2-b^2*d^2/e^3*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*
x^2+1))-b^2*arctanh(c*x)^2*d/e^2*x+c*b^2*d^3/e^3/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-
c*d+e))+c*b^2*d^3/e^3/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+1/2*I*b^2/e^3*d^
2*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*(1+(c*x+1)^2/(-c^2*x^2+1)))/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctan
h(c*x)^2+a*b/e^3*d^2*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))+b^2*d^2/e^2/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x
+1)^2/(-c^2*x^2+1)/(-c*d+e))+b^2*d^2/e^2/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e
))+2*a*b*arctanh(c*x)*d^2/e^3*ln(c*e*x+c*d)-a*b/e^3*d^2*ln(c*e*x+c*d)*ln((c*e*x+e)/(-c*d+e))-1/2*c*b^2*d^3/e^3
/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/c*a*b/e^2*ln(c*e*x+e)*d-1/c*a*b/e^2*ln(c*e*x-e)*
d+2/c*b^2/e^2*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))*arctanh(c*x)*d+2/c*b^2/e^2*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))
*arctanh(c*x)*d-2*a*b*arctanh(c*x)*d/e^2*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 1/8*(b^2*e*x^2 - 2*b^2*d*x)*log(-c*x + 1)^2/e^2 - int
egrate(-1/4*((b^2*c*e^2*x^3 - b^2*e^2*x^2)*log(c*x + 1)^2 + 4*(a*b*c*e^2*x^3 - a*b*e^2*x^2)*log(c*x + 1) + (2*
b^2*c*d^2*x - (4*a*b*c*e^2 + b^2*c*e^2)*x^3 + (b^2*c*d*e + 4*a*b*e^2)*x^2 - 2*(b^2*c*e^2*x^3 - b^2*e^2*x^2)*lo
g(c*x + 1))*log(-c*x + 1))/(c*e^3*x^2 - d*e^2 + (c*d*e^2 - e^3)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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