Integrand size = 16, antiderivative size = 97 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=-\frac {1}{2} i c \left (a+b \arctan \left (c x^2\right )\right )^2-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{2 x^2}+b c \left (a+b \arctan \left (c x^2\right )\right ) \log \left (2-\frac {2}{1-i c x^2}\right )-\frac {1}{2} i b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x^2}\right ) \]
-1/2*I*c*(a+b*arctan(c*x^2))^2-1/2*(a+b*arctan(c*x^2))^2/x^2+b*c*(a+b*arct an(c*x^2))*ln(2-2/(1-I*c*x^2))-1/2*I*b^2*c*polylog(2,-1+2/(1-I*c*x^2))
Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=-\frac {a^2}{2 x^2}+a b c \left (-\frac {\arctan \left (c x^2\right )}{c x^2}+\log \left (c x^2\right )-\frac {1}{2} \log \left (1+c^2 x^4\right )\right )+\frac {1}{2} b^2 c \left (-\frac {\arctan \left (c x^2\right )^2}{c x^2}+2 \arctan \left (c x^2\right ) \log \left (1-e^{2 i \arctan \left (c x^2\right )}\right )-i \left (\arctan \left (c x^2\right )^2+\operatorname {PolyLog}\left (2,e^{2 i \arctan \left (c x^2\right )}\right )\right )\right ) \]
-1/2*a^2/x^2 + a*b*c*(-(ArcTan[c*x^2]/(c*x^2)) + Log[c*x^2] - Log[1 + c^2* x^4]/2) + (b^2*c*(-(ArcTan[c*x^2]^2/(c*x^2)) + 2*ArcTan[c*x^2]*Log[1 - E^( (2*I)*ArcTan[c*x^2])] - I*(ArcTan[c*x^2]^2 + PolyLog[2, E^((2*I)*ArcTan[c* x^2])])))/2
Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5363, 5361, 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.
| \(\displaystyle \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx\) |
| \(\Big \downarrow \) 5363 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^4}dx^2\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {1}{2} \left (2 b c \int \frac {a+b \arctan \left (c x^2\right )}{x^2 \left (c^2 x^4+1\right )}dx^2-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}\right )\) |
| \(\Big \downarrow \) 5459 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}+2 b c \left (i \int \frac {a+b \arctan \left (c x^2\right )}{x^2 \left (c x^2+i\right )}dx^2-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )\right )\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}+2 b c \left (i \left (i b c \int \frac {\log \left (2-\frac {2}{1-i c x^2}\right )}{c^2 x^4+1}dx^2-i \log \left (2-\frac {2}{1-i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )\right )-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )\right )\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^2}+2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x^2}\right ) \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x^2}-1\right )\right )-\frac {i \left (a+b \arctan \left (c x^2\right )\right )^2}{2 b}\right )\right )\) |
(-((a + b*ArcTan[c*x^2])^2/x^2) + 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*PolyLog[ 2, -1 + 2/(1 - I*c*x^2)])/2)))/2
3.1.79.3.1 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_.)/((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 = 3.33 (sec) , antiderivative size = 339, normalized size of antiderivative = 3.49
| method | result | size |
| default | \(-\frac {a^{2}}{2 x^{2}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{2 x^{2}}+2 b^{2} c \arctan \left (c \,x^{2}\right ) \ln \left (x \right )-\frac {b^{2} \arctan \left (c \,x^{2}\right ) \ln \left (c^{2} x^{4}+1\right ) c}{2}+\frac {b^{2} \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 )}{8}-b^{2} \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 a b \left (-\frac {\arctan \left (c \,x^{2}\right )}{2 x^{2}}+c \left (\ln \left (x \right )-\frac {\ln \left (c^{2} x^{4}+1\right )}{4}\right )\right )\) | \(339\) |
| parts | \(-\frac {a^{2}}{2 x^{2}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{2 x^{2}}+2 b^{2} c \arctan \left (c \,x^{2}\right ) \ln \left (x \right )-\frac {b^{2} \arctan \left (c \,x^{2}\right ) \ln \left (c^{2} x^{4}+1\right ) c}{2}+\frac {b^{2} \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 )}{8}-b^{2} \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 a b \left (-\frac {\arctan \left (c \,x^{2}\right )}{2 x^{2}}+c \left (\ln \left (x \right )-\frac {\ln \left (c^{2} x^{4}+1\right )}{4}\right )\right )\) | \(339\) |
-1/2*a^2/x^2-1/2*b^2/x^2*arctan(c*x^2)^2+2*b^2*c*arctan(c*x^2)*ln(x)-1/2*b ^2*arctan(c*x^2)*ln(c^2*x^4+1)*c+1/8*b^2*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+_alp ha)/_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))-b^2*sum(1/_R1^2* (ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1)),_R1=RootOf(_Z^4*c^2+1))+2*a*b*( -1/2/x^2*arctan(c*x^2)+c*(ln(x)-1/4*ln(c^2*x^4+1)))
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{2}}{x^{3}}\, dx \]
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
-1/2*(c*(log(c^2*x^4 + 1) - log(x^4)) + 2*arctan(c*x^2)/x^2)*a*b + 1/32*(3 2*x^2*integrate(-1/16*(4*c^2*x^4*log(c^2*x^4 + 1) - 8*c*x^2*arctan(c*x^2) - 12*(c^2*x^4 + 1)*arctan(c*x^2)^2 - (c^2*x^4 + 1)*log(c^2*x^4 + 1)^2)/(c^ 2*x^7 + x^3), x) - 4*arctan(c*x^2)^2 + log(c^2*x^4 + 1)^2)*b^2/x^2 - 1/2*a ^2/x^2
\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^3} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^2}{x^3} \,d x \]