Integrand size = 25, antiderivative size = 88 \[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=\frac {(a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}+\frac {b^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \]
(a+b*arctan(c*x))^2*ln(2-2/(1+I*c*x))/d+I*b*(a+b*arctan(c*x))*polylog(2,-1 +2/(1+I*c*x))/d+1/2*b^2*polylog(3,-1+2/(1+I*c*x))/d
Time = 0.51 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.97 \[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=-\frac {i \left (b^2 \pi ^3+24 a^2 \arctan (c x)+48 a b \arctan (c x)^2+24 i b^2 \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )+48 i a b \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+24 i a^2 \log (c x)-12 i a^2 \log \left (1+c^2 x^2\right )-24 b^2 \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+24 a b \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )+12 i b^2 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )\right )}{24 d} \]
((-1/24*I)*(b^2*Pi^3 + 24*a^2*ArcTan[c*x] + 48*a*b*ArcTan[c*x]^2 + (24*I)* b^2*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (48*I)*a*b*ArcTan[c*x] *Log[1 - E^((2*I)*ArcTan[c*x])] + (24*I)*a^2*Log[c*x] - (12*I)*a^2*Log[1 + c^2*x^2] - 24*b^2*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + 24*a*b *PolyLog[2, E^((2*I)*ArcTan[c*x])] + (12*I)*b^2*PolyLog[3, E^((-2*I)*ArcTa n[c*x])]))/d
Time = 0.47 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5403, 5529, 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.
\(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle \frac {\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{d}\) |
\(\Big \downarrow \) 5529 |
\(\displaystyle \frac {\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {2 b c \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}\right )}{d}\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {2 b c \left (-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{4 c}\right )}{d}\) |
((a + b*ArcTan[c*x])^2*Log[2 - 2/(1 + I*c*x)])/d - (2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/c - (b*PolyLog[3, -1 + 2/(1 + I*c*x)])/(4*c)))/d
3.1.99.3.1 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[(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 = 12.26 (sec) , antiderivative size = 1449, normalized size of antiderivative = 16.47
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1449\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1451\) |
default | \(\text {Expression too large to display}\) | \(1451\) |
a^2/d*ln(x)-1/2*a^2/d*ln(c^2*x^2+1)-I*a^2/d*arctan(c*x)+b^2/d*(arctan(c*x) ^2*ln(c*x)-arctan(c*x)^2*ln(c*x-I)+arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x ^2+1))-2/3*I*arctan(c*x)^3-arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+arc tan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*arctan(c*x)*polylog(2,(1+ I*c*x)/(c^2*x^2+1)^(1/2))+2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+arctan( c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*arctan(c*x)*polylog(2,-(1+I*c *x)/(c^2*x^2+1)^(1/2))+2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*Pi* (csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+ I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))-csgn(((1+I*c*x)^2/(c^ 2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2+csgn(I*((1+I*c*x)^2/(c^2*x^2+1) -1))*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1 )/(1+(1+I*c*x)^2/(c^2*x^2+1)))-csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I* ((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2-csgn(I/(1+(1+I* c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x ^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))+csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csg n((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2-csgn(I*((1+I*c*x) ^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2* x^2+1)))^2+csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1))) ^3-csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((( 1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2+csgn(((1+I*c*x...
Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=-\frac {b^{2} \log \left (\frac {2 \, c x}{c x - i}\right ) \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 2 \, b^{2} {\rm Li}_2\left (-\frac {2 \, c x}{c x - i} + 1\right ) \log \left (-\frac {c x + i}{c x - i}\right ) + 4 i \, a b {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) - 4 \, a^{2} \log \left (x\right ) + 4 \, a^{2} \log \left (\frac {c x - i}{c}\right ) - 2 \, b^{2} {\rm polylog}\left (3, -\frac {c x + i}{c x - i}\right )}{4 \, d} \]
-1/4*(b^2*log(2*c*x/(c*x - I))*log(-(c*x + I)/(c*x - I))^2 + 2*b^2*dilog(- 2*c*x/(c*x - I) + 1)*log(-(c*x + I)/(c*x - I)) + 4*I*a*b*dilog((c*x + I)/( c*x - I) + 1) - 4*a^2*log(x) + 4*a^2*log((c*x - I)/c) - 2*b^2*polylog(3, - (c*x + I)/(c*x - I)))/d
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=- \frac {i \left (\int \frac {a^{2}}{c x^{2} - i x}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{2} - i x}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c x^{2} - i x}\, dx\right )}{d} \]
-I*(Integral(a**2/(c*x**2 - I*x), x) + Integral(b**2*atan(c*x)**2/(c*x**2 - I*x), x) + Integral(2*a*b*atan(c*x)/(c*x**2 - I*x), x))/d
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x} \,d x } \]
-a^2*(log(I*c*x + 1)/d - log(x)/d) + 1/96*(-24*I*b^2*arctan(c*x)^3 + 12*b^ 2*arctan(c*x)^2*log(c^2*x^2 + 1) - 6*I*b^2*arctan(c*x)*log(c^2*x^2 + 1)^2 + 3*b^2*log(c^2*x^2 + 1)^3 - 2*(384*b^2*c^2*integrate(1/16*x^2*arctan(c*x) ^2/(c^2*d*x^3 + d*x), x) + 192*b^2*c*integrate(1/16*x*arctan(c*x)*log(c^2* x^2 + 1)/(c^2*d*x^3 + d*x), x) + b^2*log(c^2*x^2 + 1)^3/d - 576*b^2*integr ate(1/16*arctan(c*x)^2/(c^2*d*x^3 + d*x), x) - 48*b^2*integrate(1/16*log(c ^2*x^2 + 1)^2/(c^2*d*x^3 + d*x), x) - 1536*a*b*integrate(1/16*arctan(c*x)/ (c^2*d*x^3 + d*x), x))*d - 8*I*(b^2*arctan(c*x)^3/d - 12*b^2*c*integrate(1 /16*x*log(c^2*x^2 + 1)^2/(c^2*d*x^3 + d*x), x) + 12*a*b*arctan(c*x)^2/d + 48*b^2*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^3 + d*x), x))* d)/d
\[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]