Integrand size = 23, antiderivative size = 216 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=-d (a+b \arctan (c x))^2+i c d x (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+2 i b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )-b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
-d*(a+b*arctan(c*x))^2+I*c*d*x*(a+b*arctan(c*x))^2-2*d*(a+b*arctan(c*x))^2 *arctanh(-1+2/(1+I*c*x))+2*I*b*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))-b^2*d*p olylog(2,1-2/(1+I*c*x))-I*b*d*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))+I *b*d*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-1/2*b^2*d*polylog(3,1-2/( 1+I*c*x))+1/2*b^2*d*polylog(3,-1+2/(1+I*c*x))
Time = 0.30 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.26 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=d \left (i a^2 c x+a^2 \log (c x)+i a b \left (2 c x \arctan (c x)-\log \left (1+c^2 x^2\right )\right )+b^2 \left (\arctan (c x) \left ((1+i c x) \arctan (c x)+2 i \log \left (1+e^{2 i \arctan (c x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )+i a b (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+b^2 \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \arctan (c x)^3+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-\arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )\right ) \]
d*(I*a^2*c*x + a^2*Log[c*x] + I*a*b*(2*c*x*ArcTan[c*x] - Log[1 + c^2*x^2]) + b^2*(ArcTan[c*x]*((1 + I*c*x)*ArcTan[c*x] + (2*I)*Log[1 + E^((2*I)*ArcT an[c*x])]) + PolyLog[2, -E^((2*I)*ArcTan[c*x])]) + I*a*b*(PolyLog[2, (-I)* c*x] - PolyLog[2, I*c*x]) + b^2*((-1/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^ ((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*Ar cTan[c*x])]/2 - PolyLog[3, -E^((2*I)*ArcTan[c*x])]/2))
Time = 0.61 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {5411, 2009}
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.
| \(\displaystyle \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (\frac {d (a+b \arctan (c x))^2}{x}+i c d (a+b \arctan (c x))^2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 d \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-i b d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))+i b d \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))-d (a+b \arctan (c x))^2+i c d x (a+b \arctan (c x))^2+2 i b d \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))+b^2 (-d) \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )\) |
-(d*(a + b*ArcTan[c*x])^2) + I*c*d*x*(a + b*ArcTan[c*x])^2 + 2*d*(a + b*Ar cTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (2*I)*b*d*(a + b*ArcTan[c*x])*Lo g[2/(1 + I*c*x)] - b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)] - I*b*d*(a + b*ArcT an[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d*(a + b*ArcTan[c*x])*PolyLog [2, -1 + 2/(1 + I*c*x)] - (b^2*d*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d *PolyLog[3, -1 + 2/(1 + I*c*x)])/2
3.1.72.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 26.06 (sec) , antiderivative size = 3480, normalized size of antiderivative = 16.11
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3480\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(3482\) |
| default | \(\text {Expression too large to display}\) | \(3482\) |
a^2*d*(I*c*x+ln(x))+d*b^2*(-1/2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+2*poly log(3,(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*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+arctan(c*x)^2+I*arctan(c*x)*pol ylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+di log(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x ^2+1)^(1/2))+I*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*Pi*cs gn(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))*csgn (I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*(I*arctan(c*x) *ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x ^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1-I*(1+I*c*x)/(c ^2*x^2+1)^(1/2)))-arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+1/2*Pi*arcta n(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*Pi*arctan(c*x)*ln(1-I*(1+I* c*x)/(c^2*x^2+1)^(1/2))-1/2*Pi*arctan(c*x)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))+a rctan(c*x)^2*ln(c*x)+arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+arcta n(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/4*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)))^3*(2*I*arctan(c*x)*ln(1+(1+I*c*x )^2/(c^2*x^2+1))+2*arctan(c*x)^2+polylog(2,-(1+I*c*x)^2/(c^2*x^2+1)))-1/2* I*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*(I*ar ctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)*ln(1-I*(1+I*c* x)/(c^2*x^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1-I*...
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
integral(1/4*(4*I*a^2*c*d*x + 4*a^2*d + (-I*b^2*c*d*x - b^2*d)*log(-(c*x + I)/(c*x - I))^2 - 4*(a*b*c*d*x - I*a*b*d)*log(-(c*x + I)/(c*x - I)))/x, x )
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=i d \left (\int a^{2} c\, dx + \int \left (- \frac {i a^{2}}{x}\right )\, dx + \int b^{2} c \operatorname {atan}^{2}{\left (c x \right )}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x}\right )\, dx + \int 2 a b c \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x}\right )\, dx\right ) \]
I*d*(Integral(a**2*c, x) + Integral(-I*a**2/x, x) + Integral(b**2*c*atan(c *x)**2, x) + Integral(-I*b**2*atan(c*x)**2/x, x) + Integral(2*a*b*c*atan(c *x), x) + Integral(-2*I*a*b*atan(c*x)/x, x))
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
1/4*I*b^2*c*d*x*arctan(c*x)^2 + 12*I*b^2*c^3*d*integrate(1/16*x^3*arctan(c *x)^2/(c^2*x^3 + x), x) + 4*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)*log(c ^2*x^2 + 1)/(c^2*x^3 + x), x) + I*b^2*c^3*d*integrate(1/16*x^3*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 8*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)/(c^ 2*x^3 + x), x) + 4*I*b^2*c^3*d*integrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^ 3 + x), x) - 1/4*b^2*c*d*x*arctan(c*x)*log(c^2*x^2 + 1) - 1/16*I*b^2*c*d*x *log(c^2*x^2 + 1)^2 + 1/4*I*b^2*d*arctan(c*x)^3 + 12*b^2*c^2*d*integrate(1 /16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) - 4*I*b^2*c^2*d*integrate(1/16*x^2 *arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 32*a*b*c^2*d*integrate(1 /16*x^2*arctan(c*x)/(c^2*x^3 + x), x) - 8*I*b^2*c^2*d*integrate(1/16*x^2*a rctan(c*x)/(c^2*x^3 + x), x) + 1/96*b^2*d*log(c^2*x^2 + 1)^3 + I*a^2*c*d*x + 4*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + I*b^2*c*d*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 1/1 6*b^2*d*log(c^2*x^2 + 1)^2 + I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b* d + 12*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) - 4*I*b^2*d*in tegrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + b^2*d*integr ate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*d*integrate(1/16*ar ctan(c*x)/(c^2*x^3 + x), x) + a^2*d*log(x)
Timed out. \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x} \,d x \]