Integrand size = 16, antiderivative size = 149 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=-\frac {3}{4} i b c^2 \left (a+b \arctan \left (c x^2\right )\right )^2-\frac {3 b c \left (a+b \arctan \left (c x^2\right )\right )^2}{4 x^2}-\frac {1}{4} c^2 \left (a+b \arctan \left (c x^2\right )\right )^3-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{4 x^4}+\frac {3}{2} b^2 c^2 \left (a+b \arctan \left (c x^2\right )\right ) \log \left (2-\frac {2}{1-i c x^2}\right )-\frac {3}{4} i b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x^2}\right ) \]
-3/4*I*b*c^2*(a+b*arctan(c*x^2))^2-3/4*b*c*(a+b*arctan(c*x^2))^2/x^2-1/4*c ^2*(a+b*arctan(c*x^2))^3-1/4*(a+b*arctan(c*x^2))^3/x^4+3/2*b^2*c^2*(a+b*ar ctan(c*x^2))*ln(2-2/(1-I*c*x^2))-3/4*I*b^3*c^2*polylog(2,-1+2/(1-I*c*x^2))
Time = 0.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=-\frac {3 b^2 \left (a+a c^2 x^4+b c x^2 \left (1+i c x^2\right )\right ) \arctan \left (c x^2\right )^2+b^3 \left (1+c^2 x^4\right ) \arctan \left (c x^2\right )^3+3 b \arctan \left (c x^2\right ) \left (a \left (a+2 b c x^2+a c^2 x^4\right )-2 b^2 c^2 x^4 \log \left (1-e^{2 i \arctan \left (c x^2\right )}\right )\right )+a \left (a \left (a+3 b c x^2\right )-6 b^2 c^2 x^4 \log \left (\frac {c x^2}{\sqrt {1+c^2 x^4}}\right )\right )+3 i b^3 c^2 x^4 \operatorname {PolyLog}\left (2,e^{2 i \arctan \left (c x^2\right )}\right )}{4 x^4} \]
-1/4*(3*b^2*(a + a*c^2*x^4 + b*c*x^2*(1 + 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*c^2*x^4*Log[1 - E^((2*I)*ArcTan[c*x^2])]) + a*(a*(a + 3*b*c *x^2) - 6*b^2*c^2*x^4*Log[(c*x^2)/Sqrt[1 + c^2*x^4]]) + (3*I)*b^3*c^2*x^4* PolyLog[2, E^((2*I)*ArcTan[c*x^2])])/x^4
Time = 0.88 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5363, 5361, 5453, 5361, 5419, 5459, 5403, 2897}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx\) |
| \(\Big \downarrow \) 5363 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^6}dx^2\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^4 \left (c^2 x^4+1\right )}dx^2-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 5453 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (\int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^4}dx^2-c^2 \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{c^2 x^4+1}dx^2\right )-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (c^2 \left (-\int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{c^2 x^4+1}dx^2\right )+2 b c \int \frac {a+b \arctan \left (c x^2\right )}{x^2 \left (c^2 x^4+1\right )}dx^2-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 5419 |
| \(\displaystyle \frac {1}{2} \left (\frac {3}{2} b c \left (2 b c \int \frac {a+b \arctan \left (c x^2\right )}{x^2 \left (c^2 x^4+1\right )}dx^2-\frac {c \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}\right )\) |
| \(\Big \downarrow \) 5459 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}+\frac {3}{2} b c \left (2 b c \left (i \int \frac {a+b \arctan \left (c x^2\right )}{x^2 \left (c x^2+i\right )}dx^2-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )-\frac {c \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}+\frac {3}{2} b c \left (2 b c \left (i \left (i b c \int \frac {\log \left (2-\frac {2}{1-i c x^2}\right )}{c^2 x^4+1}dx^2-i \log \left (2-\frac {2}{1-i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )\right )-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )-\frac {c \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{2 x^4}+\frac {3}{2} b c \left (2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x^2}-1\right )\right )-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )-\frac {c \left (a+b \arctan \left (c x^2\right )\right )^3}{3 b}-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )\right )\) |
(-1/2*(a + b*ArcTan[c*x^2])^3/x^4 + (3*b*c*(-((a + b*ArcTan[c*x^2])^2/x^2) - (c*(a + b*ArcTan[c*x^2])^3)/(3*b) + 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x^ 2])^2)/b + I*((-I)*(a + b*ArcTan[c*x^2])*Log[2 - 2/(1 - I*c*x^2)] - (b*Pol yLog[2, -1 + 2/(1 - I*c*x^2)])/2))))/2)/2
3.1.90.3.1 Defintions of rubi rules used
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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] - Simp[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]
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[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[Simplif y[(m + 1)/n]]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> 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]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Simp[e/(d*f^2) 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] && LtQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si mp[I/d Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.42 (sec) , antiderivative size = 465, normalized size of antiderivative = 3.12
| method | result | size |
| default | \(-\frac {a^{3}}{4 x^{4}}-\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3}}{4 x^{4}}-\frac {3 b^{3} c \arctan \left (c \,x^{2}\right )^{2}}{4 x^{2}}-\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3} c^{2}}{4}+3 b^{3} c^{2} \arctan \left (c \,x^{2}\right ) \ln \left (x \right )-\frac {3 b^{3} c^{2} \arctan \left (c \,x^{2}\right ) \ln \left (c^{2} x^{4}+1\right )}{4}+\frac {3 b^{3} c \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}-\frac {3 b^{3} c \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}}\right )}{2}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 x^{4}}-\frac {3 a \,b^{2} c \arctan \left (c \,x^{2}\right )}{2 x^{2}}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2} c^{2}}{4}+3 a \,b^{2} c^{2} \ln \left (x \right )-\frac {3 a \,b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right )}{4}-\frac {3 a^{2} b \arctan \left (c \,x^{2}\right )}{4 x^{4}}-\frac {3 a^{2} b c}{4 x^{2}}-\frac {3 a^{2} b \,c^{2} \arctan \left (c \,x^{2}\right )}{4}\) | \(465\) |
| parts | \(-\frac {a^{3}}{4 x^{4}}-\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3}}{4 x^{4}}-\frac {3 b^{3} c \arctan \left (c \,x^{2}\right )^{2}}{4 x^{2}}-\frac {b^{3} \arctan \left (c \,x^{2}\right )^{3} c^{2}}{4}+3 b^{3} c^{2} \arctan \left (c \,x^{2}\right ) \ln \left (x \right )-\frac {3 b^{3} c^{2} \arctan \left (c \,x^{2}\right ) \ln \left (c^{2} x^{4}+1\right )}{4}+\frac {3 b^{3} c \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}-\frac {3 b^{3} c \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}}\right )}{2}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 x^{4}}-\frac {3 a \,b^{2} c \arctan \left (c \,x^{2}\right )}{2 x^{2}}-\frac {3 a \,b^{2} \arctan \left (c \,x^{2}\right )^{2} c^{2}}{4}+3 a \,b^{2} c^{2} \ln \left (x \right )-\frac {3 a \,b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right )}{4}-\frac {3 a^{2} b \arctan \left (c \,x^{2}\right )}{4 x^{4}}-\frac {3 a^{2} b c}{4 x^{2}}-\frac {3 a^{2} b \,c^{2} \arctan \left (c \,x^{2}\right )}{4}\) | \(465\) |
-1/4*a^3/x^4-1/4*b^3/x^4*arctan(c*x^2)^3-3/4*b^3*c*arctan(c*x^2)^2/x^2-1/4 *b^3*arctan(c*x^2)^3*c^2+3*b^3*c^2*arctan(c*x^2)*ln(x)-3/4*b^3*c^2*arctan( c*x^2)*ln(c^2*x^4+1)+3/16*b^3*c*sum(1/_alpha^2*(2*ln(x-_alpha)*ln(c^2*x^4+ 1)-c*(1/c/_alpha^3*ln(x-_alpha)^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)/_alph a)*c-dilog((_alpha^3*c+x)/_alpha/(_alpha^2*c+1))+dilog((_alpha^3*c-x)/_alp ha/(_alpha^2*c-1))))),_alpha=RootOf(_Z^4*c^2+1))-3/2*b^3*c*sum(1/_R1^2*(ln (x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1)),_R1=RootOf(_Z^4*c^2+1))-3/4*a*b^2/ x^4*arctan(c*x^2)^2-3/2*a*b^2*c*arctan(c*x^2)/x^2-3/4*a*b^2*arctan(c*x^2)^ 2*c^2+3*a*b^2*c^2*ln(x)-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)
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
integral((b^3*arctan(c*x^2)^3 + 3*a*b^2*arctan(c*x^2)^2 + 3*a^2*b*arctan(c *x^2) + a^3)/x^5, x)
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{3}}{x^{5}}\, dx \]
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
-3/4*((c*arctan(c*x^2) + 1/x^2)*c + arctan(c*x^2)/x^4)*a^2*b + 3/4*((arcta n(c*x^2)^2 - log(c^2*x^4 + 1) + 4*log(x))*c^2 - 2*(c*arctan(c*x^2) + 1/x^2 )*c*arctan(c*x^2))*a*b^2 - 3/4*a*b^2*arctan(c*x^2)^2/x^4 + 1/128*(128*x^4* integrate(-1/64*(12*c^2*x^4*arctan(c*x^2)*log(c^2*x^4 + 1) - 12*c*x^2*arct an(c*x^2)^2 - 56*(c^2*x^4 + 1)*arctan(c*x^2)^3 + 3*(c*x^2 - 2*(c^2*x^4 + 1 )*arctan(c*x^2))*log(c^2*x^4 + 1)^2)/(c^2*x^9 + x^5), x) - 4*arctan(c*x^2) ^3 + 3*arctan(c*x^2)*log(c^2*x^4 + 1)^2)*b^3/x^4 - 1/4*a^3/x^4
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^3}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^3}{x^5} \,d x \]