\(\int x^{11} (a+b \arctan (c x^3))^2 \, dx\) [113]

Optimal result

Integrand size = 16, antiderivative size = 124 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {a b x^3}{6 c^3}+\frac {b^2 x^6}{36 c^2}+\frac {b^2 x^3 \arctan \left (c x^3\right )}{6 c^3}-\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {b^2 \log \left (1+c^2 x^6\right )}{9 c^4} \]

[Out]

1/6*a*b*x^3/c^3+1/36*b^2*x^6/c^2+1/6*b^2*x^3*arctan(c*x^3)/c^3-1/18*b*x^9*(a+b*arctan(c*x^3))/c-1/12*(a+b*arct
an(c*x^3))^2/c^4+1/12*x^12*(a+b*arctan(c*x^3))^2-1/9*b^2*ln(c^2*x^6+1)/c^4

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4948, 4946, 5036, 272, 45, 4930, 266, 5004} \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}+\frac {a b x^3}{6 c^3}+\frac {b^2 x^3 \arctan \left (c x^3\right )}{6 c^3}+\frac {b^2 x^6}{36 c^2}-\frac {b^2 \log \left (c^2 x^6+1\right )}{9 c^4} \]

[In]

Int[x^11*(a + b*ArcTan[c*x^3])^2,x]

[Out]

(a*b*x^3)/(6*c^3) + (b^2*x^6)/(36*c^2) + (b^2*x^3*ArcTan[c*x^3])/(6*c^3) - (b*x^9*(a + b*ArcTan[c*x^3]))/(18*c
) - (a + b*ArcTan[c*x^3])^2/(12*c^4) + (x^12*(a + b*ArcTan[c*x^3])^2)/12 - (b^2*Log[1 + c^2*x^6])/(9*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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[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 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 5004

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x^3 (a+b \arctan (c x))^2 \, dx,x,x^3\right ) \\ & = \frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {1}{6} (b c) \text {Subst}\left (\int \frac {x^4 (a+b \arctan (c x))}{1+c^2 x^2} \, dx,x,x^3\right ) \\ & = \frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {b \text {Subst}\left (\int x^2 (a+b \arctan (c x)) \, dx,x,x^3\right )}{6 c}+\frac {b \text {Subst}\left (\int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx,x,x^3\right )}{6 c} \\ & = -\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2+\frac {1}{18} b^2 \text {Subst}\left (\int \frac {x^3}{1+c^2 x^2} \, dx,x,x^3\right )+\frac {b \text {Subst}\left (\int (a+b \arctan (c x)) \, dx,x,x^3\right )}{6 c^3}-\frac {b \text {Subst}\left (\int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx,x,x^3\right )}{6 c^3} \\ & = \frac {a b x^3}{6 c^3}-\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2+\frac {1}{36} b^2 \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^6\right )+\frac {b^2 \text {Subst}\left (\int \arctan (c x) \, dx,x,x^3\right )}{6 c^3} \\ & = \frac {a b x^3}{6 c^3}+\frac {b^2 x^3 \arctan \left (c x^3\right )}{6 c^3}-\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2+\frac {1}{36} b^2 \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^6\right )-\frac {b^2 \text {Subst}\left (\int \frac {x}{1+c^2 x^2} \, dx,x,x^3\right )}{6 c^2} \\ & = \frac {a b x^3}{6 c^3}+\frac {b^2 x^6}{36 c^2}+\frac {b^2 x^3 \arctan \left (c x^3\right )}{6 c^3}-\frac {b x^9 \left (a+b \arctan \left (c x^3\right )\right )}{18 c}-\frac {\left (a+b \arctan \left (c x^3\right )\right )^2}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \arctan \left (c x^3\right )\right )^2-\frac {b^2 \log \left (1+c^2 x^6\right )}{9 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {c x^3 \left (6 a b+b^2 c x^3-2 a b c^2 x^6+3 a^2 c^3 x^9\right )-2 b \left (b c x^3 \left (-3+c^2 x^6\right )+a \left (3-3 c^4 x^{12}\right )\right ) \arctan \left (c x^3\right )+3 b^2 \left (-1+c^4 x^{12}\right ) \arctan \left (c x^3\right )^2-4 b^2 \log \left (1+c^2 x^6\right )}{36 c^4} \]

[In]

Integrate[x^11*(a + b*ArcTan[c*x^3])^2,x]

[Out]

(c*x^3*(6*a*b + b^2*c*x^3 - 2*a*b*c^2*x^6 + 3*a^2*c^3*x^9) - 2*b*(b*c*x^3*(-3 + c^2*x^6) + a*(3 - 3*c^4*x^12))
*ArcTan[c*x^3] + 3*b^2*(-1 + c^4*x^12)*ArcTan[c*x^3]^2 - 4*b^2*Log[1 + c^2*x^6])/(36*c^4)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.22

method	result	size
default	\(\frac {a^{2} x^{12}}{12}+\frac {b^{2} x^{12} \arctan \left (c \,x^{3}\right )^{2}}{12}-\frac {b^{2} \arctan \left (c \,x^{3}\right ) x^{9}}{18 c}+\frac {b^{2} x^{3} \arctan \left (c \,x^{3}\right )}{6 c^{3}}-\frac {b^{2} \arctan \left (c \,x^{3}\right )^{2}}{12 c^{4}}+\frac {b^{2} x^{6}}{36 c^{2}}-\frac {b^{2} \ln \left (c^{2} x^{6}+1\right )}{9 c^{4}}+\frac {a b \,x^{12} \arctan \left (c \,x^{3}\right )}{6}-\frac {a b \,x^{9}}{18 c}+\frac {a b \,x^{3}}{6 c^{3}}-\frac {a b \arctan \left (c \,x^{3}\right )}{6 c^{4}}\)	\(151\)
parts	\(\frac {a^{2} x^{12}}{12}+\frac {b^{2} x^{12} \arctan \left (c \,x^{3}\right )^{2}}{12}-\frac {b^{2} \arctan \left (c \,x^{3}\right ) x^{9}}{18 c}+\frac {b^{2} x^{3} \arctan \left (c \,x^{3}\right )}{6 c^{3}}-\frac {b^{2} \arctan \left (c \,x^{3}\right )^{2}}{12 c^{4}}+\frac {b^{2} x^{6}}{36 c^{2}}-\frac {b^{2} \ln \left (c^{2} x^{6}+1\right )}{9 c^{4}}+\frac {a b \,x^{12} \arctan \left (c \,x^{3}\right )}{6}-\frac {a b \,x^{9}}{18 c}+\frac {a b \,x^{3}}{6 c^{3}}-\frac {a b \arctan \left (c \,x^{3}\right )}{6 c^{4}}\)	\(151\)
parallelrisch	\(-\frac {-3 b^{2} \arctan \left (c \,x^{3}\right )^{2} x^{12} c^{4}-6 a b \arctan \left (c \,x^{3}\right ) x^{12} c^{4}-3 c^{4} a^{2} x^{12}+2 b^{2} \arctan \left (c \,x^{3}\right ) x^{9} c^{3}+2 a b \,c^{3} x^{9}-x^{6} b^{2} c^{2}-6 b^{2} \arctan \left (c \,x^{3}\right ) x^{3} c -6 a b c \,x^{3}+3 b^{2} \arctan \left (c \,x^{3}\right )^{2}+4 b^{2} \ln \left (c^{2} x^{6}+1\right )+6 a b \arctan \left (c \,x^{3}\right )+b^{2}}{36 c^{4}}\)	\(155\)
risch	\(-\frac {b^{2} \left (c^{4} x^{12}-1\right ) \ln \left (i c \,x^{3}+1\right )^{2}}{48 c^{4}}-\frac {i b \left (6 a \,c^{4} x^{12}+3 i b \,c^{4} x^{12} \ln \left (-i c \,x^{3}+1\right )-2 b \,c^{3} x^{9}+6 b c \,x^{3}-3 i b \ln \left (-i c \,x^{3}+1\right )\right ) \ln \left (i c \,x^{3}+1\right )}{72 c^{4}}+\frac {i a b \,x^{12} \ln \left (-i c \,x^{3}+1\right )}{12}-\frac {b^{2} x^{12} \ln \left (-i c \,x^{3}+1\right )^{2}}{48}+\frac {a^{2} x^{12}}{12}-\frac {i b^{2} x^{9} \ln \left (-i c \,x^{3}+1\right )}{36 c}-\frac {a b \,x^{9}}{18 c}+\frac {b^{2} x^{6}}{36 c^{2}}+\frac {i b^{2} x^{3} \ln \left (-i c \,x^{3}+1\right )}{12 c^{3}}+\frac {a b \,x^{3}}{6 c^{3}}+\frac {b^{2} \ln \left (-i c \,x^{3}+1\right )^{2}}{48 c^{4}}-\frac {a b \arctan \left (c \,x^{3}\right )}{6 c^{4}}-\frac {b^{2} \ln \left (c^{2} x^{6}+1\right )}{9 c^{4}}\)	\(280\)

[In]

int(x^11*(a+b*arctan(c*x^3))^2,x,method=_RETURNVERBOSE)

[Out]

1/12*a^2*x^12+1/12*b^2*x^12*arctan(c*x^3)^2-1/18*b^2*arctan(c*x^3)/c*x^9+1/6*b^2*x^3*arctan(c*x^3)/c^3-1/12*b^
2/c^4*arctan(c*x^3)^2+1/36*b^2*x^6/c^2-1/9*b^2*ln(c^2*x^6+1)/c^4+1/6*a*b*x^12*arctan(c*x^3)-1/18*a*b/c*x^9+1/6
*a*b*x^3/c^3-1/6*a*b/c^4*arctan(c*x^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {3 \, a^{2} c^{4} x^{12} - 2 \, a b c^{3} x^{9} + b^{2} c^{2} x^{6} + 6 \, a b c x^{3} + 3 \, {\left (b^{2} c^{4} x^{12} - b^{2}\right )} \arctan \left (c x^{3}\right )^{2} - 4 \, b^{2} \log \left (c^{2} x^{6} + 1\right ) + 2 \, {\left (3 \, a b c^{4} x^{12} - b^{2} c^{3} x^{9} + 3 \, b^{2} c x^{3} - 3 \, a b\right )} \arctan \left (c x^{3}\right )}{36 \, c^{4}} \]

[In]

integrate(x^11*(a+b*arctan(c*x^3))^2,x, algorithm="fricas")

[Out]

1/36*(3*a^2*c^4*x^12 - 2*a*b*c^3*x^9 + b^2*c^2*x^6 + 6*a*b*c*x^3 + 3*(b^2*c^4*x^12 - b^2)*arctan(c*x^3)^2 - 4*
b^2*log(c^2*x^6 + 1) + 2*(3*a*b*c^4*x^12 - b^2*c^3*x^9 + 3*b^2*c*x^3 - 3*a*b)*arctan(c*x^3))/c^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (112) = 224\).

Time = 147.67 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.96 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{12}}{12} + \frac {a b x^{12} \operatorname {atan}{\left (c x^{3} \right )}}{6} - \frac {a b x^{9}}{18 c} + \frac {a b x^{3}}{6 c^{3}} - \frac {a b \operatorname {atan}{\left (c x^{3} \right )}}{6 c^{4}} + \frac {b^{2} x^{12} \operatorname {atan}^{2}{\left (c x^{3} \right )}}{12} - \frac {b^{2} x^{9} \operatorname {atan}{\left (c x^{3} \right )}}{18 c} + \frac {b^{2} x^{6}}{36 c^{2}} + \frac {b^{2} x^{3} \operatorname {atan}{\left (c x^{3} \right )}}{6 c^{3}} + \frac {2 b^{2} \sqrt {- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{3} \right )}}{9 c^{3}} - \frac {2 b^{2} \log {\left (x - \sqrt [6]{- \frac {1}{c^{2}}} \right )}}{9 c^{4}} - \frac {2 b^{2} \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{9 c^{4}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x^{3} \right )}}{12 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{12}}{12} & \text {otherwise} \end {cases} \]

[In]

integrate(x**11*(a+b*atan(c*x**3))**2,x)

[Out]

Piecewise((a**2*x**12/12 + a*b*x**12*atan(c*x**3)/6 - a*b*x**9/(18*c) + a*b*x**3/(6*c**3) - a*b*atan(c*x**3)/(
6*c**4) + b**2*x**12*atan(c*x**3)**2/12 - b**2*x**9*atan(c*x**3)/(18*c) + b**2*x**6/(36*c**2) + b**2*x**3*atan
(c*x**3)/(6*c**3) + 2*b**2*sqrt(-1/c**2)*atan(c*x**3)/(9*c**3) - 2*b**2*log(x - (-1/c**2)**(1/6))/(9*c**4) - 2
*b**2*log(4*x**2 + 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3))/(9*c**4) - b**2*atan(c*x**3)**2/(12*c**4), Ne(c,
0)), (a**2*x**12/12, True))

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.36 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {1}{12} \, b^{2} x^{12} \arctan \left (c x^{3}\right )^{2} + \frac {1}{12} \, a^{2} x^{12} + \frac {1}{18} \, {\left (3 \, x^{12} \arctan \left (c x^{3}\right ) - c {\left (\frac {c^{2} x^{9} - 3 \, x^{3}}{c^{4}} + \frac {3 \, \arctan \left (c x^{3}\right )}{c^{5}}\right )}\right )} a b - \frac {1}{36} \, {\left (2 \, c {\left (\frac {c^{2} x^{9} - 3 \, x^{3}}{c^{4}} + \frac {3 \, \arctan \left (c x^{3}\right )}{c^{5}}\right )} \arctan \left (c x^{3}\right ) - \frac {c^{2} x^{6} + 3 \, \arctan \left (c x^{3}\right )^{2} - 3 \, \log \left (18 \, c^{7} x^{6} + 18 \, c^{5}\right ) - \log \left (c^{2} x^{6} + 1\right )}{c^{4}}\right )} b^{2} \]

[In]

integrate(x^11*(a+b*arctan(c*x^3))^2,x, algorithm="maxima")

[Out]

1/12*b^2*x^12*arctan(c*x^3)^2 + 1/12*a^2*x^12 + 1/18*(3*x^12*arctan(c*x^3) - c*((c^2*x^9 - 3*x^3)/c^4 + 3*arct
an(c*x^3)/c^5))*a*b - 1/36*(2*c*((c^2*x^9 - 3*x^3)/c^4 + 3*arctan(c*x^3)/c^5)*arctan(c*x^3) - (c^2*x^6 + 3*arc
tan(c*x^3)^2 - 3*log(18*c^7*x^6 + 18*c^5) - log(c^2*x^6 + 1))/c^4)*b^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {3 \, a^{2} c x^{12} + 2 \, {\left (3 \, c x^{12} \arctan \left (c x^{3}\right ) - \frac {3 \, \arctan \left (c x^{3}\right )}{c^{3}} - \frac {c^{9} x^{9} - 3 \, c^{7} x^{3}}{c^{9}}\right )} a b + {\left (3 \, c x^{12} \arctan \left (c x^{3}\right )^{2} - \frac {2 \, c^{3} x^{9} \arctan \left (c x^{3}\right ) - c^{2} x^{6} - 6 \, c x^{3} \arctan \left (c x^{3}\right ) + 3 \, \arctan \left (c x^{3}\right )^{2} + 4 \, \log \left (c^{2} x^{6} + 1\right )}{c^{3}}\right )} b^{2}}{36 \, c} \]

[In]

integrate(x^11*(a+b*arctan(c*x^3))^2,x, algorithm="giac")

[Out]

1/36*(3*a^2*c*x^12 + 2*(3*c*x^12*arctan(c*x^3) - 3*arctan(c*x^3)/c^3 - (c^9*x^9 - 3*c^7*x^3)/c^9)*a*b + (3*c*x
^12*arctan(c*x^3)^2 - (2*c^3*x^9*arctan(c*x^3) - c^2*x^6 - 6*c*x^3*arctan(c*x^3) + 3*arctan(c*x^3)^2 + 4*log(c
^2*x^6 + 1))/c^3)*b^2)/c

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.21 \[ \int x^{11} \left (a+b \arctan \left (c x^3\right )\right )^2 \, dx=\frac {a^2\,x^{12}}{12}-\frac {b^2\,{\mathrm {atan}\left (c\,x^3\right )}^2}{12\,c^4}+\frac {b^2\,x^{12}\,{\mathrm {atan}\left (c\,x^3\right )}^2}{12}-\frac {b^2\,\ln \left (c^2\,x^6+1\right )}{9\,c^4}+\frac {b^2\,x^6}{36\,c^2}+\frac {b^2\,x^3\,\mathrm {atan}\left (c\,x^3\right )}{6\,c^3}-\frac {b^2\,x^9\,\mathrm {atan}\left (c\,x^3\right )}{18\,c}+\frac {a\,b\,x^3}{6\,c^3}-\frac {a\,b\,x^9}{18\,c}-\frac {a\,b\,\mathrm {atan}\left (c\,x^3\right )}{6\,c^4}+\frac {a\,b\,x^{12}\,\mathrm {atan}\left (c\,x^3\right )}{6} \]

[In]

int(x^11*(a + b*atan(c*x^3))^2,x)

[Out]

(a^2*x^12)/12 - (b^2*atan(c*x^3)^2)/(12*c^4) + (b^2*x^12*atan(c*x^3)^2)/12 - (b^2*log(c^2*x^6 + 1))/(9*c^4) +
(b^2*x^6)/(36*c^2) + (b^2*x^3*atan(c*x^3))/(6*c^3) - (b^2*x^9*atan(c*x^3))/(18*c) + (a*b*x^3)/(6*c^3) - (a*b*x
^9)/(18*c) - (a*b*atan(c*x^3))/(6*c^4) + (a*b*x^12*atan(c*x^3))/6