3.2.14 \(\int x^8 (a+b \text {ArcTan}(c x^3))^2 \, dx\) [114]

Optimal. Leaf size=154 \[ \frac {b^2 x^3}{9 c^2}-\frac {b^2 \text {ArcTan}\left (c x^3\right )}{9 c^3}-\frac {b x^6 \left (a+b \text {ArcTan}\left (c x^3\right )\right )}{9 c}-\frac {i \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2}{9 c^3}+\frac {1}{9} x^9 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2-\frac {2 b \left (a+b \text {ArcTan}\left (c x^3\right )\right ) \log \left (\frac {2}{1+i c x^3}\right )}{9 c^3}-\frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x^3}\right )}{9 c^3} \]

[Out]

1/9*b^2*x^3/c^2-1/9*b^2*arctan(c*x^3)/c^3-1/9*b*x^6*(a+b*arctan(c*x^3))/c-1/9*I*(a+b*arctan(c*x^3))^2/c^3+1/9*
x^9*(a+b*arctan(c*x^3))^2-2/9*b*(a+b*arctan(c*x^3))*ln(2/(1+I*c*x^3))/c^3-1/9*I*b^2*polylog(2,1-2/(1+I*c*x^3))
/c^3

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {4948, 4946, 5036, 327, 209, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {i \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2}{9 c^3}-\frac {2 b \log \left (\frac {2}{1+i c x^3}\right ) \left (a+b \text {ArcTan}\left (c x^3\right )\right )}{9 c^3}+\frac {1}{9} x^9 \left (a+b \text {ArcTan}\left (c x^3\right )\right )^2-\frac {b x^6 \left (a+b \text {ArcTan}\left (c x^3\right )\right )}{9 c}-\frac {b^2 \text {ArcTan}\left (c x^3\right )}{9 c^3}-\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i c x^3+1}\right )}{9 c^3}+\frac {b^2 x^3}{9 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*ArcTan[c*x^3])^2,x]

[Out]

(b^2*x^3)/(9*c^2) - (b^2*ArcTan[c*x^3])/(9*c^3) - (b*x^6*(a + b*ArcTan[c*x^3]))/(9*c) - ((I/9)*(a + b*ArcTan[c
*x^3])^2)/c^3 + (x^9*(a + b*ArcTan[c*x^3])^2)/9 - (2*b*(a + b*ArcTan[c*x^3])*Log[2/(1 + I*c*x^3)])/(9*c^3) - (
(I/9)*b^2*PolyLog[2, 1 - 2/(1 + I*c*x^3)])/c^3

Rule 209

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

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[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 4948

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[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 5036

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

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^8 \left (a+b \tan ^{-1}\left (c x^3\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^8 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{2} b x^8 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {1}{4} b^2 x^8 \log ^2\left (1+i c x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int x^8 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2 \, dx+\frac {1}{2} b \int x^8 \left (-2 i a+b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right ) \, dx-\frac {1}{4} b^2 \int x^8 \log ^2\left (1+i c x^3\right ) \, dx\\ &=\frac {1}{12} \text {Subst}\left (\int x^2 (2 a+i b \log (1-i c x))^2 \, dx,x,x^3\right )+\frac {1}{6} b \text {Subst}\left (\int x^2 (-2 i a+b \log (1-i c x)) \log (1+i c x) \, dx,x,x^3\right )-\frac {1}{12} b^2 \text {Subst}\left (\int x^2 \log ^2(1+i c x) \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )-\frac {1}{18} (i b c) \text {Subst}\left (\int \frac {x^3 (-2 i a+b \log (1-i c x))}{1+i c x} \, dx,x,x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {x^3 (2 a+i b \log (1-i c x))}{1-i c x} \, dx,x,x^3\right )+\frac {1}{18} \left (i b^2 c\right ) \text {Subst}\left (\int \frac {x^3 \log (1+i c x)}{1-i c x} \, dx,x,x^3\right )+\frac {1}{18} \left (i b^2 c\right ) \text {Subst}\left (\int \frac {x^3 \log (1+i c x)}{1+i c x} \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )-\frac {1}{18} (i b) \text {Subst}\left (\int \frac {\left (-\frac {i}{c}+\frac {i x}{c}\right )^3 (2 a+i b \log (x))}{x} \, dx,x,1-i c x^3\right )-\frac {1}{18} (i b c) \text {Subst}\left (\int \left (\frac {i (-2 i a+b \log (1-i c x))}{c^3}+\frac {x (-2 i a+b \log (1-i c x))}{c^2}-\frac {i x^2 (-2 i a+b \log (1-i c x))}{c}-\frac {-2 i a+b \log (1-i c x)}{c^3 (-i+c x)}\right ) \, dx,x,x^3\right )+\frac {1}{18} \left (i b^2 c\right ) \text {Subst}\left (\int \left (\frac {i \log (1+i c x)}{c^3}+\frac {x \log (1+i c x)}{c^2}-\frac {i x^2 \log (1+i c x)}{c}-\frac {\log (1+i c x)}{c^3 (-i+c x)}\right ) \, dx,x,x^3\right )+\frac {1}{18} \left (i b^2 c\right ) \text {Subst}\left (\int \left (-\frac {i \log (1+i c x)}{c^3}+\frac {x \log (1+i c x)}{c^2}+\frac {i x^2 \log (1+i c x)}{c}-\frac {\log (1+i c x)}{c^3 (i+c x)}\right ) \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{108} i b \left (2 a+i b \log \left (1-i c x^3\right )\right ) \left (\frac {18 i \left (1-i c x^3\right )}{c^3}-\frac {9 i \left (1-i c x^3\right )^2}{c^3}-\frac {2 \left (i+c x^3\right )^3}{c^3}-\frac {6 i \log \left (1-i c x^3\right )}{c^3}\right )-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )-\frac {1}{18} b \text {Subst}\left (\int x^2 (-2 i a+b \log (1-i c x)) \, dx,x,x^3\right )-\frac {1}{18} b^2 \text {Subst}\left (\int -\frac {i \left (x \left (18-9 x+2 x^2\right )-6 \log (x)\right )}{6 c^3 x} \, dx,x,1-i c x^3\right )+\frac {(i b) \text {Subst}\left (\int \frac {-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^3\right )}{18 c^2}+\frac {b \text {Subst}\left (\int (-2 i a+b \log (1-i c x)) \, dx,x,x^3\right )}{18 c^2}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log (1+i c x)}{-i+c x} \, dx,x,x^3\right )}{18 c^2}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log (1+i c x)}{i+c x} \, dx,x,x^3\right )}{18 c^2}-\frac {(i b) \text {Subst}\left (\int x (-2 i a+b \log (1-i c x)) \, dx,x,x^3\right )}{18 c}+2 \frac {\left (i b^2\right ) \text {Subst}\left (\int x \log (1+i c x) \, dx,x,x^3\right )}{18 c}\\ &=-\frac {i a b x^3}{9 c^2}+\frac {i b x^6 \left (2 i a-b \log \left (1-i c x^3\right )\right )}{36 c}+\frac {1}{54} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{108} i b \left (2 a+i b \log \left (1-i c x^3\right )\right ) \left (\frac {18 i \left (1-i c x^3\right )}{c^3}-\frac {9 i \left (1-i c x^3\right )^2}{c^3}-\frac {2 \left (i+c x^3\right )^3}{c^3}-\frac {6 i \log \left (1-i c x^3\right )}{c^3}\right )-\frac {i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{18 c^3}-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )+\frac {1}{36} b^2 \text {Subst}\left (\int \frac {x^2}{1-i c x} \, dx,x,x^3\right )+2 \left (\frac {i b^2 x^6 \log \left (1+i c x^3\right )}{36 c}+\frac {1}{36} b^2 \text {Subst}\left (\int \frac {x^2}{1+i c x} \, dx,x,x^3\right )\right )+\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {x \left (18-9 x+2 x^2\right )-6 \log (x)}{x} \, dx,x,1-i c x^3\right )}{108 c^3}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+i c x^3\right )}{18 c^3}+\frac {b^2 \text {Subst}\left (\int \log (1-i c x) \, dx,x,x^3\right )}{18 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^3\right )}{18 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (-\frac {1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^3\right )}{18 c^2}-\frac {1}{54} \left (i b^2 c\right ) \text {Subst}\left (\int \frac {x^3}{1-i c x} \, dx,x,x^3\right )\\ &=-\frac {i a b x^3}{9 c^2}+\frac {i b x^6 \left (2 i a-b \log \left (1-i c x^3\right )\right )}{36 c}+\frac {1}{54} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{108} i b \left (2 a+i b \log \left (1-i c x^3\right )\right ) \left (\frac {18 i \left (1-i c x^3\right )}{c^3}-\frac {9 i \left (1-i c x^3\right )^2}{c^3}-\frac {2 \left (i+c x^3\right )^3}{c^3}-\frac {6 i \log \left (1-i c x^3\right )}{c^3}\right )-\frac {i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{18 c^3}-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {i b^2 \log ^2\left (1+i c x^3\right )}{36 c^3}-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )+2 \left (\frac {i b^2 x^6 \log \left (1+i c x^3\right )}{36 c}+\frac {1}{36} b^2 \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {i x}{c}+\frac {i}{c^2 (-i+c x)}\right ) \, dx,x,x^3\right )\right )+\frac {1}{36} b^2 \text {Subst}\left (\int \left (\frac {1}{c^2}+\frac {i x}{c}-\frac {i}{c^2 (i+c x)}\right ) \, dx,x,x^3\right )+\frac {\left (i b^2\right ) \text {Subst}\left (\int \left (18-9 x+2 x^2-\frac {6 \log (x)}{x}\right ) \, dx,x,1-i c x^3\right )}{108 c^3}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-i c x^3\right )}{18 c^3}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+i c x^3\right )}{18 c^3}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-i c x^3\right )}{18 c^3}-\frac {1}{54} \left (i b^2 c\right ) \text {Subst}\left (\int \left (-\frac {i}{c^3}+\frac {x}{c^2}+\frac {i x^2}{c}-\frac {1}{c^3 (i+c x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {i a b x^3}{9 c^2}+\frac {13 b^2 x^3}{108 c^2}+\frac {i b^2 x^6}{216 c}+\frac {b^2 x^9}{162}-\frac {i b^2 \left (1-i c x^3\right )^2}{24 c^3}+\frac {i b^2 \left (1-i c x^3\right )^3}{162 c^3}+\frac {i b^2 \left (1-i c x^3\right ) \log \left (1-i c x^3\right )}{18 c^3}+\frac {i b x^6 \left (2 i a-b \log \left (1-i c x^3\right )\right )}{36 c}+\frac {1}{54} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{108} i b \left (2 a+i b \log \left (1-i c x^3\right )\right ) \left (\frac {18 i \left (1-i c x^3\right )}{c^3}-\frac {9 i \left (1-i c x^3\right )^2}{c^3}-\frac {2 \left (i+c x^3\right )^3}{c^3}-\frac {6 i \log \left (1-i c x^3\right )}{c^3}\right )-\frac {i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{18 c^3}-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {i b^2 \log ^2\left (1+i c x^3\right )}{36 c^3}-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )+2 \left (\frac {b^2 x^3}{36 c^2}-\frac {i b^2 x^6}{72 c}+\frac {i b^2 \log \left (i-c x^3\right )}{36 c^3}+\frac {i b^2 x^6 \log \left (1+i c x^3\right )}{36 c}\right )-\frac {i b^2 \log \left (i+c x^3\right )}{108 c^3}+\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1-i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-i c x^3\right )}{18 c^3}\\ &=-\frac {i a b x^3}{9 c^2}+\frac {13 b^2 x^3}{108 c^2}+\frac {i b^2 x^6}{216 c}+\frac {b^2 x^9}{162}-\frac {i b^2 \left (1-i c x^3\right )^2}{24 c^3}+\frac {i b^2 \left (1-i c x^3\right )^3}{162 c^3}+\frac {i b^2 \left (1-i c x^3\right ) \log \left (1-i c x^3\right )}{18 c^3}-\frac {i b^2 \log ^2\left (1-i c x^3\right )}{36 c^3}+\frac {i b x^6 \left (2 i a-b \log \left (1-i c x^3\right )\right )}{36 c}+\frac {1}{54} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a+i b \log \left (1-i c x^3\right )\right )^2+\frac {1}{108} i b \left (2 a+i b \log \left (1-i c x^3\right )\right ) \left (\frac {18 i \left (1-i c x^3\right )}{c^3}-\frac {9 i \left (1-i c x^3\right )^2}{c^3}-\frac {2 \left (i+c x^3\right )^3}{c^3}-\frac {6 i \log \left (1-i c x^3\right )}{c^3}\right )-\frac {i b \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )}{18 c^3}-\frac {1}{18} b x^9 \left (2 i a-b \log \left (1-i c x^3\right )\right ) \log \left (1+i c x^3\right )-\frac {i b^2 \log ^2\left (1+i c x^3\right )}{36 c^3}-\frac {1}{36} b^2 x^9 \log ^2\left (1+i c x^3\right )+2 \left (\frac {b^2 x^3}{36 c^2}-\frac {i b^2 x^6}{72 c}+\frac {i b^2 \log \left (i-c x^3\right )}{36 c^3}+\frac {i b^2 x^6 \log \left (1+i c x^3\right )}{36 c}\right )-\frac {i b^2 \log \left (i+c x^3\right )}{108 c^3}+\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1-i c x^3\right )\right )}{18 c^3}-\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1+i c x^3\right )\right )}{18 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 141, normalized size = 0.92 \begin {gather*} \frac {b^2 c x^3-a b c^2 x^6+a^2 c^3 x^9+b^2 \left (i+c^3 x^9\right ) \text {ArcTan}\left (c x^3\right )^2-b \text {ArcTan}\left (c x^3\right ) \left (b+b c^2 x^6-2 a c^3 x^9+2 b \log \left (1+e^{2 i \text {ArcTan}\left (c x^3\right )}\right )\right )+a b \log \left (1+c^2 x^6\right )+i b^2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}\left (c x^3\right )}\right )}{9 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*ArcTan[c*x^3])^2,x]

[Out]

(b^2*c*x^3 - a*b*c^2*x^6 + a^2*c^3*x^9 + b^2*(I + c^3*x^9)*ArcTan[c*x^3]^2 - b*ArcTan[c*x^3]*(b + b*c^2*x^6 -
2*a*c^3*x^9 + 2*b*Log[1 + E^((2*I)*ArcTan[c*x^3])]) + a*b*Log[1 + c^2*x^6] + I*b^2*PolyLog[2, -E^((2*I)*ArcTan
[c*x^3])])/(9*c^3)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{8} \left (a +b \arctan \left (c \,x^{3}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a+b*arctan(c*x^3))^2,x)

[Out]

int(x^8*(a+b*arctan(c*x^3))^2,x)

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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^8*(a+b*arctan(c*x^3))^2,x, algorithm="maxima")

[Out]

1/9*a^2*x^9 + 1/9*(2*x^9*arctan(c*x^3) - (x^6/c^2 - log(c^2*x^6 + 1)/c^4)*c)*a*b + 1/144*(4*x^9*arctan(c*x^3)^
2 - x^9*log(c^2*x^6 + 1)^2 + 144*integrate(1/48*(4*c^2*x^14*log(c^2*x^6 + 1) - 8*c*x^11*arctan(c*x^3) + 36*(c^
2*x^14 + x^8)*arctan(c*x^3)^2 + 3*(c^2*x^14 + x^8)*log(c^2*x^6 + 1)^2)/(c^2*x^6 + 1), x))*b^2

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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^8*(a+b*arctan(c*x^3))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^8*arctan(c*x^3)^2 + 2*a*b*x^8*arctan(c*x^3) + a^2*x^8, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{8} \left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(a+b*atan(c*x**3))**2,x)

[Out]

Integral(x**8*(a + b*atan(c*x**3))**2, 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^8*(a+b*arctan(c*x^3))^2,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)^2*x^8, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^8\,{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*atan(c*x^3))^2,x)

[Out]

int(x^8*(a + b*atan(c*x^3))^2, x)

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