3.1.96 \(\int \frac {x^2 (a+b \tanh ^{-1}(c x))^2}{d+c d x} \, dx\) [96]

Optimal. Leaf size=247 \[ \frac {a b x}{c^2 d}+\frac {b^2 x \tanh ^{-1}(c x)}{c^2 d}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d}-\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d}+\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^3 d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^3 d}+\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^3 d}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^3 d}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^3 d} \]

[Out]

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

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Rubi [A]
time = 0.39, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {6077, 6037, 6127, 6021, 266, 6095, 6131, 6055, 2449, 2352, 6203, 6745} \begin {gather*} \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d}+\frac {a b x}{c^2 d}-\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3 d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^3 d}+\frac {b^2 x \tanh ^{-1}(c x)}{c^2 d}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6055

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 6077

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d*(f/e), Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d
+ e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 0]

Rule 6095

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 6127

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 6131

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 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6745

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

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Mathematica [A]
time = 0.30, size = 260, normalized size = 1.05 \begin {gather*} \frac {-2 a^2 c x+2 a b c x+a^2 c^2 x^2-2 a b \tanh ^{-1}(c x)-4 a b c x \tanh ^{-1}(c x)+2 b^2 c x \tanh ^{-1}(c x)+2 a b c^2 x^2 \tanh ^{-1}(c x)+b^2 \tanh ^{-1}(c x)^2-2 b^2 c x \tanh ^{-1}(c x)^2+b^2 c^2 x^2 \tanh ^{-1}(c x)^2-4 a b \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+4 b^2 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-2 b^2 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+2 a^2 \log (1+c x)-2 a b \log \left (1-c^2 x^2\right )+b^2 \log \left (1-c^2 x^2\right )+2 b \left (a-b+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+b^2 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )}{2 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 7.85, size = 1099, normalized size = 4.45

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1099\)
default	\(\text {Expression too large to display}\)	\(1099\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x))^2/(c*d*x+d),x,method=_RETURNVERBOSE)

[Out]

1/c^3*(a^2/d*ln(c*x+1)-b^2/d*arctanh(c*x)^2*c*x-a*b/d*dilog(1/2*c*x+1/2)-1/2*a*b/d*ln(c*x+1)^2-1/2*a*b/d*ln(c*
x-1)-3/2*a*b/d*ln(c*x+1)+2*b^2/d*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d*arctanh(c*x)*ln(1+I*(
c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d*arctanh(c*x)^2*ln(c*x+1)-b^2/d*arctanh(c*x)^2*ln(2)-b^2/d*arctanh(c*x)*polylo
g(2,-(c*x+1)^2/(-c^2*x^2+1))-2*b^2/d*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+a*b/d*c*x+b^2/d*arctanh(c*x
)*c*x+1/2*b^2/d*arctanh(c*x)^2*c^2*x^2+2*a*b/d*arctanh(c*x)*ln(c*x+1)-a*b/d*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+a
*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)-b^2/d*ln(1+(c*x+1)^2/(-c^2*x^2+1))+2*b^2/d*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2
))+1/2*b^2/d*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+2*b^2/d*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d*arctanh(c*
x)-3/2*b^2/d*arctanh(c*x)^2+2/3*b^2/d*arctanh(c*x)^3+1/2*I*b^2/d*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*
(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*arctanh(c*x)^2-a^2/d*c*x+1/2*a
^2/d*c^2*x^2-1/2*I*b^2/d*arctanh(c*x)^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+
(c*x+1)^2/(-c^2*x^2+1)))^2+1/2*I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x
^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-I*b^2/d*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1
)^2/(c^2*x^2-1))^2-1/2*I*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*arctanh(c
*x)^2+a*b/d+a*b/d*arctanh(c*x)*c^2*x^2-2*a*b/d*arctanh(c*x)*c*x-1/2*I*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1
+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2-1/2*I*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*((c*x^2 - 2*x)/(c^2*d) + 2*log(c*x + 1)/(c^3*d)) + 1/8*(b^2*c^2*x^2 - 2*b^2*c*x + 2*b^2*log(c*x + 1))*
log(-c*x + 1)^2/(c^3*d) - integrate(-1/4*((b^2*c^3*x^3 - b^2*c^2*x^2)*log(c*x + 1)^2 + 4*(a*b*c^3*x^3 - a*b*c^
2*x^2)*log(c*x + 1) + (2*b^2*c*x - (4*a*b*c^3 + b^2*c^3)*x^3 + (4*a*b*c^2 + b^2*c^2)*x^2 - 2*(b^2*c^3*x^3 - b^
2*c^2*x^2 + b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1))/(c^4*d*x^2 - c^2*d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} x^{2}}{c x + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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