Integrand size = 25, antiderivative size = 383 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {b^2}{16 c^4 d^3 (i-c x)^2}+\frac {21 i b^2}{16 c^4 d^3 (i-c x)}-\frac {21 i b^2 \arctan (c x)}{16 c^4 d^3}+\frac {i b (a+b \arctan (c x))}{4 c^4 d^3 (i-c x)^2}-\frac {11 b (a+b \arctan (c x))}{4 c^4 d^3 (i-c x)}+\frac {3 (a+b \arctan (c x))^2}{8 c^4 d^3}+\frac {i x (a+b \arctan (c x))^2}{c^3 d^3}-\frac {(a+b \arctan (c x))^2}{2 c^4 d^3 (i-c x)^2}-\frac {3 i (a+b \arctan (c x))^2}{c^4 d^3 (i-c x)}+\frac {2 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d^3} \]
1/16*b^2/c^4/d^3/(I-c*x)^2+21/16*I*b^2/c^4/d^3/(I-c*x)-21/16*I*b^2*arctan( c*x)/c^4/d^3+1/4*I*b*(a+b*arctan(c*x))/c^4/d^3/(I-c*x)^2-11/4*b*(a+b*arcta n(c*x))/c^4/d^3/(I-c*x)+3/8*(a+b*arctan(c*x))^2/c^4/d^3+I*x*(a+b*arctan(c* x))^2/c^3/d^3-1/2*(a+b*arctan(c*x))^2/c^4/d^3/(I-c*x)^2-3*I*(a+b*arctan(c* x))^2/c^4/d^3/(I-c*x)+2*I*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4/d^3+3*(a +b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^4/d^3-b^2*polylog(2,1-2/(1+I*c*x))/c^4 /d^3+3*I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^4/d^3+3/2*b^2*poly log(3,1-2/(1+I*c*x))/c^4/d^3
Time = 1.25 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {64 i a^2 c x-\frac {32 a^2}{(-i+c x)^2}+\frac {192 i a^2}{-i+c x}-192 i a^2 \arctan (c x)-96 a^2 \log \left (1+c^2 x^2\right )+4 i a b \left (-96 \arctan (c x)^2+20 \cos (2 \arctan (c x))-\cos (4 \arctan (c x))-16 \log \left (1+c^2 x^2\right )-48 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-20 i \sin (2 \arctan (c x))+4 \arctan (c x) \left (8 c x+10 i \cos (2 \arctan (c x))-i \cos (4 \arctan (c x))-24 i \log \left (1+e^{2 i \arctan (c x)}\right )+10 \sin (2 \arctan (c x))-\sin (4 \arctan (c x))\right )+i \sin (4 \arctan (c x))\right )+i b^2 \left (-64 i \arctan (c x)^2+64 c x \arctan (c x)^2-128 \arctan (c x)^3-40 i \cos (2 \arctan (c x))+80 \arctan (c x) \cos (2 \arctan (c x))+80 i \arctan (c x)^2 \cos (2 \arctan (c x))+i \cos (4 \arctan (c x))-4 \arctan (c x) \cos (4 \arctan (c x))-8 i \arctan (c x)^2 \cos (4 \arctan (c x))+128 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-192 i \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-64 (i+3 \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-96 i \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )-40 \sin (2 \arctan (c x))-80 i \arctan (c x) \sin (2 \arctan (c x))+80 \arctan (c x)^2 \sin (2 \arctan (c x))+\sin (4 \arctan (c x))+4 i \arctan (c x) \sin (4 \arctan (c x))-8 \arctan (c x)^2 \sin (4 \arctan (c x))\right )}{64 c^4 d^3} \]
((64*I)*a^2*c*x - (32*a^2)/(-I + c*x)^2 + ((192*I)*a^2)/(-I + c*x) - (192* I)*a^2*ArcTan[c*x] - 96*a^2*Log[1 + c^2*x^2] + (4*I)*a*b*(-96*ArcTan[c*x]^ 2 + 20*Cos[2*ArcTan[c*x]] - Cos[4*ArcTan[c*x]] - 16*Log[1 + c^2*x^2] - 48* PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (20*I)*Sin[2*ArcTan[c*x]] + 4*ArcTan[ c*x]*(8*c*x + (10*I)*Cos[2*ArcTan[c*x]] - I*Cos[4*ArcTan[c*x]] - (24*I)*Lo g[1 + E^((2*I)*ArcTan[c*x])] + 10*Sin[2*ArcTan[c*x]] - Sin[4*ArcTan[c*x]]) + I*Sin[4*ArcTan[c*x]]) + I*b^2*((-64*I)*ArcTan[c*x]^2 + 64*c*x*ArcTan[c* x]^2 - 128*ArcTan[c*x]^3 - (40*I)*Cos[2*ArcTan[c*x]] + 80*ArcTan[c*x]*Cos[ 2*ArcTan[c*x]] + (80*I)*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + I*Cos[4*ArcTan[ c*x]] - 4*ArcTan[c*x]*Cos[4*ArcTan[c*x]] - (8*I)*ArcTan[c*x]^2*Cos[4*ArcTa n[c*x]] + 128*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (192*I)*ArcTan[ c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - 64*(I + 3*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (96*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])] - 40*S in[2*ArcTan[c*x]] - (80*I)*ArcTan[c*x]*Sin[2*ArcTan[c*x]] + 80*ArcTan[c*x] ^2*Sin[2*ArcTan[c*x]] + Sin[4*ArcTan[c*x]] + (4*I)*ArcTan[c*x]*Sin[4*ArcTa n[c*x]] - 8*ArcTan[c*x]^2*Sin[4*ArcTan[c*x]]))/(64*c^4*d^3)
Time = 0.85 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
\(\displaystyle \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx\) |
\(\Big \downarrow \) 5411 |
\(\displaystyle \int \left (-\frac {3 (a+b \arctan (c x))^2}{c^3 d^3 (c x-i)}-\frac {3 i (a+b \arctan (c x))^2}{c^3 d^3 (c x-i)^2}+\frac {i (a+b \arctan (c x))^2}{c^3 d^3}+\frac {(a+b \arctan (c x))^2}{c^3 d^3 (c x-i)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^4 d^3}-\frac {11 b (a+b \arctan (c x))}{4 c^4 d^3 (-c x+i)}+\frac {i b (a+b \arctan (c x))}{4 c^4 d^3 (-c x+i)^2}-\frac {3 i (a+b \arctan (c x))^2}{c^4 d^3 (-c x+i)}-\frac {(a+b \arctan (c x))^2}{2 c^4 d^3 (-c x+i)^2}+\frac {3 (a+b \arctan (c x))^2}{8 c^4 d^3}+\frac {2 i b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^4 d^3}+\frac {3 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^4 d^3}+\frac {i x (a+b \arctan (c x))^2}{c^3 d^3}-\frac {21 i b^2 \arctan (c x)}{16 c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^4 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 c^4 d^3}+\frac {21 i b^2}{16 c^4 d^3 (-c x+i)}+\frac {b^2}{16 c^4 d^3 (-c x+i)^2}\) |
b^2/(16*c^4*d^3*(I - c*x)^2) + (((21*I)/16)*b^2)/(c^4*d^3*(I - c*x)) - ((( 21*I)/16)*b^2*ArcTan[c*x])/(c^4*d^3) + ((I/4)*b*(a + b*ArcTan[c*x]))/(c^4* d^3*(I - c*x)^2) - (11*b*(a + b*ArcTan[c*x]))/(4*c^4*d^3*(I - c*x)) + (3*( a + b*ArcTan[c*x])^2)/(8*c^4*d^3) + (I*x*(a + b*ArcTan[c*x])^2)/(c^3*d^3) - (a + b*ArcTan[c*x])^2/(2*c^4*d^3*(I - c*x)^2) - ((3*I)*(a + b*ArcTan[c*x ])^2)/(c^4*d^3*(I - c*x)) + ((2*I)*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x) ])/(c^4*d^3) + (3*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d^3) - (b ^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^3) + ((3*I)*b*(a + b*ArcTan[c*x]) *PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^3) + (3*b^2*PolyLog[3, 1 - 2/(1 + I *c*x)])/(2*c^4*d^3)
3.2.12.3.1 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 = 8.35 (sec) , antiderivative size = 4306, normalized size of antiderivative = 11.24
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(4306\) |
default | \(\text {Expression too large to display}\) | \(4306\) |
parts | \(\text {Expression too large to display}\) | \(4365\) |
1/c^4*(3*I*a*b/d^3*dilog(-1/2*I*(c*x+I))-1/2*a^2/d^3/(c*x-I)^2-3/2*I*a*b/d ^3*ln(c*x-I)^2-3/2*a^2/d^3*ln(c^2*x^2+1)+3/32*I*a*b/d^3*ln(c^4*x^4+10*c^2* x^2+9)+b^2/d^3*(3/2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+19/16*polylog(2,-( 1+I*c*x)^2/(c^2*x^2+1))+19/8*arctan(c*x)^2-3*I*arctan(c*x)*polylog(2,-(1+I *c*x)^2/(c^2*x^2+1))-2*I*arctan(c*x)^3-3/8*dilog(1+I*(1+I*c*x)/(c^2*x^2+1) ^(1/2))-3/8*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*Pi*arctan(c*x)*ln(1+I *(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*Pi*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1 )^(1/2))-3*Pi*arctan(c*x)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))+I*arctan(c*x)^2*c* x+3*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-3*arctan(c*x)^2*ln(c*x-I )-5/8*(c*x+I)/(c*x-I)-3/2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/( c^2*x^2+1)))^3*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*Pi*csgn ((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)*ln(1+( 1+I*c*x)^2/(c^2*x^2+1))-3*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/( c^2*x^2+1)))^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*Pi*csgn(( 1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)*ln(1+(1+ I*c*x)^2/(c^2*x^2+1))-3*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^ 2*x^2+1)))^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*Pi*csgn(( 1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)*ln(1+I*( 1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*Pi*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*arctan(c*x)*ln...
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \]
integral(1/4*(-I*b^2*x^3*log(-(c*x + I)/(c*x - I))^2 - 4*a*b*x^3*log(-(c*x + I)/(c*x - I)) + 4*I*a^2*x^3)/(c^3*d^3*x^3 - 3*I*c^2*d^3*x^2 - 3*c*d^3*x + I*d^3), x)
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \]
1/128*(128*I*a^2*c^3*x^3 + 32*a^2*c^2*x^2*(3*I*arctan2(1, c*x) + 8) + 64*a ^2*c*x*(3*arctan2(1, c*x) + 4*I) + 96*(-I*b^2*c^2*x^2 - 2*b^2*c*x + I*b^2) *arctan(c*x)^3 + 12*(b^2*c^2*x^2 - 2*I*b^2*c*x - b^2)*log(c^2*x^2 + 1)^3 + 32*a^2*(-3*I*arctan2(1, c*x) + 10) + 16*(2*I*b^2*c^3*x^3 + 4*b^2*c^2*x^2 + 4*I*b^2*c*x + 5*b^2)*arctan(c*x)^2 - 4*(2*I*b^2*c^3*x^3 + 4*b^2*c^2*x^2 + 4*I*b^2*c*x + 5*b^2 - 6*(-I*b^2*c^2*x^2 - 2*b^2*c*x + I*b^2)*arctan(c*x) )*log(c^2*x^2 + 1)^2 - 18*(b^2*c^7*d^3*x^2 - 2*I*b^2*c^6*d^3*x - b^2*c^5*d ^3)*(((8*c^2*x^2 + 7)*c^2/(c^15*d^3*x^4 + 2*c^13*d^3*x^2 + c^11*d^3) + 2*( 4*c^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^13*d^3*x^4 + 2*c^11*d^3*x^2 + c^9*d^3)) *c^4 + 2*(2*c^2*x^2 + 1)*c^2*log(c^2*x^2 + 1)^2/(c^11*d^3*x^4 + 2*c^9*d^3* x^2 + c^7*d^3) - c^2*(c^2/(c^13*d^3*x^4 + 2*c^11*d^3*x^2 + c^9*d^3) + 2*lo g(c^2*x^2 + 1)/(c^11*d^3*x^4 + 2*c^9*d^3*x^2 + c^7*d^3)) - 4096*c^2*integr ate(1/128*x^3*arctan(c*x)^2/(c^9*d^3*x^6 + 3*c^7*d^3*x^4 + 3*c^5*d^3*x^2 + c^3*d^3), x) - 2*log(c^2*x^2 + 1)^2/(c^9*d^3*x^4 + 2*c^7*d^3*x^2 + c^5*d^ 3) + 4096*integrate(1/128*x*arctan(c*x)^2/(c^9*d^3*x^6 + 3*c^7*d^3*x^4 + 3 *c^5*d^3*x^2 + c^3*d^3), x)) - 8*(b^2*c^9*d^3*x^2 - 2*I*b^2*c^8*d^3*x - b^ 2*c^7*d^3)*(((8*c^2*x^2 + 7)*c^2/(c^15*d^3*x^4 + 2*c^13*d^3*x^2 + c^11*d^3 ) + 2*(4*c^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^13*d^3*x^4 + 2*c^11*d^3*x^2 + c^ 9*d^3))*c^2 + 4096*c^2*integrate(1/128*x^5*arctan(c*x)^2/(c^9*d^3*x^6 + 3* c^7*d^3*x^4 + 3*c^5*d^3*x^2 + c^3*d^3), x) + 1024*c^2*integrate(1/128*x...
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]