3.1.69 \(\int x^2 (d+c d x) (a+b \tanh ^{-1}(c x))^2 \, dx\) [69]

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

[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 = 0.39, 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 = {6087, 6037, 6127, 327, 212, 6131, 6055, 2449, 2352, 272, 45, 6021, 266, 6095} \begin {gather*} \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} \end {gather*}

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 45

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 272

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 327

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 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 6087

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 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]

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 ) \text {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 ) \text {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 ) \text {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 = 0.37, size = 234, normalized size = 0.99 \begin {gather*} \frac {d \left (-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 \left (-7+4 c^3 x^3+3 c^4 x^4\right ) \tanh ^{-1}(c x)^2+2 b \tanh ^{-1}(c x) \left (a c^3 x^3 (4+3 c x)+b \left (-2+3 c x+2 c^2 x^2+c^3 x^3\right )-4 b \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+3 a b \log (1-c x)-3 a b \log (1+c x)+4 b^2 \log \left (1-c^2 x^2\right )+4 a b \log \left (-1+c^2 x^2\right )+4 b^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.31, size = 364, normalized size = 1.54

method	result	size
derivativedivides	\(\frac {d \,a^{2} \left (\frac {1}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+\frac {d a b \arctanh \left (c x \right ) c^{4} x^{4}}{2}+\frac {2 d a b \arctanh \left (c x \right ) c^{3} x^{3}}{3}+\frac {7 d a b \ln \left (c x -1\right )}{12}+\frac {d \,b^{2} \arctanh \left (c x \right )^{2} c^{3} x^{3}}{3}+\frac {d \,b^{2} \arctanh \left (c x \right )^{2} c^{4} x^{4}}{4}+\frac {d \,b^{2} \arctanh \left (c x \right ) c^{3} x^{3}}{6}+\frac {d \,b^{2} \arctanh \left (c x \right ) c x}{2}+\frac {d \,b^{2} \arctanh \left (c x \right ) c^{2} x^{2}}{3}+\frac {7 d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{12}+\frac {d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{12}+\frac {d a b \ln \left (c x +1\right )}{12}-\frac {7 d \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{24}+\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{24}-\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{24}+\frac {7 d \,b^{2} \ln \left (c x -1\right )^{2}}{48}-\frac {d \,b^{2} \ln \left (c x +1\right )^{2}}{48}+\frac {d \,b^{2} \ln \left (c x -1\right )}{2}+\frac {d \,b^{2} \ln \left (c x +1\right )}{6}-\frac {d \,b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}+\frac {d \,b^{2} c^{2} x^{2}}{12}+\frac {d \,b^{2} c x}{3}+\frac {d a b \,c^{3} x^{3}}{6}+\frac {d a b \,c^{2} x^{2}}{3}+\frac {d a b c x}{2}}{c^{3}}\)	\(364\)
default	\(\frac {d \,a^{2} \left (\frac {1}{4} c^{4} x^{4}+\frac {1}{3} x^{3} c^{3}\right )+\frac {d a b \arctanh \left (c x \right ) c^{4} x^{4}}{2}+\frac {2 d a b \arctanh \left (c x \right ) c^{3} x^{3}}{3}+\frac {7 d a b \ln \left (c x -1\right )}{12}+\frac {d \,b^{2} \arctanh \left (c x \right )^{2} c^{3} x^{3}}{3}+\frac {d \,b^{2} \arctanh \left (c x \right )^{2} c^{4} x^{4}}{4}+\frac {d \,b^{2} \arctanh \left (c x \right ) c^{3} x^{3}}{6}+\frac {d \,b^{2} \arctanh \left (c x \right ) c x}{2}+\frac {d \,b^{2} \arctanh \left (c x \right ) c^{2} x^{2}}{3}+\frac {7 d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{12}+\frac {d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{12}+\frac {d a b \ln \left (c x +1\right )}{12}-\frac {7 d \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{24}+\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{24}-\frac {d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{24}+\frac {7 d \,b^{2} \ln \left (c x -1\right )^{2}}{48}-\frac {d \,b^{2} \ln \left (c x +1\right )^{2}}{48}+\frac {d \,b^{2} \ln \left (c x -1\right )}{2}+\frac {d \,b^{2} \ln \left (c x +1\right )}{6}-\frac {d \,b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}+\frac {d \,b^{2} c^{2} x^{2}}{12}+\frac {d \,b^{2} c x}{3}+\frac {d a b \,c^{3} x^{3}}{6}+\frac {d a b \,c^{2} x^{2}}{3}+\frac {d a b c x}{2}}{c^{3}}\)	\(364\)
risch	\(-\frac {d c a b \ln \left (-c x +1\right ) 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 {d c \,b^{2} \ln \left (-c x +1\right ) x^{4}}{32}-\frac {7 d \,b^{2} \ln \left (-c x +1\right ) x^{2}}{48 c}-\frac {7 d \,b^{2} \ln \left (-c x +1\right ) x}{24 c^{2}}+\frac {7 d a b \ln \left (-c x +1\right )}{12 c^{3}}+\frac {d c \,b^{2} \ln \left (-c x +1\right )^{2} x^{4}}{16}+\frac {d \,b^{2} \left (3 c^{4} x^{4}+4 x^{3} c^{3}+1\right ) \ln \left (c x +1\right )^{2}}{48 c^{3}}-\frac {d a b \ln \left (-c x +1\right ) x^{3}}{3}-\frac {5 d \,b^{2} \ln \left (-c x +1\right ) \left (-c x +1\right )^{3}}{36 c^{3}}+\frac {5 d \,b^{2} \ln \left (-c x +1\right ) \left (-c x +1\right )^{2}}{24 c^{3}}-\frac {d \,b^{2} \ln \left (-c x +1\right ) \left (-c x +1\right )}{6 c^{3}}+\frac {d \,b^{2} \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{3 c^{3}}-\frac {d \,b^{2} \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{3 c^{3}}+\frac {d b \ln \left (-c x -1\right ) a}{12 c^{3}}+\frac {d \,b^{2} \ln \left (-c x +1\right ) \left (-c x +1\right )^{4}}{32 c^{3}}+\frac {d b a \,x^{2}}{3 c}+\frac {d \,b^{2} \dilog \left (-\frac {c x}{2}+\frac {1}{2}\right )}{3 c^{3}}+\frac {d c \,x^{4} a^{2}}{4}-\frac {7 d \,b^{2} \ln \left (-c x +1\right ) x^{3}}{72}-\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 )^{2} x^{3}}{12}-\frac {7 d \,a^{2}}{12 c^{3}}+\frac {d \,x^{3} a^{2}}{3}+\left (-\frac {d \,b^{2} x^{3} \left (3 c x +4\right ) \ln \left (-c x +1\right )}{24}+\frac {d b \left (6 c^{4} x^{4} a +8 c^{3} x^{3} a +2 b \,c^{3} x^{3}+4 b \,c^{2} x^{2}+6 b c x +7 b \ln \left (-c x +1\right )\right )}{24 c^{3}}\right ) \ln \left (c x +1\right )+\frac {d b a \,x^{3}}{6}+\frac {d \,b^{2} \ln \left (-c x -1\right )}{6 c^{3}}-\frac {d b a}{c^{3}}-\frac {5 d \,b^{2}}{12 c^{3}}+\frac {163 d \,b^{2} \ln \left (-c x +1\right )}{288 c^{3}}\)	\(594\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 402, normalized size = 1.70 \begin {gather*} \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}} \end {gather*}

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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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*(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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