\(\int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx\) [123]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {i (a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c d} \] Output:

I*(a+b*arctan(c*x))^3*ln(2/(1+I*c*x))/c/d-3/2*b*(a+b*arctan(c*x))^2*polylo 
g(2,1-2/(1+I*c*x))/c/d+3/2*I*b^2*(a+b*arctan(c*x))*polylog(3,1-2/(1+I*c*x) 
)/c/d+3/4*b^3*polylog(4,1-2/(1+I*c*x))/c/d
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {i \left (4 (a+b \arctan (c x))^3 \log \left (\frac {2 d}{d+i c d x}\right )+3 i b \left (2 (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )-b \left (2 i (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,\frac {i+c x}{-i+c x}\right )+b \operatorname {PolyLog}\left (4,\frac {i+c x}{-i+c x}\right )\right )\right )\right )}{4 c d} \] Input:

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

Output:

((I/4)*(4*(a + b*ArcTan[c*x])^3*Log[(2*d)/(d + I*c*d*x)] + (3*I)*b*(2*(a + 
 b*ArcTan[c*x])^2*PolyLog[2, (I + c*x)/(-I + c*x)] - b*((2*I)*(a + b*ArcTa 
n[c*x])*PolyLog[3, (I + c*x)/(-I + c*x)] + b*PolyLog[4, (I + c*x)/(-I + c* 
x)]))))/(c*d)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5379, 5529, 5533, 7164}

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.

\(\displaystyle \int \frac {(a+b \arctan (c x))^3}{d+i c d x} \, dx\)

\(\Big \downarrow \) 5379

\(\displaystyle \frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c d}-\frac {3 i b \int \frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{d}\)

\(\Big \downarrow \) 5529

\(\displaystyle \frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c d}-\frac {3 i b \left (i b \int \frac {(a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c}\right )}{d}\)

\(\Big \downarrow \) 5533

\(\displaystyle \frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c d}-\frac {3 i b \left (i b \left (\frac {i \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c}\right )}{d}\)

\(\Big \downarrow \) 7164

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

Input:

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

Output:

(I*(a + b*ArcTan[c*x])^3*Log[2/(1 + I*c*x)])/(c*d) - ((3*I)*b*(((-1/2*I)*( 
a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c + I*b*(((I/2)*(a + b 
*ArcTan[c*x])*PolyLog[3, 1 - 2/(1 + I*c*x)])/c + (b*PolyLog[4, 1 - 2/(1 + 
I*c*x)])/(4*c))))/d
 

Defintions of rubi rules used

rule 5379

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] 
 :> Simp[(-(a + b*ArcTan[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c*( 
p/e)   Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x^2)) 
, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0 
]
 

rule 5529

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 
), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) 
, x] + Simp[b*p*(I/2)   Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/ 
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c 
^2*d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I - c*x)))^2, 0]
 

rule 5533

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_. 
)*(x_)^2), x_Symbol] :> Simp[I*(a + b*ArcTan[c*x])^p*(PolyLog[k + 1, u]/(2* 
c*d)), x] - Simp[b*p*(I/2)   Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[k + 1 
, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && 
EqQ[e, c^2*d] && EqQ[u^2 - (1 - 2*(I/(I - c*x)))^2, 0]
 

rule 7164

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, 
x]}, Simp[w*PolyLog[n + 1, v], x] /;  !FalseQ[w]] /; FreeQ[n, x]
 

Maple [C] (warning: unable to verify)

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

Time = 11.05 (sec) , antiderivative size = 1629, normalized size of antiderivative = 11.72

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1629\)
default	\(\text {Expression too large to display}\)	\(1629\)
parts	\(\text {Expression too large to display}\)	\(1640\)

Input:

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

Output:

1/c*(-1/2*I*a^3/d*ln(c^2*x^2+1)+a^3/d*arctan(c*x)+b^3/d*(-I*ln(1+I*c*x)*ar 
ctan(c*x)^3+3*I*(1/3*arctan(c*x)^3*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/6*I*a 
rctan(c*x)^4+1/6*I*Pi*(csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^ 
2+1)))^3+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-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-csgn((1+I*c*x)^2/(c^2*x^2+1 
)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x) 
^2/(c^2*x^2+1)))^2-csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*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)))-csgn( 
I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3+csgn((1+I*c*x)^2/ 
(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1 
+(1+I*c*x)^2/(c^2*x^2+1)))+csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/( 
c^2*x^2+1)))^2-1)*arctan(c*x)^3-1/2*I*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2 
/(c^2*x^2+1))+1/2*arctan(c*x)*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+1/4*I*po 
lylog(4,-(1+I*c*x)^2/(c^2*x^2+1))))+3*a*b^2/d*(-I*ln(1+I*c*x)*arctan(c*x)^ 
2+2*I*(1/2*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+1/4*I*Pi*(csgn((1 
+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3+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 
-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-csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1...
 

Fricas [F]

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

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

Output:

integral(-1/8*(b^3*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*log(-(c*x + I)/ 
(c*x - I))^2 - 12*a^2*b*log(-(c*x + I)/(c*x - I)) + 8*I*a^3)/(c*d*x - I*d) 
, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-I*a^3*log(I*c*d*x + d)/(c*d) + 1/128*(16*b^3*arctan(c*x)^4 + 16*I*b^3*arc 
tan(c*x)^3*log(c^2*x^2 + 1) + 4*I*b^3*arctan(c*x)*log(c^2*x^2 + 1)^3 - b^3 
*log(c^2*x^2 + 1)^4 + 16*(b^3*arctan(c*x)^4/(c*d) + 8*b^3*c*integrate(1/16 
*x*log(c^2*x^2 + 1)^3/(c^2*d*x^2 + d), x) + 8*a*b^2*arctan(c*x)^3/(c*d) + 
12*a^2*b*arctan(c*x)^2/(c*d))*c*d - 128*I*c*d*integrate(1/32*(40*b^3*c*x*a 
rctan(c*x)^3 + 6*b^3*c*x*arctan(c*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*c*x*arc 
tan(c*x)^2 + 96*a^2*b*c*x*arctan(c*x) + 12*b^3*arctan(c*x)^2*log(c^2*x^2 + 
 1) + b^3*log(c^2*x^2 + 1)^3)/(c^2*d*x^2 + d), x))/(c*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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