Integrand size = 20, antiderivative size = 220 \[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\frac {3 b d (a+b \arctan (c x))^2}{2 c}-\frac {3}{2} i b d x (a+b \arctan (c x))^2-\frac {i d (1+i c x)^2 (a+b \arctan (c x))^3}{2 c}+\frac {3 b d (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {3 i b^2 d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^2 d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c}+\frac {3 b^3 d \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c}+\frac {3 b^3 d \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 c} \]
3/2*b*d*(a+b*arctan(c*x))^2/c-3/2*I*b*d*x*(a+b*arctan(c*x))^2-1/2*I*d*(1+I *c*x)^2*(a+b*arctan(c*x))^3/c+3*b*d*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/c- 3*I*b^2*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c-3*I*b^2*d*(a+b*arctan(c*x))* polylog(2,1-2/(1-I*c*x))/c+3/2*b^3*d*polylog(2,1-2/(1+I*c*x))/c+3/2*b^3*d* polylog(3,1-2/(1-I*c*x))/c
Time = 0.54 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.67 \[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\frac {i d \left (-2 i a^3 c x-3 a^2 b c x+a^3 c^2 x^2+3 a^2 b \arctan (c x)-6 i a^2 b c x \arctan (c x)-6 a b^2 c x \arctan (c x)+3 a^2 b c^2 x^2 \arctan (c x)-3 a b^2 \arctan (c x)^2+3 i b^3 \arctan (c x)^2-6 i a b^2 c x \arctan (c x)^2-3 b^3 c x \arctan (c x)^2+3 a b^2 c^2 x^2 \arctan (c x)^2-b^3 \arctan (c x)^3-2 i b^3 c x \arctan (c x)^3+b^3 c^2 x^2 \arctan (c x)^3-12 i a b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-6 b^3 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-6 i b^3 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+3 i a^2 b \log \left (1+c^2 x^2\right )+3 a b^2 \log \left (1+c^2 x^2\right )-3 b^2 (2 a-i b+2 b \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-3 i b^3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{2 c} \]
((I/2)*d*((-2*I)*a^3*c*x - 3*a^2*b*c*x + a^3*c^2*x^2 + 3*a^2*b*ArcTan[c*x] - (6*I)*a^2*b*c*x*ArcTan[c*x] - 6*a*b^2*c*x*ArcTan[c*x] + 3*a^2*b*c^2*x^2 *ArcTan[c*x] - 3*a*b^2*ArcTan[c*x]^2 + (3*I)*b^3*ArcTan[c*x]^2 - (6*I)*a*b ^2*c*x*ArcTan[c*x]^2 - 3*b^3*c*x*ArcTan[c*x]^2 + 3*a*b^2*c^2*x^2*ArcTan[c* x]^2 - b^3*ArcTan[c*x]^3 - (2*I)*b^3*c*x*ArcTan[c*x]^3 + b^3*c^2*x^2*ArcTa n[c*x]^3 - (12*I)*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 6*b^3 *ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (6*I)*b^3*ArcTan[c*x]^2*Log[ 1 + E^((2*I)*ArcTan[c*x])] + (3*I)*a^2*b*Log[1 + c^2*x^2] + 3*a*b^2*Log[1 + c^2*x^2] - 3*b^2*(2*a - I*b + 2*b*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcT an[c*x])] - (3*I)*b^3*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/c
Time = 0.52 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int (d+i c d x) (a+b \arctan (c x))^3 \, dx\) |
| \(\Big \downarrow \) 5389 |
| \(\displaystyle \frac {3 i b \int \left (-d^2 (a+b \arctan (c x))^2-\frac {2 i d^2 (i-c x) (a+b \arctan (c x))^2}{c^2 x^2+1}\right )dx}{2 d}-\frac {i d (1+i c x)^2 (a+b \arctan (c x))^3}{2 c}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 i b \left (-\frac {2 b d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{c}+d^2 (-x) (a+b \arctan (c x))^2-\frac {i d^2 (a+b \arctan (c x))^2}{c}-\frac {2 i d^2 \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{c}-\frac {2 b d^2 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{c}\right )}{2 d}-\frac {i d (1+i c x)^2 (a+b \arctan (c x))^3}{2 c}\) |
((-1/2*I)*d*(1 + I*c*x)^2*(a + b*ArcTan[c*x])^3)/c + (((3*I)/2)*b*(((-I)*d ^2*(a + b*ArcTan[c*x])^2)/c - d^2*x*(a + b*ArcTan[c*x])^2 - ((2*I)*d^2*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/c - (2*b*d^2*(a + b*ArcTan[c*x])*Lo g[2/(1 + I*c*x)])/c - (2*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I *c*x)])/c - (I*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (I*b^2*d^2*PolyL og[3, 1 - 2/(1 - I*c*x)])/c))/d
3.2.22.3.1 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 = 2.04 (sec) , antiderivative size = 3777, normalized size of antiderivative = 17.17
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(3777\) |
| default | \(\text {Expression too large to display}\) | \(3777\) |
| parts | \(\text {Expression too large to display}\) | \(3779\) |
1/c*(-I*d*a^3*(-1/2*c^2*x^2+I*c*x)+d*b^3*(1/2*I*arctan(c*x)^3*c^2*x^2+arct an(c*x)^3*c*x+3/2*I*(-1/4*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*(2*I*arctan (c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*arctan(c*x)^2+polylog(2,-(1+I*c*x)^2 /(c^2*x^2+1)))-1/4*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*(2*I*arctan(c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*arctan(c*x)^2 +polylog(2,-(1+I*c*x)^2/(c^2*x^2+1)))+1/4*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+ 1)+1)^2)^3*(2*I*arctan(c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*arctan(c*x)^2+ polylog(2,-(1+I*c*x)^2/(c^2*x^2+1)))-1/2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1 )+1)^2)^3*(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)))+1/2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x ^2+1))^3*(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)))-2*ln(2)*arctan(c*x)*ln(1+I*(1+I*c *x)/(c^2*x^2+1)^(1/2))-2*ln(2)*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1 /2))+2*ln(2)*arctan(c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*I*ln(2)*dilog(1+I *(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I*ln(2)*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1 /2))-2*I*arctan(c*x)^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))-I*ln(2)*polylog(2,- (1+I*c*x)^2/(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)^3*(I*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2)...
\[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
1/16*(b^3*c*d*x^2 - 2*I*b^3*d*x)*log(-(c*x + I)/(c*x - I))^3 + integral(1/ 8*(8*I*a^3*c^3*d*x^3 + 8*a^3*c^2*d*x^2 + 8*I*a^3*c*d*x + 8*a^3*d - 3*(2*I* a*b^2*c^3*d*x^3 + (2*a*b^2 - I*b^3)*c^2*d*x^2 + 2*a*b^2*d + 2*(I*a*b^2 - b ^3)*c*d*x)*log(-(c*x + I)/(c*x - I))^2 - 12*(a^2*b*c^3*d*x^3 - I*a^2*b*c^2 *d*x^2 + a^2*b*c*d*x - I*a^2*b*d)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1) , x)
Exception generated. \[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\text {Exception raised: TypeError} \]
\[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
12*b^3*c^3*d*integrate(1/64*x^3*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c^3*d*integrate(1/64*x^3*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) + 12*b^3*c^3*d*integrate(1/64*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) - 3*b^3 *c^3*d*integrate(1/64*x^3*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 1/2*I*a^3 *c*d*x^2 + 7/32*b^3*d*arctan(c*x)^4/c + 56*b^3*c^2*d*integrate(1/64*x^2*ar ctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x) *log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*d*integrate(1/64*x^2 *arctan(c*x)^2/(c^2*x^2 + 1), x) + 36*b^3*c^2*d*integrate(1/64*x^2*arctan( c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3/2*I*(x^2*arctan(c*x) - c*(x/c^ 2 - arctan(c*x)/c^3))*a^2*b*c*d + a*b^2*d*arctan(c*x)^3/c + 12*b^3*c*d*int egrate(1/64*x*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c*d*i ntegrate(1/64*x*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) - 24*b^3*c*d*integrat e(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*c*d*integrate(1/64*x*log( c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + a^3*d*x + 6*b^3*d*integrate(1/64*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/c + 1/16*(I*b^3*c*d*x^2 + 2*b^3*d*x)*arctan(c*x)^3 - 3/32*(b^3*c*d*x^2 - 2*I*b^3*d*x)*arctan(c*x)^2*log(c^2*x^2 + 1) + 3/64*(- I*b^3*c*d*x^2 - 2*b^3*d*x)*arctan(c*x)*log(c^2*x^2 + 1)^2 + 1/128*(b^3*c*d *x^2 - 2*I*b^3*d*x)*log(c^2*x^2 + 1)^3 + I*integrate(1/64*(56*(b^3*c^3*d*x ^3 + b^3*c*d*x)*arctan(c*x)^3 + (b^3*c^2*d*x^2 + b^3*d)*log(c^2*x^2 + 1...
\[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (d+i c d x) (a+b \arctan (c x))^3 \, dx=\int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3\,\left (d+c\,d\,x\,1{}\mathrm {i}\right ) \,d x \]