\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx\) [36]

Optimal result

Integrand size = 23, antiderivative size = 190 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=-6 a c^2 d^4 x+2 i b c^2 d^4 x-\frac {1}{6} b c^3 d^4 x^2-2 i b c d^4 \arctan (c x)-6 b c^2 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{x}-2 i c^3 d^4 x^2 (a+b \arctan (c x))+\frac {1}{3} c^4 d^4 x^3 (a+b \arctan (c x))+4 i a c d^4 \log (x)+b c d^4 \log (x)+\frac {8}{3} b c d^4 \log \left (1+c^2 x^2\right )-2 b c d^4 \operatorname {PolyLog}(2,-i c x)+2 b c d^4 \operatorname {PolyLog}(2,i c x) \] Output:

-6*a*c^2*d^4*x+2*I*b*c^2*d^4*x-1/6*b*c^3*d^4*x^2-2*I*b*c*d^4*arctan(c*x)-6 
*b*c^2*d^4*x*arctan(c*x)-d^4*(a+b*arctan(c*x))/x-2*I*c^3*d^4*x^2*(a+b*arct 
an(c*x))+1/3*c^4*d^4*x^3*(a+b*arctan(c*x))+4*I*a*c*d^4*ln(x)+b*c*d^4*ln(x) 
+8/3*b*c*d^4*ln(c^2*x^2+1)-2*b*c*d^4*polylog(2,-I*c*x)+2*b*c*d^4*polylog(2 
,I*c*x)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {d^4 \left (-6 a-36 a c^2 x^2+12 i b c^2 x^2-12 i a c^3 x^3-b c^3 x^3+2 a c^4 x^4-6 b \arctan (c x)-12 i b c x \arctan (c x)-36 b c^2 x^2 \arctan (c x)-12 i b c^3 x^3 \arctan (c x)+2 b c^4 x^4 \arctan (c x)+24 i a c x \log (x)+6 b c x \log (c x)+16 b c x \log \left (1+c^2 x^2\right )-12 b c x \operatorname {PolyLog}(2,-i c x)+12 b c x \operatorname {PolyLog}(2,i c x)\right )}{6 x} \] Input:

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^2,x]
 

Output:

(d^4*(-6*a - 36*a*c^2*x^2 + (12*I)*b*c^2*x^2 - (12*I)*a*c^3*x^3 - b*c^3*x^ 
3 + 2*a*c^4*x^4 - 6*b*ArcTan[c*x] - (12*I)*b*c*x*ArcTan[c*x] - 36*b*c^2*x^ 
2*ArcTan[c*x] - (12*I)*b*c^3*x^3*ArcTan[c*x] + 2*b*c^4*x^4*ArcTan[c*x] + ( 
24*I)*a*c*x*Log[x] + 6*b*c*x*Log[c*x] + 16*b*c*x*Log[1 + c^2*x^2] - 12*b*c 
*x*PolyLog[2, (-I)*c*x] + 12*b*c*x*PolyLog[2, I*c*x]))/(6*x)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 190, 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.

\(\displaystyle \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx\)

\(\Big \downarrow \) 5411

\(\displaystyle \int \left (c^4 d^4 x^2 (a+b \arctan (c x))-4 i c^3 d^4 x (a+b \arctan (c x))-6 c^2 d^4 (a+b \arctan (c x))+\frac {d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c d^4 (a+b \arctan (c x))}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} c^4 d^4 x^3 (a+b \arctan (c x))-2 i c^3 d^4 x^2 (a+b \arctan (c x))-\frac {d^4 (a+b \arctan (c x))}{x}-6 a c^2 d^4 x+4 i a c d^4 \log (x)-6 b c^2 d^4 x \arctan (c x)-2 i b c d^4 \arctan (c x)-\frac {1}{6} b c^3 d^4 x^2+\frac {8}{3} b c d^4 \log \left (c^2 x^2+1\right )+2 i b c^2 d^4 x-2 b c d^4 \operatorname {PolyLog}(2,-i c x)+2 b c d^4 \operatorname {PolyLog}(2,i c x)+b c d^4 \log (x)\)

Input:

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^2,x]
 

Output:

-6*a*c^2*d^4*x + (2*I)*b*c^2*d^4*x - (b*c^3*d^4*x^2)/6 - (2*I)*b*c*d^4*Arc 
Tan[c*x] - 6*b*c^2*d^4*x*ArcTan[c*x] - (d^4*(a + b*ArcTan[c*x]))/x - (2*I) 
*c^3*d^4*x^2*(a + b*ArcTan[c*x]) + (c^4*d^4*x^3*(a + b*ArcTan[c*x]))/3 + ( 
4*I)*a*c*d^4*Log[x] + b*c*d^4*Log[x] + (8*b*c*d^4*Log[1 + c^2*x^2])/3 - 2* 
b*c*d^4*PolyLog[2, (-I)*c*x] + 2*b*c*d^4*PolyLog[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 = 0.80 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.99

method	result	size
parts	\(d^{4} a \left (\frac {c^{4} x^{3}}{3}-2 i c^{3} x^{2}-6 c^{2} x -\frac {1}{x}+4 i c \ln \left (x \right )\right )+d^{4} b c \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\)	\(189\)
derivativedivides	\(c \left (d^{4} a \left (-6 c x +\frac {c^{3} x^{3}}{3}-2 i c^{2} x^{2}+4 i \ln \left (c x \right )-\frac {1}{c x}\right )+d^{4} b \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(192\)
default	\(c \left (d^{4} a \left (-6 c x +\frac {c^{3} x^{3}}{3}-2 i c^{2} x^{2}+4 i \ln \left (c x \right )-\frac {1}{c x}\right )+d^{4} b \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(192\)
risch	\(-b \,c^{3} d^{4} \ln \left (i c x +1\right ) x^{2}+\frac {11 d^{4} c b \ln \left (-i c x +1\right )}{3}+\frac {d^{4} c^{4} a \,x^{3}}{3}-\frac {25 i d^{4} c a}{3}+d^{4} c^{3} b \ln \left (-i c x +1\right ) x^{2}-2 i d^{4} c^{3} a \,x^{2}-\frac {i b \,c^{4} d^{4} \ln \left (i c x +1\right ) x^{3}}{6}+\frac {i d^{4} c^{4} b \,x^{3} \ln \left (-i c x +1\right )}{6}+3 i b \,c^{2} d^{4} \ln \left (i c x +1\right ) x +2 d^{4} c b \operatorname {dilog}\left (-i c x +1\right )+\frac {d^{4} c b \ln \left (-i c x \right )}{2}-\frac {d^{4} a}{x}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{2 x}+2 i b \,c^{2} d^{4} x -\frac {i d^{4} b \ln \left (-i c x +1\right )}{2 x}-\frac {119 b c \,d^{4}}{18}+\frac {5 b c \,d^{4} \ln \left (i c x +1\right )}{3}+\frac {b c \,d^{4} \ln \left (i c x \right )}{2}-2 b c \,d^{4} \operatorname {dilog}\left (i c x +1\right )+4 i d^{4} c a \ln \left (-i c x \right )-3 i d^{4} c^{2} b x \ln \left (-i c x +1\right )-6 a \,c^{2} d^{4} x -\frac {b \,c^{3} d^{4} x^{2}}{6}\)	\(339\)

Input:

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x,method=_RETURNVERBOSE)
 

Output:

d^4*a*(1/3*c^4*x^3-2*I*c^3*x^2-6*c^2*x-1/x+4*I*c*ln(x))+d^4*b*c*(-6*c*x*ar 
ctan(c*x)+1/3*c^3*x^3*arctan(c*x)-2*I*arctan(c*x)*c^2*x^2+4*I*arctan(c*x)* 
ln(c*x)-1/c/x*arctan(c*x)-2*ln(c*x)*ln(1+I*c*x)+2*ln(c*x)*ln(1-I*c*x)-2*di 
log(1+I*c*x)+2*dilog(1-I*c*x)+2*I*c*x-1/6*c^2*x^2+ln(c*x)+8/3*ln(c^2*x^2+1 
)-2*I*arctan(c*x))
 

Fricas [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \] Input:

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x, algorithm="fricas")
 

Output:

integral(1/2*(2*a*c^4*d^4*x^4 - 8*I*a*c^3*d^4*x^3 - 12*a*c^2*d^4*x^2 + 8*I 
*a*c*d^4*x + 2*a*d^4 + (I*b*c^4*d^4*x^4 + 4*b*c^3*d^4*x^3 - 6*I*b*c^2*d^4* 
x^2 - 4*b*c*d^4*x + I*b*d^4)*log(-(c*x + I)/(c*x - I)))/x^2, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\text {Timed out} \] Input:

integrate((d+I*c*d*x)**4*(a+b*atan(c*x))/x**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.26 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {1}{3} \, a c^{4} d^{4} x^{3} - 2 i \, a c^{3} d^{4} x^{2} - \frac {1}{6} \, b c^{3} d^{4} x^{2} - 6 \, a c^{2} d^{4} x + 2 i \, b c^{2} d^{4} x - \frac {1}{6} \, {\left (6 i \, \pi - 1\right )} b c d^{4} \log \left (c^{2} x^{2} + 1\right ) + 4 i \, b c d^{4} \arctan \left (c x\right ) \log \left (c x\right ) - 3 \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c d^{4} + 2 \, b c d^{4} {\rm Li}_2\left (i \, c x + 1\right ) - 2 \, b c d^{4} {\rm Li}_2\left (-i \, c x + 1\right ) + 4 i \, a c d^{4} \log \left (x\right ) - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{4} - \frac {a d^{4}}{x} + \frac {1}{3} \, {\left (b c^{4} d^{4} x^{3} - 6 i \, b c^{3} d^{4} x^{2} - 6 i \, b c d^{4}\right )} \arctan \left (c x\right ) \] Input:

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x, algorithm="maxima")
 

Output:

1/3*a*c^4*d^4*x^3 - 2*I*a*c^3*d^4*x^2 - 1/6*b*c^3*d^4*x^2 - 6*a*c^2*d^4*x 
+ 2*I*b*c^2*d^4*x - 1/6*(6*I*pi - 1)*b*c*d^4*log(c^2*x^2 + 1) + 4*I*b*c*d^ 
4*arctan(c*x)*log(c*x) - 3*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*c*d^4 
+ 2*b*c*d^4*dilog(I*c*x + 1) - 2*b*c*d^4*dilog(-I*c*x + 1) + 4*I*a*c*d^4*l 
og(x) - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d^4 - a* 
d^4/x + 1/3*(b*c^4*d^4*x^3 - 6*I*b*c^3*d^4*x^2 - 6*I*b*c*d^4)*arctan(c*x)
 

Giac [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \] Input:

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x, algorithm="giac")
 

Output:

integrate((I*c*d*x + d)^4*(b*arctan(c*x) + a)/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.33 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{x} & \text {\ if\ \ }c=0\\ \frac {a\,c^4\,d^4\,x^3}{3}-\frac {a\,d^4}{x}+\frac {b\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+2\,b\,c\,d^4\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )+3\,b\,c\,d^4\,\ln \left (c^2\,x^2+1\right )-6\,a\,c^2\,d^4\,x-\frac {b\,c^3\,d^4\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )}{3}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{x}-6\,b\,c^2\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+\frac {b\,c^4\,d^4\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}-a\,c^3\,d^4\,x^2\,2{}\mathrm {i}+b\,c^2\,d^4\,x\,2{}\mathrm {i}+a\,c\,d^4\,\ln \left (x\right )\,4{}\mathrm {i}-b\,c^3\,d^4\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\,4{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \] Input:

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

Output:

piecewise(c == 0, -(a*d^4)/x, c ~= 0, - (a*d^4)/x - a*c^3*d^4*x^2*2i + (a* 
c^4*d^4*x^3)/3 + (b*d^4*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2))/c + 2*b*c 
*d^4*(dilog(- c*x*1i + 1) - dilog(c*x*1i + 1)) + 3*b*c*d^4*log(c^2*x^2 + 1 
) - 6*a*c^2*d^4*x + b*c^2*d^4*x*2i - (b*c^3*d^4*(x^2/2 - log(c^2*x^2 + 1)/ 
(2*c^2)))/3 + a*c*d^4*log(x)*4i - (b*d^4*atan(c*x))/x - 6*b*c^2*d^4*x*atan 
(c*x) - b*c^3*d^4*atan(c*x)*(1/(2*c^2) + x^2/2)*4i + (b*c^4*d^4*x^3*atan(c 
*x))/3)
 

Reduce [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {d^{4} \left (2 \mathit {atan} \left (c x \right ) b \,c^{4} x^{4}-12 \mathit {atan} \left (c x \right ) b \,c^{3} i \,x^{3}-36 \mathit {atan} \left (c x \right ) b \,c^{2} x^{2}-12 \mathit {atan} \left (c x \right ) b c i x -6 \mathit {atan} \left (c x \right ) b +24 \left (\int \frac {\mathit {atan} \left (c x \right )}{x}d x \right ) b c i x +16 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b c x +24 \,\mathrm {log}\left (x \right ) a c i x +6 \,\mathrm {log}\left (x \right ) b c x +2 a \,c^{4} x^{4}-12 a \,c^{3} i \,x^{3}-36 a \,c^{2} x^{2}-6 a -b \,c^{3} x^{3}+12 b \,c^{2} i \,x^{2}\right )}{6 x} \] Input:

int((d+I*c*d*x)^4*(a+b*atan(c*x))/x^2,x)
 

Output:

(d**4*(2*atan(c*x)*b*c**4*x**4 - 12*atan(c*x)*b*c**3*i*x**3 - 36*atan(c*x) 
*b*c**2*x**2 - 12*atan(c*x)*b*c*i*x - 6*atan(c*x)*b + 24*int(atan(c*x)/x,x 
)*b*c*i*x + 16*log(c**2*x**2 + 1)*b*c*x + 24*log(x)*a*c*i*x + 6*log(x)*b*c 
*x + 2*a*c**4*x**4 - 12*a*c**3*i*x**3 - 36*a*c**2*x**2 - 6*a - b*c**3*x**3 
 + 12*b*c**2*i*x**2))/(6*x)