\(\int x^3 (a+b \arctan (c x^2))^3 \, dx\) [86]

Optimal result

Integrand size = 16, antiderivative size = 149 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=-\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{2 c^2}-\frac {3 i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )}{4 c^2} \]

[Out]

-3/4*I*b*(a+b*arctan(c*x^2))^2/c^2-3/4*b*x^2*(a+b*arctan(c*x^2))^2/c+1/4*(a+b*arctan(c*x^2))^3/c^2+1/4*x^4*(a+
b*arctan(c*x^2))^3-3/2*b^2*(a+b*arctan(c*x^2))*ln(2/(1+I*c*x^2))/c^2-3/4*I*b^3*polylog(2,1-2/(1+I*c*x^2))/c^2

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {4948, 4946, 5036, 4930, 5040, 4964, 2449, 2352, 5004} \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=-\frac {3 b^2 \log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )}{2 c^2}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}-\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {3 i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^2+1}\right )}{4 c^2} \]

[In]

Int[x^3*(a + b*ArcTan[c*x^2])^3,x]

[Out]

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

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 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 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 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]

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x (a+b \arctan (c x))^3 \, dx,x,x^2\right ) \\ & = \frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {1}{4} (3 b c) \text {Subst}\left (\int \frac {x^2 (a+b \arctan (c x))^2}{1+c^2 x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {(3 b) \text {Subst}\left (\int (a+b \arctan (c x))^2 \, dx,x,x^2\right )}{4 c}+\frac {(3 b) \text {Subst}\left (\int \frac {(a+b \arctan (c x))^2}{1+c^2 x^2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3+\frac {1}{2} \left (3 b^2\right ) \text {Subst}\left (\int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx,x,x^2\right ) \\ & = -\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {a+b \arctan (c x)}{i-c x} \, dx,x,x^2\right )}{2 c} \\ & = -\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{2 c^2}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )}{2 c} \\ & = -\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{2 c^2}-\frac {\left (3 i b^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x^2}\right )}{2 c^2} \\ & = -\frac {3 i b \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c^2}-\frac {3 b x^2 \left (a+b \arctan \left (c x^2\right )\right )^2}{4 c}+\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {3 b^2 \left (a+b \arctan \left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{2 c^2}-\frac {3 i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )}{4 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.14 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\frac {3 b^2 \left (a+a c^2 x^4+b \left (i-c x^2\right )\right ) \arctan \left (c x^2\right )^2+b^3 \left (1+c^2 x^4\right ) \arctan \left (c x^2\right )^3+3 b \arctan \left (c x^2\right ) \left (a \left (a-2 b c x^2+a c^2 x^4\right )-2 b^2 \log \left (1+e^{2 i \arctan \left (c x^2\right )}\right )\right )+a \left (a c x^2 \left (-3 b+a c x^2\right )+3 b^2 \log \left (1+c^2 x^4\right )\right )+3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^2\right )}\right )}{4 c^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.66 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.68

method	result	size
default	\(\frac {a^{3} x^{4}}{4}+\frac {b^{3} x^{4} \arctan \left (c \,x^{2}\right )^{3}}{4}-\frac {3 b^{3} \arctan \left (c \,x^{2}\right )^{2} x^{2}}{4 c}+\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3}}{4 c^{2}}+\frac {3 b^{3} \ln \left (c^{2} x^{4}+1\right ) \arctan \left (c \,x^{2}\right )}{4 c^{2}}-\frac {3 b^{3} \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (c^{2} x^{4}+1\right )-c \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{c \,\underline {\hspace {1.25 ex}}\alpha ^{3}}+\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2} \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right ) c -\ln \left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}+x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c +1\right )}\right )+\ln \left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}-x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c -1\right )}\right )\right )}{\underline {\hspace {1.25 ex}}\alpha }+\frac {2 \underline {\hspace {1.25 ex}}\alpha ^{2} \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right ) c -2 \operatorname {dilog}\left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}+x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c +1\right )}\right )+2 \operatorname {dilog}\left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}-x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c -1\right )}\right )}{\underline {\hspace {1.25 ex}}\alpha }\right )}{\underline {\hspace {1.25 ex}}\alpha ^{2}}\right )}{16 c^{3}}+\frac {3 a \,b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right ) x^{2}}{2 c}+\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 c^{2}}+\frac {3 a \,b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {3 a^{2} b \,x^{4} \arctan \left (c \,x^{2}\right )}{4}-\frac {3 a^{2} b \,x^{2}}{4 c}+\frac {3 a^{2} b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\)	\(399\)
parts	\(\frac {a^{3} x^{4}}{4}+\frac {b^{3} x^{4} \arctan \left (c \,x^{2}\right )^{3}}{4}-\frac {3 b^{3} \arctan \left (c \,x^{2}\right )^{2} x^{2}}{4 c}+\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3}}{4 c^{2}}+\frac {3 b^{3} \ln \left (c^{2} x^{4}+1\right ) \arctan \left (c \,x^{2}\right )}{4 c^{2}}-\frac {3 b^{3} \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (c^{2} x^{4}+1\right )-c \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{c \,\underline {\hspace {1.25 ex}}\alpha ^{3}}+\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2} \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right ) c -\ln \left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}+x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c +1\right )}\right )+\ln \left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}-x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c -1\right )}\right )\right )}{\underline {\hspace {1.25 ex}}\alpha }+\frac {2 \underline {\hspace {1.25 ex}}\alpha ^{2} \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right ) c -2 \operatorname {dilog}\left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}+x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c +1\right )}\right )+2 \operatorname {dilog}\left (\frac {c \,\underline {\hspace {1.25 ex}}\alpha ^{3}-x}{\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{2} c -1\right )}\right )}{\underline {\hspace {1.25 ex}}\alpha }\right )}{\underline {\hspace {1.25 ex}}\alpha ^{2}}\right )}{16 c^{3}}+\frac {3 a \,b^{2} x^{4} \arctan \left (c \,x^{2}\right )^{2}}{4}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right ) x^{2}}{2 c}+\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 c^{2}}+\frac {3 a \,b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {3 a^{2} b \,x^{4} \arctan \left (c \,x^{2}\right )}{4}-\frac {3 a^{2} b \,x^{2}}{4 c}+\frac {3 a^{2} b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\)	\(399\)
risch	\(\frac {3 a \,b^{2} \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}+\frac {3 i b^{2} \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (c \,\textit {\_Z}^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1, \operatorname {index} =1\right )\right )}{\sum }\frac {\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-i c \,x^{2}+1\right )+2 c \left (-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \left (\ln \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i \sqrt {\frac {i}{c}}+\sqrt {\frac {i}{c}}+x -\underline {\hspace {1.25 ex}}\alpha \right )}{\sqrt {\frac {i}{c}}}\right )+\ln \left (\frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i \sqrt {\frac {i}{c}}-\sqrt {\frac {i}{c}}-x +\underline {\hspace {1.25 ex}}\alpha \right )}{\sqrt {\frac {i}{c}}}\right )\right )}{2 c}-\frac {\operatorname {dilog}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (i \sqrt {\frac {i}{c}}+\sqrt {\frac {i}{c}}+x -\underline {\hspace {1.25 ex}}\alpha \right )}{\sqrt {\frac {i}{c}}}\right )+\operatorname {dilog}\left (\frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i \sqrt {\frac {i}{c}}-\sqrt {\frac {i}{c}}-x +\underline {\hspace {1.25 ex}}\alpha \right )}{\sqrt {\frac {i}{c}}}\right )}{2 c}\right )\right ) b}{c}\right )}{4 c}+\frac {a^{3} x^{4}}{4}-\frac {3 a \,b^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )^{2}}{16}-\frac {3 a \,b^{2} \ln \left (-i c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {3 i b \,a^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )}{8}+\frac {3 i b^{3} \ln \left (-i c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {3 b^{3} x^{2} \ln \left (-i c \,x^{2}+1\right )^{2}}{16 c}+\frac {i b^{3} \left (c^{2} x^{4}+1\right ) \ln \left (i c \,x^{2}+1\right )^{3}}{32 c^{2}}+\frac {3 i b^{3} \ln \left (c^{2} x^{4}+1\right )}{16 c^{2}}-\frac {3 a^{2} b \,x^{2}}{4 c}+\frac {3 a^{2} b \arctan \left (c \,x^{2}\right )}{4 c^{2}}-\frac {3 b^{3} \arctan \left (c \,x^{2}\right )}{8 c^{2}}-\frac {3 b^{2} \left (i b \,c^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )+2 a \,c^{2} x^{4}-2 b c \,x^{2}+i b \ln \left (-i c \,x^{2}+1\right )+2 i b +2 a \right ) \ln \left (i c \,x^{2}+1\right )^{2}}{32 c^{2}}-\frac {3 i a \,b^{2} x^{2} \ln \left (-i c \,x^{2}+1\right )}{4 c}-\frac {i b^{3} \ln \left (-i c \,x^{2}+1\right )^{3}}{32 c^{2}}-\frac {i b^{3} x^{4} \ln \left (-i c \,x^{2}+1\right )^{3}}{32}+\left (\frac {3 i b^{3} \left (c^{2} x^{4}+1\right ) \ln \left (-i c \,x^{2}+1\right )^{2}}{32 c^{2}}+\frac {3 b^{2} \left (2 c \,x^{2} a -b \right )^{2} \ln \left (-i c \,x^{2}+1\right )}{32 c^{2} a}-\frac {3 b \left (4 i a^{3} c^{2} x^{4}-8 i a^{2} b c \,x^{2}+4 i \ln \left (-i c \,x^{2}+1\right ) a \,b^{2}+4 i a \,b^{2}-4 \ln \left (-i c \,x^{2}+1\right ) a^{2} b +\ln \left (-i c \,x^{2}+1\right ) b^{3}\right )}{32 a \,c^{2}}\right ) \ln \left (i c \,x^{2}+1\right )\)	\(758\)

[In]

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

[Out]

1/4*a^3*x^4+1/4*b^3*x^4*arctan(c*x^2)^3-3/4*b^3*arctan(c*x^2)^2/c*x^2+1/4*b^3*arctan(c*x^2)^3/c^2+3/4*b^3/c^2*
ln(c^2*x^4+1)*arctan(c*x^2)-3/16*b^3/c^3*sum(1/_alpha^2*(2*ln(x-_alpha)*ln(c^2*x^4+1)-c*(1/c/_alpha^3*ln(x-_al
pha)^2+2/_alpha*ln(x-_alpha)*(_alpha^2*ln(1/2*(x+_alpha)/_alpha)*c-ln((_alpha^3*c+x)/_alpha/(_alpha^2*c+1))+ln
((_alpha^3*c-x)/_alpha/(_alpha^2*c-1)))+2/_alpha*(_alpha^2*dilog(1/2*(x+_alpha)/_alpha)*c-dilog((_alpha^3*c+x)
/_alpha/(_alpha^2*c+1))+dilog((_alpha^3*c-x)/_alpha/(_alpha^2*c-1))))),_alpha=RootOf(_Z^4*c^2+1))+3/4*a*b^2*x^
4*arctan(c*x^2)^2-3/2*a*b^2*arctan(c*x^2)/c*x^2+3/4*a*b^2/c^2*arctan(c*x^2)^2+3/4*a*b^2/c^2*ln(c^2*x^4+1)+3/4*
a^2*b*x^4*arctan(c*x^2)-3/4*a^2*b/c*x^2+3/4*a^2*b/c^2*arctan(c*x^2)

Fricas [F]

\[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int x^{3} \left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{3}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]

[In]

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

[Out]

3/4*a*b^2*x^4*arctan(c*x^2)^2 + 1/4*a^3*x^4 + 3/4*(x^4*arctan(c*x^2) - c*(x^2/c^2 - arctan(c*x^2)/c^3))*a^2*b
- 3/4*(2*c*(x^2/c^2 - arctan(c*x^2)/c^3)*arctan(c*x^2) + (arctan(c*x^2)^2 - log(4*c^5*x^4 + 4*c^3))/c^2)*a*b^2
+ 1/128*(4*x^4*arctan(c*x^2)^3 - 3*x^4*arctan(c*x^2)*log(c^2*x^4 + 1)^2 + 128*integrate(1/64*(12*c^2*x^7*arct
an(c*x^2)*log(c^2*x^4 + 1) - 12*c*x^5*arctan(c*x^2)^2 + 56*(c^2*x^7 + x^3)*arctan(c*x^2)^3 + 3*(c*x^5 + 2*(c^2
*x^7 + x^3)*arctan(c*x^2))*log(c^2*x^4 + 1)^2)/(c^2*x^4 + 1), x))*b^3

Giac [F]

\[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right )^3 \, dx=\int x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^3 \,d x \]

[In]

int(x^3*(a + b*atan(c*x^2))^3,x)

[Out]

int(x^3*(a + b*atan(c*x^2))^3, x)