Integrand size = 22, antiderivative size = 298 \[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=-a b^2 d^2 x-b^3 d^2 x \arctan (c x)+\frac {7 b d^2 (a+b \arctan (c x))^2}{2 c}-3 i b d^2 x (a+b \arctan (c x))^2+\frac {1}{2} b c d^2 x^2 (a+b \arctan (c x))^2-\frac {i d^2 (1+i c x)^3 (a+b \arctan (c x))^3}{3 c}+\frac {4 b d^2 (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {6 i b^2 d^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {b^3 d^2 \log \left (1+c^2 x^2\right )}{2 c}-\frac {4 i b^2 d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c}+\frac {3 b^3 d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}+\frac {2 b^3 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{c} \]
-a*b^2*d^2*x-b^3*d^2*x*arctan(c*x)+7/2*b*d^2*(a+b*arctan(c*x))^2/c-3*I*b*d ^2*x*(a+b*arctan(c*x))^2+1/2*b*c*d^2*x^2*(a+b*arctan(c*x))^2-1/3*I*d^2*(1+ I*c*x)^3*(a+b*arctan(c*x))^3/c+4*b*d^2*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x)) /c-6*I*b^2*d^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c+1/2*b^3*d^2*ln(c^2*x^2+ 1)/c-4*I*b^2*d^2*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/c+3*b^3*d^2*po lylog(2,1-2/(1+I*c*x))/c+2*b^3*d^2*polylog(3,1-2/(1-I*c*x))/c
Time = 1.70 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.77 \[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=-\frac {d^2 \left (-6 a^3 c x+18 i a^2 b c x+6 a b^2 c x-6 i a^3 c^2 x^2-3 a^2 b c^2 x^2+2 a^3 c^3 x^3-18 i a^2 b \arctan (c x)-6 a b^2 \arctan (c x)-18 a^2 b c x \arctan (c x)+36 i a b^2 c x \arctan (c x)+6 b^3 c x \arctan (c x)-18 i a^2 b c^2 x^2 \arctan (c x)-6 a b^2 c^2 x^2 \arctan (c x)+6 a^2 b c^3 x^3 \arctan (c x)+6 i a b^2 \arctan (c x)^2+15 b^3 \arctan (c x)^2-18 a b^2 c x \arctan (c x)^2+18 i b^3 c x \arctan (c x)^2-18 i a b^2 c^2 x^2 \arctan (c x)^2-3 b^3 c^2 x^2 \arctan (c x)^2+6 a b^2 c^3 x^3 \arctan (c x)^2+2 i b^3 \arctan (c x)^3-6 b^3 c x \arctan (c x)^3-6 i b^3 c^2 x^2 \arctan (c x)^3+2 b^3 c^3 x^3 \arctan (c x)^3-48 a b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+36 i b^3 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-24 b^3 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+12 a^2 b \log \left (1+c^2 x^2\right )-18 i a b^2 \log \left (1+c^2 x^2\right )-3 b^3 \log \left (1+c^2 x^2\right )+6 b^2 (4 i a+3 b+4 i b \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-12 b^3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c} \]
-1/6*(d^2*(-6*a^3*c*x + (18*I)*a^2*b*c*x + 6*a*b^2*c*x - (6*I)*a^3*c^2*x^2 - 3*a^2*b*c^2*x^2 + 2*a^3*c^3*x^3 - (18*I)*a^2*b*ArcTan[c*x] - 6*a*b^2*Ar cTan[c*x] - 18*a^2*b*c*x*ArcTan[c*x] + (36*I)*a*b^2*c*x*ArcTan[c*x] + 6*b^ 3*c*x*ArcTan[c*x] - (18*I)*a^2*b*c^2*x^2*ArcTan[c*x] - 6*a*b^2*c^2*x^2*Arc Tan[c*x] + 6*a^2*b*c^3*x^3*ArcTan[c*x] + (6*I)*a*b^2*ArcTan[c*x]^2 + 15*b^ 3*ArcTan[c*x]^2 - 18*a*b^2*c*x*ArcTan[c*x]^2 + (18*I)*b^3*c*x*ArcTan[c*x]^ 2 - (18*I)*a*b^2*c^2*x^2*ArcTan[c*x]^2 - 3*b^3*c^2*x^2*ArcTan[c*x]^2 + 6*a *b^2*c^3*x^3*ArcTan[c*x]^2 + (2*I)*b^3*ArcTan[c*x]^3 - 6*b^3*c*x*ArcTan[c* x]^3 - (6*I)*b^3*c^2*x^2*ArcTan[c*x]^3 + 2*b^3*c^3*x^3*ArcTan[c*x]^3 - 48* a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (36*I)*b^3*ArcTan[c*x]* Log[1 + E^((2*I)*ArcTan[c*x])] - 24*b^3*ArcTan[c*x]^2*Log[1 + E^((2*I)*Arc Tan[c*x])] + 12*a^2*b*Log[1 + c^2*x^2] - (18*I)*a*b^2*Log[1 + c^2*x^2] - 3 *b^3*Log[1 + c^2*x^2] + 6*b^2*((4*I)*a + 3*b + (4*I)*b*ArcTan[c*x])*PolyLo g[2, -E^((2*I)*ArcTan[c*x])] - 12*b^3*PolyLog[3, -E^((2*I)*ArcTan[c*x])])) /c
Time = 0.68 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5389, 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.
| \(\displaystyle \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx\) |
| \(\Big \downarrow \) 5389 |
| \(\displaystyle \frac {i b \int \left (-i c x (a+b \arctan (c x))^2 d^3-\frac {4 i (i-c x) (a+b \arctan (c x))^2 d^3}{c^2 x^2+1}-3 (a+b \arctan (c x))^2 d^3\right )dx}{d}-\frac {i d^2 (1+i c x)^3 (a+b \arctan (c x))^3}{3 c}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {i b \left (-\frac {4 b d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{c}-\frac {1}{2} i c d^3 x^2 (a+b \arctan (c x))^2-3 d^3 x (a+b \arctan (c x))^2-\frac {7 i d^3 (a+b \arctan (c x))^2}{2 c}-\frac {4 i d^3 \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{c}-\frac {6 b d^3 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}+i a b d^3 x+i b^2 d^3 x \arctan (c x)-\frac {i b^2 d^3 \log \left (c^2 x^2+1\right )}{2 c}-\frac {3 i b^2 d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c}-\frac {2 i b^2 d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{c}\right )}{d}-\frac {i d^2 (1+i c x)^3 (a+b \arctan (c x))^3}{3 c}\) |
((-1/3*I)*d^2*(1 + I*c*x)^3*(a + b*ArcTan[c*x])^3)/c + (I*b*(I*a*b*d^3*x + I*b^2*d^3*x*ArcTan[c*x] - (((7*I)/2)*d^3*(a + b*ArcTan[c*x])^2)/c - 3*d^3 *x*(a + b*ArcTan[c*x])^2 - (I/2)*c*d^3*x^2*(a + b*ArcTan[c*x])^2 - ((4*I)* d^3*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/c - (6*b*d^3*(a + b*ArcTan[c *x])*Log[2/(1 + I*c*x)])/c - ((I/2)*b^2*d^3*Log[1 + c^2*x^2])/c - (4*b*d^3 *(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/c - ((3*I)*b^2*d^3*Pol yLog[2, 1 - 2/(1 + I*c*x)])/c - ((2*I)*b^2*d^3*PolyLog[3, 1 - 2/(1 - I*c*x )])/c))/d
3.2.21.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Sy mbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTan[c*x])^p/(e*(q + 1))), x] - S imp[b*c*(p/(e*(q + 1))) Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p - 1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 13.64 (sec) , antiderivative size = 1384, normalized size of antiderivative = 4.64
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1384\) |
| default | \(\text {Expression too large to display}\) | \(1384\) |
| parts | \(\text {Expression too large to display}\) | \(1392\) |
1/c*(-1/3*I*d^2*a^3*(1+I*c*x)^3+d^2*b^3*(-1/3*arctan(c*x)^3*c^3*x^3+I*arct an(c*x)^3*c^2*x^2+arctan(c*x)^3*c*x-1/3*I*arctan(c*x)^3+I*(-3*arctan(c*x)^ 2*c*x-6*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*arctan(c*x)*ln(1 -I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-4*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2* x^2+1))-2*I*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+5/2*I*arctan(c*x)^2+I*ln(( 1+I*c*x)^2/(c^2*x^2+1)+1)+2*I*arctan(c*x)^2*ln(c^2*x^2+1)-Pi*csgn(I*(1+I*c *x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1) +1)^2)*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)^2-4*I*arctan(c*x) ^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))-Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^ 2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)^2+2*Pi*csgn(I*(1+I*c*x)/(c^2 *x^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*arctan(c*x)^2-1/2*I*arcta n(c*x)^2*c^2*x^2+6*I*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)* (c*x-I)+6*I*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+Pi*csgn(I*(1+I*c*x)^2/( c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1 )+1)^2)*arctan(c*x)^2+Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^ 2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*arctan(c*x)^2-2*Pi*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)^2)^2*arctan (c*x)^2+Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2* x^2+1)+1)^2)*arctan(c*x)^2-Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*arctan(c*x )^2-Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*...
\[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
1/24*(I*b^3*c^2*d^2*x^3 + 3*b^3*c*d^2*x^2 - 3*I*b^3*d^2*x)*log(-(c*x + I)/ (c*x - I))^3 + integral(-1/4*(4*a^3*c^4*d^2*x^4 - 8*I*a^3*c^3*d^2*x^3 - 8* I*a^3*c*d^2*x - 4*a^3*d^2 - (3*a*b^2*c^4*d^2*x^4 + 3*I*b^3*c^2*d^2*x^2 + ( -6*I*a*b^2 - b^3)*c^3*d^2*x^3 - 3*a*b^2*d^2 - 3*(2*I*a*b^2 - b^3)*c*d^2*x) *log(-(c*x + I)/(c*x - I))^2 + 6*(I*a^2*b*c^4*d^2*x^4 + 2*a^2*b*c^3*d^2*x^ 3 + 2*a^2*b*c*d^2*x - I*a^2*b*d^2)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1 ), x)
Timed out. \[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=\text {Timed out} \]
\[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
-1/3*a^3*c^2*d^2*x^3 - 28*b^3*c^4*d^2*integrate(1/32*x^4*arctan(c*x)^3/(c^ 2*x^2 + 1), x) - 3*b^3*c^4*d^2*integrate(1/32*x^4*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) - 96*a*b^2*c^4*d^2*integrate(1/32*x^4*arctan(c*x) ^2/(c^2*x^2 + 1), x) - 4*b^3*c^4*d^2*integrate(1/32*x^4*arctan(c*x)*log(c^ 2*x^2 + 1)/(c^2*x^2 + 1), x) + 12*b^3*c^3*d^2*integrate(1/32*x^3*arctan(c* x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c^3*d^2*integrate(1/32*x^3*l og(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) + 16*b^3*c^3*d^2*integrate(1/32*x^3*ar ctan(c*x)^2/(c^2*x^2 + 1), x) - 4*b^3*c^3*d^2*integrate(1/32*x^3*log(c^2*x ^2 + 1)^2/(c^2*x^2 + 1), x) - 1/2*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^ 2*x^2 + 1)/c^4))*a^2*b*c^2*d^2 + I*a^3*c*d^2*x^2 + 7/32*b^3*d^2*arctan(c*x )^4/c + 24*b^3*c^2*d^2*integrate(1/32*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^ 2*x^2 + 1), x) + 3*I*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a^2*b *c*d^2 + a*b^2*d^2*arctan(c*x)^3/c + 12*b^3*c*d^2*integrate(1/32*x*arctan( c*x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c*d^2*integrate(1/32*x*log (c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) - 12*b^3*c*d^2*integrate(1/32*x*arctan(c *x)^2/(c^2*x^2 + 1), x) + 3*b^3*c*d^2*integrate(1/32*x*log(c^2*x^2 + 1)^2/ (c^2*x^2 + 1), x) + a^3*d^2*x + 3*b^3*d^2*integrate(1/32*arctan(c*x)*log(c ^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1 ))*a^2*b*d^2/c - 1/24*(b^3*c^2*d^2*x^3 - 3*I*b^3*c*d^2*x^2 - 3*b^3*d^2*x)* arctan(c*x)^3 + 1/16*(-I*b^3*c^2*d^2*x^3 - 3*b^3*c*d^2*x^2 + 3*I*b^3*d^...
\[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (d+i c d x)^2 (a+b \arctan (c x))^3 \, dx=\int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2 \,d x \]