\(\int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx\) [24]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=3 i a c d^3 x+\frac {3}{2} b c d^3 x+\frac {1}{6} i b c^2 d^3 x^2-\frac {3}{2} b d^3 \arctan (c x)+3 i b c d^3 x \arctan (c x)-\frac {3}{2} c^2 d^3 x^2 (a+b \arctan (c x))-\frac {1}{3} i c^3 d^3 x^3 (a+b \arctan (c x))+a d^3 \log (x)-\frac {5}{3} i b d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x) \] Output:

3*I*a*c*d^3*x+3/2*b*c*d^3*x+1/6*I*b*c^2*d^3*x^2-3/2*b*d^3*arctan(c*x)+3*I* 
b*c*d^3*x*arctan(c*x)-3/2*c^2*d^3*x^2*(a+b*arctan(c*x))-1/3*I*c^3*d^3*x^3* 
(a+b*arctan(c*x))+a*d^3*ln(x)-5/3*I*b*d^3*ln(c^2*x^2+1)+1/2*I*b*d^3*polylo 
g(2,-I*c*x)-1/2*I*b*d^3*polylog(2,I*c*x)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=-\frac {1}{6} i d^3 \left (-18 a c x+9 i b c x-9 i a c^2 x^2-b c^2 x^2+2 a c^3 x^3-9 i b \arctan (c x)-18 b c x \arctan (c x)-9 i b c^2 x^2 \arctan (c x)+2 b c^3 x^3 \arctan (c x)+6 i a \log (x)+10 b \log \left (1+c^2 x^2\right )-3 b \operatorname {PolyLog}(2,-i c x)+3 b \operatorname {PolyLog}(2,i c x)\right ) \] Input:

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

Output:

(-1/6*I)*d^3*(-18*a*c*x + (9*I)*b*c*x - (9*I)*a*c^2*x^2 - b*c^2*x^2 + 2*a* 
c^3*x^3 - (9*I)*b*ArcTan[c*x] - 18*b*c*x*ArcTan[c*x] - (9*I)*b*c^2*x^2*Arc 
Tan[c*x] + 2*b*c^3*x^3*ArcTan[c*x] + (6*I)*a*Log[x] + 10*b*Log[1 + c^2*x^2 
] - 3*b*PolyLog[2, (-I)*c*x] + 3*b*PolyLog[2, I*c*x])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.

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

\(\Big \downarrow \) 5411

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{3} i c^3 d^3 x^3 (a+b \arctan (c x))-\frac {3}{2} c^2 d^3 x^2 (a+b \arctan (c x))+3 i a c d^3 x+a d^3 \log (x)-\frac {3}{2} b d^3 \arctan (c x)+3 i b c d^3 x \arctan (c x)+\frac {1}{6} i b c^2 d^3 x^2-\frac {5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x)+\frac {3}{2} b c d^3 x\)

Input:

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

Output:

(3*I)*a*c*d^3*x + (3*b*c*d^3*x)/2 + (I/6)*b*c^2*d^3*x^2 - (3*b*d^3*ArcTan[ 
c*x])/2 + (3*I)*b*c*d^3*x*ArcTan[c*x] - (3*c^2*d^3*x^2*(a + b*ArcTan[c*x]) 
)/2 - (I/3)*c^3*d^3*x^3*(a + b*ArcTan[c*x]) + a*d^3*Log[x] - ((5*I)/3)*b*d 
^3*Log[1 + c^2*x^2] + (I/2)*b*d^3*PolyLog[2, (-I)*c*x] - (I/2)*b*d^3*PolyL 
og[2, I*c*x]
 

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 [A] (verified)

Time = 0.51 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.97

method	result	size
parts	\(d^{3} a \left (-\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+3 i c x +\ln \left (x \right )\right )+d^{3} b \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\ln \left (c x \right ) \arctan \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(165\)
derivativedivides	\(d^{3} a \left (3 i c x -\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+\ln \left (c x \right )\right )+d^{3} b \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\ln \left (c x \right ) \arctan \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(167\)
default	\(d^{3} a \left (3 i c x -\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+\ln \left (c x \right )\right )+d^{3} b \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\ln \left (c x \right ) \arctan \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(167\)
risch	\(-\frac {d^{3} b \ln \left (i c x +1\right ) c^{3} x^{3}}{6}-\frac {3 i d^{3} x^{2} b \ln \left (-i c x +1\right ) c^{2}}{4}+\frac {3 d^{3} b \ln \left (i c x +1\right ) c x}{2}+\frac {65 i d^{3} b}{18}-\frac {11 i d^{3} b \ln \left (i c x +1\right )}{12}+\frac {3 b c \,d^{3} x}{2}+\frac {i x^{2} b \,c^{2} d^{3}}{6}+3 i a c \,d^{3} x +\frac {3 i d^{3} b \ln \left (i c x +1\right ) c^{2} x^{2}}{4}-\frac {3 a \,c^{2} d^{3} x^{2}}{2}+\frac {i d^{3} b \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {29 d^{3} a}{6}+d^{3} a \ln \left (-i c x \right )+\frac {d^{3} b \ln \left (-i c x +1\right ) c^{3} x^{3}}{6}-\frac {i x^{3} a \,c^{3} d^{3}}{3}-\frac {3 d^{3} b \ln \left (-i c x +1\right ) c x}{2}-\frac {29 i d^{3} b \ln \left (-i c x +1\right )}{12}-\frac {i d^{3} b \operatorname {dilog}\left (-i c x +1\right )}{2}\)	\(255\)

Input:

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

Output:

d^3*a*(-1/3*I*c^3*x^3-3/2*c^2*x^2+3*I*c*x+ln(x))+d^3*b*(3*I*arctan(c*x)*c* 
x-1/3*I*arctan(c*x)*c^3*x^3-3/2*c^2*x^2*arctan(c*x)+ln(c*x)*arctan(c*x)+1/ 
2*I*ln(c*x)*ln(1+I*c*x)-1/2*I*ln(c*x)*ln(1-I*c*x)+1/2*I*dilog(1+I*c*x)-1/2 
*I*dilog(1-I*c*x)+3/2*c*x+1/6*I*c^2*x^2-5/3*I*ln(c^2*x^2+1)-3/2*arctan(c*x 
))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.08 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=-\frac {1}{3} i \, a c^{3} d^{3} x^{3} - \frac {3}{2} \, a c^{2} d^{3} x^{2} + \frac {1}{6} i \, b c^{2} d^{3} x^{2} + 3 i \, a c d^{3} x + \frac {3}{2} \, b c d^{3} x - \frac {1}{12} \, {\left (3 \, \pi + 2 i\right )} b d^{3} \log \left (c^{2} x^{2} + 1\right ) + b d^{3} \arctan \left (c x\right ) \log \left (c x\right ) + \frac {3}{2} i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3} - \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (i \, c x + 1\right ) + \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + a d^{3} \log \left (x\right ) + \frac {1}{6} \, {\left (-2 i \, b c^{3} d^{3} x^{3} - 9 \, b c^{2} d^{3} x^{2} - 9 \, b d^{3}\right )} \arctan \left (c x\right ) \] Input:

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

Output:

-1/3*I*a*c^3*d^3*x^3 - 3/2*a*c^2*d^3*x^2 + 1/6*I*b*c^2*d^3*x^2 + 3*I*a*c*d 
^3*x + 3/2*b*c*d^3*x - 1/12*(3*pi + 2*I)*b*d^3*log(c^2*x^2 + 1) + b*d^3*ar 
ctan(c*x)*log(c*x) + 3/2*I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^3 - 
1/2*I*b*d^3*dilog(I*c*x + 1) + 1/2*I*b*d^3*dilog(-I*c*x + 1) + a*d^3*log(x 
) + 1/6*(-2*I*b*c^3*d^3*x^3 - 9*b*c^2*d^3*x^2 - 9*b*d^3)*arctan(c*x)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.15 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} a\,d^3\,\ln \left (x\right ) & \text {\ if\ \ }c=0\\ a\,d^3\,\ln \left (x\right )-\frac {b\,d^3\,\ln \left (c^2\,x^2+1\right )\,3{}\mathrm {i}}{2}-\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {3\,a\,c^2\,d^3\,x^2}{2}-\frac {a\,c^3\,d^3\,x^3\,1{}\mathrm {i}}{3}+a\,c\,d^3\,x\,3{}\mathrm {i}+\frac {3\,b\,c\,d^3\,x}{2}+\frac {b\,c^2\,d^3\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )\,1{}\mathrm {i}}{3}-3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )-\frac {b\,c^3\,d^3\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{3}+b\,c\,d^3\,x\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \] Input:

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

Output:

piecewise(c == 0, a*d^3*log(x), c ~= 0, - (b*d^3*log(c^2*x^2 + 1)*3i)/2 + 
a*d^3*log(x) - (b*d^3*dilog(- c*x*1i + 1)*1i)/2 + (b*d^3*dilog(c*x*1i + 1) 
*1i)/2 - (3*a*c^2*d^3*x^2)/2 - (a*c^3*d^3*x^3*1i)/3 + a*c*d^3*x*3i + (3*b* 
c*d^3*x)/2 + (b*c^2*d^3*(x^2/2 - log(c^2*x^2 + 1)/(2*c^2))*1i)/3 - 3*b*c^2 
*d^3*atan(c*x)*(1/(2*c^2) + x^2/2) - (b*c^3*d^3*x^3*atan(c*x)*1i)/3 + b*c* 
d^3*x*atan(c*x)*3i)
 

Reduce [F]

\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=\frac {d^{3} \left (-2 \mathit {atan} \left (c x \right ) b \,c^{3} i \,x^{3}-9 \mathit {atan} \left (c x \right ) b \,c^{2} x^{2}+18 \mathit {atan} \left (c x \right ) b c i x -9 \mathit {atan} \left (c x \right ) b +6 \left (\int \frac {\mathit {atan} \left (c x \right )}{x}d x \right ) b -10 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b i +6 \,\mathrm {log}\left (x \right ) a -2 a \,c^{3} i \,x^{3}-9 a \,c^{2} x^{2}+18 a c i x +b \,c^{2} i \,x^{2}+9 b c x \right )}{6} \] Input:

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

Output:

(d**3*( - 2*atan(c*x)*b*c**3*i*x**3 - 9*atan(c*x)*b*c**2*x**2 + 18*atan(c* 
x)*b*c*i*x - 9*atan(c*x)*b + 6*int(atan(c*x)/x,x)*b - 10*log(c**2*x**2 + 1 
)*b*i + 6*log(x)*a - 2*a*c**3*i*x**3 - 9*a*c**2*x**2 + 18*a*c*i*x + b*c**2 
*i*x**2 + 9*b*c*x))/6