\(\int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx\) [112]

Optimal result

Integrand size = 25, antiderivative size = 383 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {b^2}{16 c^4 d^3 (i-c x)^2}+\frac {21 i b^2}{16 c^4 d^3 (i-c x)}-\frac {21 i b^2 \arctan (c x)}{16 c^4 d^3}+\frac {i b (a+b \arctan (c x))}{4 c^4 d^3 (i-c x)^2}-\frac {11 b (a+b \arctan (c x))}{4 c^4 d^3 (i-c x)}+\frac {3 (a+b \arctan (c x))^2}{8 c^4 d^3}+\frac {i x (a+b \arctan (c x))^2}{c^3 d^3}-\frac {(a+b \arctan (c x))^2}{2 c^4 d^3 (i-c x)^2}-\frac {3 i (a+b \arctan (c x))^2}{c^4 d^3 (i-c x)}+\frac {2 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d^3} \] Output:

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
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {64 i a^2 c x-\frac {32 a^2}{(-i+c x)^2}+\frac {192 i a^2}{-i+c x}-192 i a^2 \arctan (c x)-96 a^2 \log \left (1+c^2 x^2\right )+4 i a b \left (-96 \arctan (c x)^2+20 \cos (2 \arctan (c x))-\cos (4 \arctan (c x))-16 \log \left (1+c^2 x^2\right )-48 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-20 i \sin (2 \arctan (c x))+4 \arctan (c x) \left (8 c x+10 i \cos (2 \arctan (c x))-i \cos (4 \arctan (c x))-24 i \log \left (1+e^{2 i \arctan (c x)}\right )+10 \sin (2 \arctan (c x))-\sin (4 \arctan (c x))\right )+i \sin (4 \arctan (c x))\right )+i b^2 \left (-64 i \arctan (c x)^2+64 c x \arctan (c x)^2-128 \arctan (c x)^3-40 i \cos (2 \arctan (c x))+80 \arctan (c x) \cos (2 \arctan (c x))+80 i \arctan (c x)^2 \cos (2 \arctan (c x))+i \cos (4 \arctan (c x))-4 \arctan (c x) \cos (4 \arctan (c x))-8 i \arctan (c x)^2 \cos (4 \arctan (c x))+128 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-192 i \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-64 (i+3 \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-96 i \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )-40 \sin (2 \arctan (c x))-80 i \arctan (c x) \sin (2 \arctan (c x))+80 \arctan (c x)^2 \sin (2 \arctan (c x))+\sin (4 \arctan (c x))+4 i \arctan (c x) \sin (4 \arctan (c x))-8 \arctan (c x)^2 \sin (4 \arctan (c x))\right )}{64 c^4 d^3} \] Input:

Integrate[(x^3*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]
 

Output:

((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)
 

Rubi [A] (verified)

Time = 0.94 (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

\(\displaystyle \frac {3 i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^4 d^3}-\frac {11 b (a+b \arctan (c x))}{4 c^4 d^3 (-c x+i)}+\frac {i b (a+b \arctan (c x))}{4 c^4 d^3 (-c x+i)^2}-\frac {3 i (a+b \arctan (c x))^2}{c^4 d^3 (-c x+i)}-\frac {(a+b \arctan (c x))^2}{2 c^4 d^3 (-c x+i)^2}+\frac {3 (a+b \arctan (c x))^2}{8 c^4 d^3}+\frac {2 i b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^4 d^3}+\frac {3 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^4 d^3}+\frac {i x (a+b \arctan (c x))^2}{c^3 d^3}-\frac {21 i b^2 \arctan (c x)}{16 c^4 d^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^4 d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 c^4 d^3}+\frac {21 i b^2}{16 c^4 d^3 (-c x+i)}+\frac {b^2}{16 c^4 d^3 (-c x+i)^2}\)

Input:

Int[(x^3*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]
 

Output:

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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5411

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])
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 10.30 (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	\(\text {Expression too large to display}\)	\(4365\)

Input:

int(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x,method=_RETURNVERBOSE)
 

Output:

1/c^4*(2*I*a*b/d^3*arctan(c*x)*c*x-3/2*a^2/d^3*ln(c^2*x^2+1)+3*I*a*b/d^3*l 
n(c*x-I)*ln(-1/2*I*(c*x+I))+3*I*a^2/d^3/(c*x-I)-1/2*a^2/d^3/(c*x-I)^2+b^2/ 
d^3*(-3*I*Pi*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+19/8*arctan(c*x)^2+19/ 
16*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-3*I*arctan(c*x)*polylog(2,-(1+I*c*x 
)^2/(c^2*x^2+1))-3*I*Pi*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8*I*arcta 
n(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8*I*arctan(c*x)*ln(1-I*(1+I*c 
*x)/(c^2*x^2+1)^(1/2))+19/8*I*arctan(c*x)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))+3* 
I*Pi*arctan(c*x)^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))-3/8*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/8*dil 
og(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*arctan(c*x)^3+3/2*polylog(3,-(1+I* 
c*x)^2/(c^2*x^2+1))+3/2*I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^ 
2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*dilog(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))*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(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*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))+3/2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*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))+3/2*P 
i*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(...
 

Fricas [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="fricas")
 

Output:

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)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\text {Timed out} \] Input:

integrate(x**3*(a+b*atan(c*x))**2/(d+I*c*d*x)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="maxima")
 

Output:

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...
 

Giac [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x) + a)^2*x^3/(I*c*d*x + d)^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((x^3*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3,x)
 

Output:

int((x^3*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3, x)
 

Reduce [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {-2 \left (\int \frac {\mathit {atan} \left (c x \right ) x^{3}}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a b \,c^{4}-\left (\int \frac {\mathit {atan} \left (c x \right )^{2} x^{3}}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) b^{2} c^{4}+3 \left (\int \frac {x}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a^{2} c^{2}-2 \left (\int \frac {1}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a^{2} c i -\mathrm {log}\left (c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1\right ) a^{2}+a^{2} c i x}{c^{4} d^{3}} \] Input:

int(x^3*(a+b*atan(c*x))^2/(d+I*c*d*x)^3,x)
 

Output:

( - 2*int((atan(c*x)*x**3)/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a* 
b*c**4 - int((atan(c*x)**2*x**3)/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1) 
,x)*b**2*c**4 + 3*int(x/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a**2* 
c**2 - 2*int(1/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a**2*c*i - log 
(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1)*a**2 + a**2*c*i*x)/(c**4*d**3)