Integrand size = 23, antiderivative size = 228 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=-i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+2 b c d (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
-I*c*d*(a+b*arctan(c*x))^2-d*(a+b*arctan(c*x))^2/x-2*I*c*d*(a+b*arctan(c*x ))^2*arctanh(-1+2/(1+I*c*x))+2*b*c*d*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))-I *b^2*c*d*polylog(2,-1+2/(1-I*c*x))+b*c*d*(a+b*arctan(c*x))*polylog(2,1-2/( 1+I*c*x))-b*c*d*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-1/2*I*b^2*c*d* polylog(3,1-2/(1+I*c*x))+1/2*I*b^2*c*d*polylog(3,-1+2/(1+I*c*x))
Time = 0.32 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.27 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\frac {i d \left (i a^2+a^2 c x \log (x)+i a b \left (2 \arctan (c x)+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )\right )+i b^2 \left (\arctan (c x)^2-2 c x \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+i c x \left (\arctan (c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )\right )+i a b c x (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+\frac {1}{24} b^2 c x \left (-i \pi ^3+16 i \arctan (c x)^3+24 \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-24 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-12 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )\right )}{x} \]
(I*d*(I*a^2 + a^2*c*x*Log[x] + I*a*b*(2*ArcTan[c*x] + c*x*(-2*Log[c*x] + L og[1 + c^2*x^2])) + I*b^2*(ArcTan[c*x]^2 - 2*c*x*ArcTan[c*x]*Log[1 - E^((2 *I)*ArcTan[c*x])] + I*c*x*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x] )])) + I*a*b*c*x*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*c*x*((- I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 24*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan [c*x])] - 24*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*ArcTan[ c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E ^((2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])] - 12*PolyLog [3, -E^((2*I)*ArcTan[c*x])]))/24))/x
Time = 0.64 (sec) , antiderivative size = 228, 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.
\(\displaystyle \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx\) |
\(\Big \downarrow \) 5411 |
\(\displaystyle \int \left (\frac {d (a+b \arctan (c x))^2}{x^2}+\frac {i c d (a+b \arctan (c x))^2}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 i c d \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2+b c d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))-b c d \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))-i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-i b^2 c d \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )-\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )\) |
(-I)*c*d*(a + b*ArcTan[c*x])^2 - (d*(a + b*ArcTan[c*x])^2)/x + (2*I)*c*d*( a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + 2*b*c*d*(a + b*ArcTan[c* x])*Log[2 - 2/(1 - I*c*x)] - I*b^2*c*d*PolyLog[2, -1 + 2/(1 - I*c*x)] + b* c*d*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] - b*c*d*(a + b*ArcTa n[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - (I/2)*b^2*c*d*PolyLog[3, 1 - 2/(1 + I*c*x)] + (I/2)*b^2*c*d*PolyLog[3, -1 + 2/(1 + I*c*x)]
3.1.73.3.1 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 = 15.89 (sec) , antiderivative size = 5609, normalized size of antiderivative = 24.60
method | result | size |
parts | \(\text {Expression too large to display}\) | \(5609\) |
derivativedivides | \(\text {Expression too large to display}\) | \(5613\) |
default | \(\text {Expression too large to display}\) | \(5613\) |
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]
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^2, x)
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=i d \left (\int \left (- \frac {i a^{2}}{x^{2}}\right )\, dx + \int \frac {a^{2} c}{x}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \frac {b^{2} c \operatorname {atan}^{2}{\left (c x \right )}}{x}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \frac {2 a b c \operatorname {atan}{\left (c x \right )}}{x}\, dx\right ) \]
I*d*(Integral(-I*a**2/x**2, x) + Integral(a**2*c/x, x) + Integral(-I*b**2* atan(c*x)**2/x**2, x) + Integral(b**2*c*atan(c*x)**2/x, x) + Integral(-2*I *a*b*atan(c*x)/x**2, x) + Integral(2*a*b*c*atan(c*x)/x, x))
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]
I*a^2*c*d*log(x) - (c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b *d - a^2*d/x - 1/96*(24*b^2*d*arctan(c*x)^2 + 24*I*b^2*d*arctan(c*x)*log(c ^2*x^2 + 1) - 6*b^2*d*log(c^2*x^2 + 1)^2 - 24*(b^2*c*d*arctan(c*x)^3 + 16* b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 4*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x ) - 16*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 16*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 32*b^2*c*d*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^2), x) + 48*b^2 *d*integrate(1/16*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 4*b^2*d*integrate(1/ 16*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x))*x - I*(1152*b^2*c^3*d*integrate (1/16*x^3*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 3072*a*b*c^3*d*integrate(1/1 6*x^3*arctan(c*x)/(c^2*x^4 + x^2), x) + b^2*c*d*log(c^2*x^2 + 1)^3 + 24*b^ 2*c*d*arctan(c*x)^2 - 384*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)*log(c^2 *x^2 + 1)/(c^2*x^4 + x^2), x) + 1152*b^2*c*d*integrate(1/16*x*arctan(c*x)^ 2/(c^2*x^4 + x^2), x) + 96*b^2*c*d*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^ 2*x^4 + x^2), x) + 3072*a*b*c*d*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^ 2), x) + 384*b^2*c*d*integrate(1/16*x*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 384*b^2*d*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x))*x)/x
Timed out. \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x^2} \,d x \]