3.1.64 \(\int x^7 (a+b \tanh ^{-1}(c x^2))^2 \, dx\) [64]

Optimal. Leaf size=125 \[ \frac {a b x^2}{4 c^3}+\frac {b^2 x^4}{24 c^2}+\frac {b^2 x^2 \tanh ^{-1}\left (c x^2\right )}{4 c^3}+\frac {b x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{12 c}-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{8 c^4}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^4\right )}{6 c^4} \]

[Out]

1/4*a*b*x^2/c^3+1/24*b^2*x^4/c^2+1/4*b^2*x^2*arctanh(c*x^2)/c^3+1/12*b*x^6*(a+b*arctanh(c*x^2))/c-1/8*(a+b*arc
tanh(c*x^2))^2/c^4+1/8*x^8*(a+b*arctanh(c*x^2))^2+1/6*b^2*ln(-c^2*x^4+1)/c^4

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Rubi [A]
time = 0.19, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6039, 6037, 6127, 272, 45, 6021, 266, 6095} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{8 c^4}+\frac {a b x^2}{4 c^3}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {b x^6 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{12 c}+\frac {b^2 x^2 \tanh ^{-1}\left (c x^2\right )}{4 c^3}+\frac {b^2 x^4}{24 c^2}+\frac {b^2 \log \left (1-c^2 x^4\right )}{6 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTanh[c*x^2])^2,x]

[Out]

(a*b*x^2)/(4*c^3) + (b^2*x^4)/(24*c^2) + (b^2*x^2*ArcTanh[c*x^2])/(4*c^3) + (b*x^6*(a + b*ArcTanh[c*x^2]))/(12
*c) - (a + b*ArcTanh[c*x^2])^2/(8*c^4) + (x^8*(a + b*ArcTanh[c*x^2])^2)/8 + (b^2*Log[1 - c^2*x^4])/(6*c^4)

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 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 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 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

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]

Rubi steps

\begin {align*} \int x^7 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^7 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{2} b x^7 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{4} b^2 x^7 \log ^2\left (1+c x^2\right )\right ) \, dx\\ &=\frac {1}{4} \int x^7 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \, dx-\frac {1}{2} b \int x^7 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right ) \, dx+\frac {1}{4} b^2 \int x^7 \log ^2\left (1+c x^2\right ) \, dx\\ &=\frac {1}{8} \text {Subst}\left (\int x^3 (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )-\frac {1}{4} b \text {Subst}\left (\int x^3 (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int x^3 \log ^2(1+c x) \, dx,x,x^2\right )\\ &=\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )-\frac {1}{16} (b c) \text {Subst}\left (\int \frac {x^4 (2 a-b \log (1-c x))}{1-c x} \, dx,x,x^2\right )+\frac {1}{16} (b c) \text {Subst}\left (\int \frac {x^4 (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^4 \log (1+c x)}{1-c x} \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^4 \log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )+\frac {1}{16} b \text {Subst}\left (\int \frac {\left (\frac {1}{c}-\frac {x}{c}\right )^4 (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )+\frac {1}{16} (b c) \text {Subst}\left (\int \left (-\frac {-2 a+b \log (1-c x)}{c^4}+\frac {x (-2 a+b \log (1-c x))}{c^3}-\frac {x^2 (-2 a+b \log (1-c x))}{c^2}+\frac {x^3 (-2 a+b \log (1-c x))}{c}+\frac {-2 a+b \log (1-c x)}{c^4 (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c^4}-\frac {x \log (1+c x)}{c^3}-\frac {x^2 \log (1+c x)}{c^2}-\frac {x^3 \log (1+c x)}{c}-\frac {\log (1+c x)}{c^4 (-1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c^4}+\frac {x \log (1+c x)}{c^3}-\frac {x^2 \log (1+c x)}{c^2}+\frac {x^3 \log (1+c x)}{c}+\frac {\log (1+c x)}{c^4 (1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{192} b \left (2 a-b \log \left (1-c x^2\right )\right ) \left (\frac {48 \left (1-c x^2\right )}{c^4}-\frac {36 \left (1-c x^2\right )^2}{c^4}+\frac {16 \left (1-c x^2\right )^3}{c^4}-\frac {3 \left (1-c x^2\right )^4}{c^4}-\frac {12 \log \left (1-c x^2\right )}{c^4}\right )+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )+\frac {1}{16} b \text {Subst}\left (\int x^3 (-2 a+b \log (1-c x)) \, dx,x,x^2\right )+\frac {1}{16} b^2 \text {Subst}\left (\int \frac {x \left (-48+36 x-16 x^2+3 x^3\right )+12 \log (x)}{12 c^4 x} \, dx,x,1-c x^2\right )-\frac {b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^2\right )}{16 c^3}+\frac {b \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )}{16 c^3}+2 \frac {b^2 \text {Subst}\left (\int \log (1+c x) \, dx,x,x^2\right )}{16 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )}{16 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{1+c x} \, dx,x,x^2\right )}{16 c^3}+\frac {b \text {Subst}\left (\int x (-2 a+b \log (1-c x)) \, dx,x,x^2\right )}{16 c^2}-\frac {b \text {Subst}\left (\int x^2 (-2 a+b \log (1-c x)) \, dx,x,x^2\right )}{16 c}+2 \frac {b^2 \text {Subst}\left (\int x^2 \log (1+c x) \, dx,x,x^2\right )}{16 c}\\ &=\frac {a b x^2}{8 c^3}-\frac {b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )}{32 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^2\right )\right )}{48 c}-\frac {1}{64} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{192} b \left (2 a-b \log \left (1-c x^2\right )\right ) \left (\frac {48 \left (1-c x^2\right )}{c^4}-\frac {36 \left (1-c x^2\right )^2}{c^4}+\frac {16 \left (1-c x^2\right )^3}{c^4}-\frac {3 \left (1-c x^2\right )^4}{c^4}-\frac {12 \log \left (1-c x^2\right )}{c^4}\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^4}+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )-\frac {1}{48} b^2 \text {Subst}\left (\int \frac {x^3}{1-c x} \, dx,x,x^2\right )+2 \left (\frac {b^2 x^6 \log \left (1+c x^2\right )}{48 c}-\frac {1}{48} b^2 \text {Subst}\left (\int \frac {x^3}{1+c x} \, dx,x,x^2\right )\right )+\frac {b^2 \text {Subst}\left (\int \frac {x \left (-48+36 x-16 x^2+3 x^3\right )+12 \log (x)}{x} \, dx,x,1-c x^2\right )}{192 c^4}+2 \frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{16 c^4}-\frac {b^2 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c x^2\right )}{16 c^4}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )}{16 c^3}-\frac {b^2 \text {Subst}\left (\int \log (1-c x) \, dx,x,x^2\right )}{16 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )}{16 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {x^2}{1-c x} \, dx,x,x^2\right )}{32 c}+\frac {1}{64} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^4}{1-c x} \, dx,x,x^2\right )\\ &=\frac {a b x^2}{8 c^3}-\frac {b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )}{32 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^2\right )\right )}{48 c}-\frac {1}{64} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{192} b \left (2 a-b \log \left (1-c x^2\right )\right ) \left (\frac {48 \left (1-c x^2\right )}{c^4}-\frac {36 \left (1-c x^2\right )^2}{c^4}+\frac {16 \left (1-c x^2\right )^3}{c^4}-\frac {3 \left (1-c x^2\right )^4}{c^4}-\frac {12 \log \left (1-c x^2\right )}{c^4}\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^4}+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \log ^2\left (1+c x^2\right )}{32 c^4}+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )+2 \left (-\frac {b^2 x^2}{16 c^3}+\frac {b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^4}\right )-\frac {1}{48} b^2 \text {Subst}\left (\int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx,x,x^2\right )+2 \left (\frac {b^2 x^6 \log \left (1+c x^2\right )}{48 c}-\frac {1}{48} b^2 \text {Subst}\left (\int \left (\frac {1}{c^3}-\frac {x}{c^2}+\frac {x^2}{c}-\frac {1}{c^3 (1+c x)}\right ) \, dx,x,x^2\right )\right )+\frac {b^2 \text {Subst}\left (\int \left (-48+36 x-16 x^2+3 x^3+\frac {12 \log (x)}{x}\right ) \, dx,x,1-c x^2\right )}{192 c^4}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{16 c^4}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{16 c^4}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{16 c^4}+\frac {b^2 \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )}{32 c}+\frac {1}{64} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {1}{c^4}-\frac {x}{c^3}-\frac {x^2}{c^2}-\frac {x^3}{c}-\frac {1}{c^4 (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {a b x^2}{8 c^3}+\frac {55 b^2 x^2}{192 c^3}-\frac {5 b^2 x^4}{384 c^2}+\frac {b^2 x^6}{576 c}-\frac {b^2 x^8}{256}+\frac {3 b^2 \left (1-c x^2\right )^2}{32 c^4}-\frac {b^2 \left (1-c x^2\right )^3}{36 c^4}+\frac {b^2 \left (1-c x^2\right )^4}{256 c^4}-\frac {5 b^2 \log \left (1-c x^2\right )}{192 c^4}+\frac {b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{16 c^4}-\frac {b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )}{32 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^2\right )\right )}{48 c}-\frac {1}{64} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{192} b \left (2 a-b \log \left (1-c x^2\right )\right ) \left (\frac {48 \left (1-c x^2\right )}{c^4}-\frac {36 \left (1-c x^2\right )^2}{c^4}+\frac {16 \left (1-c x^2\right )^3}{c^4}-\frac {3 \left (1-c x^2\right )^4}{c^4}-\frac {12 \log \left (1-c x^2\right )}{c^4}\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^4}+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \log ^2\left (1+c x^2\right )}{32 c^4}+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )+2 \left (-\frac {b^2 x^2}{48 c^3}+\frac {b^2 x^4}{96 c^2}-\frac {b^2 x^6}{144 c}+\frac {b^2 \log \left (1+c x^2\right )}{48 c^4}+\frac {b^2 x^6 \log \left (1+c x^2\right )}{48 c}\right )+2 \left (-\frac {b^2 x^2}{16 c^3}+\frac {b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^4}\right )+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{16 c^4}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}+\frac {b^2 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x^2\right )}{16 c^4}\\ &=\frac {a b x^2}{8 c^3}+\frac {55 b^2 x^2}{192 c^3}-\frac {5 b^2 x^4}{384 c^2}+\frac {b^2 x^6}{576 c}-\frac {b^2 x^8}{256}+\frac {3 b^2 \left (1-c x^2\right )^2}{32 c^4}-\frac {b^2 \left (1-c x^2\right )^3}{36 c^4}+\frac {b^2 \left (1-c x^2\right )^4}{256 c^4}-\frac {5 b^2 \log \left (1-c x^2\right )}{192 c^4}+\frac {b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{16 c^4}+\frac {b^2 \log ^2\left (1-c x^2\right )}{32 c^4}-\frac {b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )}{32 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^2\right )\right )}{48 c}-\frac {1}{64} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac {1}{32} x^8 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{192} b \left (2 a-b \log \left (1-c x^2\right )\right ) \left (\frac {48 \left (1-c x^2\right )}{c^4}-\frac {36 \left (1-c x^2\right )^2}{c^4}+\frac {16 \left (1-c x^2\right )^3}{c^4}-\frac {3 \left (1-c x^2\right )^4}{c^4}-\frac {12 \log \left (1-c x^2\right )}{c^4}\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{16 c^4}+\frac {1}{16} b x^8 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b^2 \log ^2\left (1+c x^2\right )}{32 c^4}+\frac {1}{32} b^2 x^8 \log ^2\left (1+c x^2\right )+2 \left (-\frac {b^2 x^2}{48 c^3}+\frac {b^2 x^4}{96 c^2}-\frac {b^2 x^6}{144 c}+\frac {b^2 \log \left (1+c x^2\right )}{48 c^4}+\frac {b^2 x^6 \log \left (1+c x^2\right )}{48 c}\right )+2 \left (-\frac {b^2 x^2}{16 c^3}+\frac {b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{16 c^4}\right )+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{16 c^4}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 146, normalized size = 1.17 \begin {gather*} \frac {6 a b c x^2+b^2 c^2 x^4+2 a b c^3 x^6+3 a^2 c^4 x^8+2 b c x^2 \left (3 a c^3 x^6+b \left (3+c^2 x^4\right )\right ) \tanh ^{-1}\left (c x^2\right )+3 b^2 \left (-1+c^4 x^8\right ) \tanh ^{-1}\left (c x^2\right )^2+b (3 a+4 b) \log \left (1-c x^2\right )-3 a b \log \left (1+c x^2\right )+4 b^2 \log \left (1+c x^2\right )}{24 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*ArcTanh[c*x^2])^2,x]

[Out]

(6*a*b*c*x^2 + b^2*c^2*x^4 + 2*a*b*c^3*x^6 + 3*a^2*c^4*x^8 + 2*b*c*x^2*(3*a*c^3*x^6 + b*(3 + c^2*x^4))*ArcTanh
[c*x^2] + 3*b^2*(-1 + c^4*x^8)*ArcTanh[c*x^2]^2 + b*(3*a + 4*b)*Log[1 - c*x^2] - 3*a*b*Log[1 + c*x^2] + 4*b^2*
Log[1 + c*x^2])/(24*c^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(111)=222\).
time = 0.08, size = 298, normalized size = 2.38

method	result	size
risch	\(\frac {b^{2} \left (x^{8} c^{4}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{32 c^{4}}+\frac {b \left (-3 x^{8} b \ln \left (-c \,x^{2}+1\right ) c^{4}+6 a \,c^{4} x^{8}+2 b \,c^{3} x^{6}+6 b c \,x^{2}+3 b \ln \left (-c \,x^{2}+1\right )\right ) \ln \left (c \,x^{2}+1\right )}{48 c^{4}}+\frac {b^{2} x^{8} \ln \left (-c \,x^{2}+1\right )^{2}}{32}-\frac {a b \,x^{8} \ln \left (-c \,x^{2}+1\right )}{8}+\frac {x^{8} a^{2}}{8}-\frac {b^{2} x^{6} \ln \left (-c \,x^{2}+1\right )}{24 c}+\frac {a b \,x^{6}}{12 c}+\frac {b^{2} x^{4}}{24 c^{2}}-\frac {b^{2} x^{2} \ln \left (-c \,x^{2}+1\right )}{8 c^{3}}+\frac {a b \,x^{2}}{4 c^{3}}-\frac {b^{2} \ln \left (-c \,x^{2}+1\right )^{2}}{32 c^{4}}+\frac {b \ln \left (-c \,x^{2}+1\right ) a}{8 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{2}+1\right )}{6 c^{4}}-\frac {b \ln \left (-c \,x^{2}-1\right ) a}{8 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{2}-1\right )}{6 c^{4}}\)	\(298\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a+b*arctanh(c*x^2))^2,x,method=_RETURNVERBOSE)

[Out]

1/32*b^2*(c^4*x^8-1)/c^4*ln(c*x^2+1)^2+1/48*b*(-3*x^8*b*ln(-c*x^2+1)*c^4+6*a*c^4*x^8+2*b*c^3*x^6+6*b*c*x^2+3*b
*ln(-c*x^2+1))/c^4*ln(c*x^2+1)+1/32*b^2*x^8*ln(-c*x^2+1)^2-1/8*a*b*x^8*ln(-c*x^2+1)+1/8*x^8*a^2-1/24/c*b^2*x^6
*ln(-c*x^2+1)+1/12/c*a*b*x^6+1/24*b^2*x^4/c^2-1/8/c^3*b^2*x^2*ln(-c*x^2+1)+1/4*a*b*x^2/c^3-1/32/c^4*b^2*ln(-c*
x^2+1)^2+1/8/c^4*b*ln(-c*x^2+1)*a+1/6/c^4*b^2*ln(-c*x^2+1)-1/8/c^4*b*ln(-c*x^2-1)*a+1/6/c^4*b^2*ln(-c*x^2-1)

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Maxima [A]
time = 0.26, size = 217, normalized size = 1.74 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} \operatorname {artanh}\left (c x^{2}\right )^{2} + \frac {1}{8} \, a^{2} x^{8} + \frac {1}{24} \, {\left (6 \, x^{8} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{6} + 3 \, x^{2}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{2} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{2} - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{96} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{6} + 3 \, x^{2}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{2} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{2} - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c x^{2}\right ) + \frac {4 \, c^{2} x^{4} - 2 \, {\left (3 \, \log \left (c x^{2} - 1\right ) - 8\right )} \log \left (c x^{2} + 1\right ) + 3 \, \log \left (c x^{2} + 1\right )^{2} + 3 \, \log \left (c x^{2} - 1\right )^{2} + 16 \, \log \left (c x^{2} - 1\right )}{c^{4}}\right )} b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="maxima")

[Out]

1/8*b^2*x^8*arctanh(c*x^2)^2 + 1/8*a^2*x^8 + 1/24*(6*x^8*arctanh(c*x^2) + c*(2*(c^2*x^6 + 3*x^2)/c^4 - 3*log(c
*x^2 + 1)/c^5 + 3*log(c*x^2 - 1)/c^5))*a*b + 1/96*(4*c*(2*(c^2*x^6 + 3*x^2)/c^4 - 3*log(c*x^2 + 1)/c^5 + 3*log
(c*x^2 - 1)/c^5)*arctanh(c*x^2) + (4*c^2*x^4 - 2*(3*log(c*x^2 - 1) - 8)*log(c*x^2 + 1) + 3*log(c*x^2 + 1)^2 +
3*log(c*x^2 - 1)^2 + 16*log(c*x^2 - 1))/c^4)*b^2

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Fricas [A]
time = 0.37, size = 176, normalized size = 1.41 \begin {gather*} \frac {12 \, a^{2} c^{4} x^{8} + 8 \, a b c^{3} x^{6} + 4 \, b^{2} c^{2} x^{4} + 24 \, a b c x^{2} + 3 \, {\left (b^{2} c^{4} x^{8} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{2} + 1\right ) + 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{2} - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{8} + b^{2} c^{3} x^{6} + 3 \, b^{2} c x^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{96 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="fricas")

[Out]

1/96*(12*a^2*c^4*x^8 + 8*a*b*c^3*x^6 + 4*b^2*c^2*x^4 + 24*a*b*c*x^2 + 3*(b^2*c^4*x^8 - b^2)*log(-(c*x^2 + 1)/(
c*x^2 - 1))^2 - 4*(3*a*b - 4*b^2)*log(c*x^2 + 1) + 4*(3*a*b + 4*b^2)*log(c*x^2 - 1) + 4*(3*a*b*c^4*x^8 + b^2*c
^3*x^6 + 3*b^2*c*x^2)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^4

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Sympy [A]
time = 9.25, size = 206, normalized size = 1.65 \begin {gather*} \begin {cases} \frac {a^{2} x^{8}}{8} + \frac {a b x^{8} \operatorname {atanh}{\left (c x^{2} \right )}}{4} + \frac {a b x^{6}}{12 c} + \frac {a b x^{2}}{4 c^{3}} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{4}} + \frac {b^{2} x^{8} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{8} + \frac {b^{2} x^{6} \operatorname {atanh}{\left (c x^{2} \right )}}{12 c} + \frac {b^{2} x^{4}}{24 c^{2}} + \frac {b^{2} x^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{3}} + \frac {b^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{3 c^{4}} + \frac {b^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{3 c^{4}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{8 c^{4}} - \frac {b^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{8}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(a+b*atanh(c*x**2))**2,x)

[Out]

Piecewise((a**2*x**8/8 + a*b*x**8*atanh(c*x**2)/4 + a*b*x**6/(12*c) + a*b*x**2/(4*c**3) - a*b*atanh(c*x**2)/(4
*c**4) + b**2*x**8*atanh(c*x**2)**2/8 + b**2*x**6*atanh(c*x**2)/(12*c) + b**2*x**4/(24*c**2) + b**2*x**2*atanh
(c*x**2)/(4*c**3) + b**2*log(x - sqrt(-1/c))/(3*c**4) + b**2*log(x + sqrt(-1/c))/(3*c**4) - b**2*atanh(c*x**2)
**2/(8*c**4) - b**2*atanh(c*x**2)/(3*c**4), Ne(c, 0)), (a**2*x**8/8, True))

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Giac [A]
time = 0.52, size = 175, normalized size = 1.40 \begin {gather*} \frac {1}{8} \, a^{2} x^{8} + \frac {a b x^{6}}{12 \, c} + \frac {b^{2} x^{4}}{24 \, c^{2}} + \frac {1}{32} \, {\left (b^{2} x^{8} - \frac {b^{2}}{c^{4}}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} + \frac {1}{24} \, {\left (3 \, a b x^{8} + \frac {b^{2} x^{6}}{c} + \frac {3 \, b^{2} x^{2}}{c^{3}}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {a b x^{2}}{4 \, c^{3}} - \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{2} + 1\right )}{24 \, c^{4}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{2} - 1\right )}{24 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="giac")

[Out]

1/8*a^2*x^8 + 1/12*a*b*x^6/c + 1/24*b^2*x^4/c^2 + 1/32*(b^2*x^8 - b^2/c^4)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 + 1
/24*(3*a*b*x^8 + b^2*x^6/c + 3*b^2*x^2/c^3)*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 1/4*a*b*x^2/c^3 - 1/24*(3*a*b - 4*
b^2)*log(c*x^2 + 1)/c^4 + 1/24*(3*a*b + 4*b^2)*log(c*x^2 - 1)/c^4

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Mupad [B]
time = 1.73, size = 335, normalized size = 2.68 \begin {gather*} \frac {a^2\,x^8}{8}+\frac {b^2\,\ln \left (c\,x^2-1\right )}{6\,c^4}+\frac {b^2\,\ln \left (c\,x^2+1\right )}{6\,c^4}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{32\,c^4}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{32\,c^4}+\frac {b^2\,x^4}{24\,c^2}+\frac {b^2\,x^8\,{\ln \left (c\,x^2+1\right )}^2}{32}+\frac {b^2\,x^8\,{\ln \left (1-c\,x^2\right )}^2}{32}+\frac {b^2\,x^2\,\ln \left (c\,x^2+1\right )}{8\,c^3}-\frac {b^2\,x^2\,\ln \left (1-c\,x^2\right )}{8\,c^3}+\frac {b^2\,x^6\,\ln \left (c\,x^2+1\right )}{24\,c}-\frac {b^2\,x^6\,\ln \left (1-c\,x^2\right )}{24\,c}+\frac {a\,b\,\ln \left (c\,x^2-1\right )}{8\,c^4}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{8\,c^4}+\frac {a\,b\,x^8\,\ln \left (c\,x^2+1\right )}{8}-\frac {a\,b\,x^8\,\ln \left (1-c\,x^2\right )}{8}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{16\,c^4}+\frac {a\,b\,x^2}{4\,c^3}+\frac {a\,b\,x^6}{12\,c}-\frac {b^2\,x^8\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*atanh(c*x^2))^2,x)

[Out]

(a^2*x^8)/8 + (b^2*log(c*x^2 - 1))/(6*c^4) + (b^2*log(c*x^2 + 1))/(6*c^4) - (b^2*log(c*x^2 + 1)^2)/(32*c^4) -
(b^2*log(1 - c*x^2)^2)/(32*c^4) + (b^2*x^4)/(24*c^2) + (b^2*x^8*log(c*x^2 + 1)^2)/32 + (b^2*x^8*log(1 - c*x^2)
^2)/32 + (b^2*x^2*log(c*x^2 + 1))/(8*c^3) - (b^2*x^2*log(1 - c*x^2))/(8*c^3) + (b^2*x^6*log(c*x^2 + 1))/(24*c)
- (b^2*x^6*log(1 - c*x^2))/(24*c) + (a*b*log(c*x^2 - 1))/(8*c^4) - (a*b*log(c*x^2 + 1))/(8*c^4) + (a*b*x^8*lo
g(c*x^2 + 1))/8 - (a*b*x^8*log(1 - c*x^2))/8 + (b^2*log(c*x^2 + 1)*log(1 - c*x^2))/(16*c^4) + (a*b*x^2)/(4*c^3
) + (a*b*x^6)/(12*c) - (b^2*x^8*log(c*x^2 + 1)*log(1 - c*x^2))/16

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