Integrand size = 22, antiderivative size = 91 \[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=-\frac {i \arctan (a x)^3}{3 c}+\frac {\arctan (a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {i \arctan (a x) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{c}+\frac {\operatorname {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )}{2 c} \]
-1/3*I*arctan(a*x)^3/c+arctan(a*x)^2*ln(2-2/(1-I*a*x))/c-I*arctan(a*x)*pol ylog(2,-1+2/(1-I*a*x))/c+1/2*polylog(3,-1+2/(1-I*a*x))/c
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\frac {i \arctan (a x)^3}{3 c}+\frac {\arctan (a x)^2 \log \left (1-e^{-2 i \arctan (a x)}\right )}{c}+\frac {i \arctan (a x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (a x)}\right )}{c}+\frac {\operatorname {PolyLog}\left (3,e^{-2 i \arctan (a x)}\right )}{2 c} \]
((I/3)*ArcTan[a*x]^3)/c + (ArcTan[a*x]^2*Log[1 - E^((-2*I)*ArcTan[a*x])])/ c + (I*ArcTan[a*x]*PolyLog[2, E^((-2*I)*ArcTan[a*x])])/c + PolyLog[3, E^(( -2*I)*ArcTan[a*x])]/(2*c)
Time = 0.53 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5459, 5403, 5527, 7164}
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.
\(\displaystyle \int \frac {\arctan (a x)^2}{x \left (a^2 c x^2+c\right )} \, dx\) |
\(\Big \downarrow \) 5459 |
\(\displaystyle \frac {i \int \frac {\arctan (a x)^2}{x (a x+i)}dx}{c}-\frac {i \arctan (a x)^3}{3 c}\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle \frac {i \left (2 i a \int \frac {\arctan (a x) \log \left (2-\frac {2}{1-i a x}\right )}{a^2 x^2+1}dx-i \arctan (a x)^2 \log \left (2-\frac {2}{1-i a x}\right )\right )}{c}-\frac {i \arctan (a x)^3}{3 c}\) |
\(\Big \downarrow \) 5527 |
\(\displaystyle \frac {i \left (2 i a \left (\frac {i \arctan (a x) \operatorname {PolyLog}\left (2,\frac {2}{1-i a x}-1\right )}{2 a}-\frac {1}{2} i \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{1-i a x}-1\right )}{a^2 x^2+1}dx\right )-i \arctan (a x)^2 \log \left (2-\frac {2}{1-i a x}\right )\right )}{c}-\frac {i \arctan (a x)^3}{3 c}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {i \left (2 i a \left (\frac {i \arctan (a x) \operatorname {PolyLog}\left (2,\frac {2}{1-i a x}-1\right )}{2 a}-\frac {\operatorname {PolyLog}\left (3,\frac {2}{1-i a x}-1\right )}{4 a}\right )-i \arctan (a x)^2 \log \left (2-\frac {2}{1-i a x}\right )\right )}{c}-\frac {i \arctan (a x)^3}{3 c}\) |
((-1/3*I)*ArcTan[a*x]^3)/c + (I*((-I)*ArcTan[a*x]^2*Log[2 - 2/(1 - I*a*x)] + (2*I)*a*(((I/2)*ArcTan[a*x]*PolyLog[2, -1 + 2/(1 - I*a*x)])/a - PolyLog [3, -1 + 2/(1 - I*a*x)]/(4*a))))/c
3.3.87.3.1 Defintions of rubi rules used
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]
Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 ), x_Symbol] :> Simp[I*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x ] - Simp[b*p*(I/2) Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2* d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I + c*x)))^2, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 11.53 (sec) , antiderivative size = 1578, normalized size of antiderivative = 17.34
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1578\) |
default | \(\text {Expression too large to display}\) | \(1578\) |
parts | \(\text {Expression too large to display}\) | \(1989\) |
1/c*arctan(a*x)^2*ln(a*x)-1/2/c*arctan(a*x)^2*ln(a^2*x^2+1)-1/c*(-arctan(a *x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+arctan(a*x)^2*ln((1+I*a*x)^2/(a^2*x^ 2+1)-1)+1/3*I*arctan(a*x)^3-1/4*(-I*Pi*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2)) ^2*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))-I*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1 +I*a*x)^2/(a^2*x^2+1)+1)^2)^3-2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/(( 1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/( a^2*x^2+1)+1))^2-2*I*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2 *x^2+1)+1))^2-2*I*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x) ^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2+2*I*Pi*csgn(I*((1+I*a*x)^ 2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3+2*I*Pi*csgn(((1+I*a*x)^2/( a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3+2*I*Pi*csgn(I*((1+I*a*x)^2/(a ^2*x^2+1)-1))*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2 *x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))+2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^ 2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+ I*a*x)^2/(a^2*x^2+1)+1))+I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3+I*Pi *csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x) ^2/(a^2*x^2+1)+1)^2)^2+I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))^2*csgn(I*( (1+I*a*x)^2/(a^2*x^2+1)+1)^2)+2*I*Pi-2*I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1 )+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2+I*Pi*csgn(I/((1+I*a*x)^2/(a^ 2*x^2+1)+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+...
\[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
\[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{2} x^{3} + x}\, dx}{c} \]
\[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
\[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
Timed out. \[ \int \frac {\arctan (a x)^2}{x \left (c+a^2 c x^2\right )} \, dx=\int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \]