\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx\) [38]

Optimal result

Integrand size = 23, antiderivative size = 201 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d^4}{6 x^2}-\frac {2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \arctan (c x)+b c^4 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)-\frac {19}{3} b c^3 d^4 \log (x)+\frac {8}{3} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x) \] Output:

-1/6*b*c*d^4/x^2-2*I*b*c^2*d^4/x+a*c^4*d^4*x-2*I*b*c^3*d^4*arctan(c*x)+b*c 
^4*d^4*x*arctan(c*x)-1/3*d^4*(a+b*arctan(c*x))/x^3-2*I*c*d^4*(a+b*arctan(c 
*x))/x^2+6*c^2*d^4*(a+b*arctan(c*x))/x-4*I*a*c^3*d^4*ln(x)-19/3*b*c^3*d^4* 
ln(x)+8/3*b*c^3*d^4*ln(c^2*x^2+1)+2*b*c^3*d^4*polylog(2,-I*c*x)-2*b*c^3*d^ 
4*polylog(2,I*c*x)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\frac {d^4 \left (-2 a-12 i a c x-b c x+36 a c^2 x^2-12 i b c^2 x^2+6 a c^4 x^4-2 b \arctan (c x)-12 i b c x \arctan (c x)+36 b c^2 x^2 \arctan (c x)-12 i b c^3 x^3 \arctan (c x)+6 b c^4 x^4 \arctan (c x)-24 i a c^3 x^3 \log (x)-38 b c^3 x^3 \log (c x)+16 b c^3 x^3 \log \left (1+c^2 x^2\right )+12 b c^3 x^3 \operatorname {PolyLog}(2,-i c x)-12 b c^3 x^3 \operatorname {PolyLog}(2,i c x)\right )}{6 x^3} \] Input:

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

Output:

(d^4*(-2*a - (12*I)*a*c*x - b*c*x + 36*a*c^2*x^2 - (12*I)*b*c^2*x^2 + 6*a* 
c^4*x^4 - 2*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] + 6*b*c^4*x^4*ArcTan[c*x] - (24*I)*a* 
c^3*x^3*Log[x] - 38*b*c^3*x^3*Log[c*x] + 16*b*c^3*x^3*Log[1 + c^2*x^2] + 1 
2*b*c^3*x^3*PolyLog[2, (-I)*c*x] - 12*b*c^3*x^3*PolyLog[2, I*c*x]))/(6*x^3 
)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 201, 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.

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

\(\Big \downarrow \) 5411

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+a c^4 d^4 x-4 i a c^3 d^4 \log (x)+b c^4 d^4 x \arctan (c x)-2 i b c^3 d^4 \arctan (c x)+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x)-\frac {19}{3} b c^3 d^4 \log (x)-\frac {2 i b c^2 d^4}{x}+\frac {8}{3} b c^3 d^4 \log \left (c^2 x^2+1\right )-\frac {b c d^4}{6 x^2}\)

Input:

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

Output:

-1/6*(b*c*d^4)/x^2 - ((2*I)*b*c^2*d^4)/x + a*c^4*d^4*x - (2*I)*b*c^3*d^4*A 
rcTan[c*x] + b*c^4*d^4*x*ArcTan[c*x] - (d^4*(a + b*ArcTan[c*x]))/(3*x^3) - 
 ((2*I)*c*d^4*(a + b*ArcTan[c*x]))/x^2 + (6*c^2*d^4*(a + b*ArcTan[c*x]))/x 
 - (4*I)*a*c^3*d^4*Log[x] - (19*b*c^3*d^4*Log[x])/3 + (8*b*c^3*d^4*Log[1 + 
 c^2*x^2])/3 + 2*b*c^3*d^4*PolyLog[2, (-I)*c*x] - 2*b*c^3*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 = 1.09 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97

method	result	size
parts	\(d^{4} a \left (c^{4} x -\frac {2 i c}{x^{2}}-\frac {1}{3 x^{3}}+\frac {6 c^{2}}{x}-4 i c^{3} \ln \left (x \right )\right )+d^{4} b \,c^{3} \left (c x \arctan \left (c x \right )-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+\frac {6 \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}\right )\)	\(195\)
derivativedivides	\(c^{3} \left (d^{4} a \left (c x -\frac {2 i}{c^{2} x^{2}}+\frac {6}{c x}-\frac {1}{3 c^{3} x^{3}}-4 i \ln \left (c x \right )\right )+d^{4} b \left (c x \arctan \left (c x \right )-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+\frac {6 \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}\right )\right )\)	\(198\)
default	\(c^{3} \left (d^{4} a \left (c x -\frac {2 i}{c^{2} x^{2}}+\frac {6}{c x}-\frac {1}{3 c^{3} x^{3}}-4 i \ln \left (c x \right )\right )+d^{4} b \left (c x \arctan \left (c x \right )-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+\frac {6 \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}\right )\right )\)	\(198\)
risch	\(b \,c^{3} d^{4}-\frac {i d^{4} b \ln \left (-i c x +1\right )}{6 x^{3}}-\frac {i b \,d^{4} c^{4} \ln \left (i c x +1\right ) x}{2}-2 i b \,c^{3} d^{4} \arctan \left (c x \right )+\frac {3 i d^{4} c^{2} b \ln \left (-i c x +1\right )}{x}-\frac {2 i d^{4} c a}{x^{2}}-\frac {b \,d^{4} c \ln \left (i c x +1\right )}{x^{2}}+i d^{4} c^{3} a +\frac {i d^{4} c^{4} b \ln \left (-i c x +1\right ) x}{2}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{6 x^{3}}-\frac {2 i b \,c^{2} d^{4}}{x}+\frac {d^{4} c b \ln \left (-i c x +1\right )}{x^{2}}-\frac {13 b \,d^{4} c^{3} \ln \left (i c x \right )}{6}+2 b \,d^{4} c^{3} \operatorname {dilog}\left (i c x +1\right )-\frac {25 d^{4} c^{3} b \ln \left (-i c x \right )}{6}-2 d^{4} c^{3} b \operatorname {dilog}\left (-i c x +1\right )-\frac {d^{4} a}{3 x^{3}}+\frac {6 d^{4} c^{2} a}{x}+a \,c^{4} d^{4} x -\frac {b c \,d^{4}}{6 x^{2}}+\frac {8 b \,c^{3} d^{4} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 i b \,d^{4} c^{2} \ln \left (i c x +1\right )}{x}-4 i d^{4} c^{3} a \ln \left (-i c x \right )\)	\(348\)

Input:

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

Output:

d^4*a*(c^4*x-2*I*c/x^2-1/3/x^3+6*c^2/x-4*I*c^3*ln(x))+d^4*b*c^3*(c*x*arcta 
n(c*x)-2*I*arctan(c*x)/c^2/x^2+6/c/x*arctan(c*x)-1/3/c^3/x^3*arctan(c*x)-4 
*I*arctan(c*x)*ln(c*x)+2*ln(c*x)*ln(1+I*c*x)-2*ln(c*x)*ln(1-I*c*x)+2*dilog 
(1+I*c*x)-2*dilog(1-I*c*x)+8/3*ln(c^2*x^2+1)-2*I*arctan(c*x)-1/6/c^2/x^2-2 
*I/c/x-19/3*ln(c*x))
 

Fricas [F]

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

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.13 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.30 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{3\,x^3} & \text {\ if\ \ }c=0\\ \frac {b\,c^3\,d^4\,\ln \left (-\frac {3\,c^6\,x^2}{2}-\frac {3\,c^4}{2}\right )}{6}-\frac {b\,c^3\,d^4\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c^3\,d^4\,\ln \left (x\right )}{3}-2\,b\,c^3\,d^4\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )-6\,b\,c\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )-\frac {b\,c\,d^4}{6\,x^2}-\frac {a\,d^4\,\left (1-3\,c^4\,x^4-18\,c^2\,x^2+c\,x\,6{}\mathrm {i}+c^3\,x^3\,\ln \left (x\right )\,12{}\mathrm {i}\right )}{3\,x^3}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}+b\,c^4\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+\frac {6\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )}{x}-b\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )\,2{}\mathrm {i}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,2{}\mathrm {i}}{x^2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \] Input:

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

Output:

piecewise(c == 0, -(a*d^4)/(3*x^3), c ~= 0, - b*d^4*(c^3*atan(c*x) + c^2/x 
)*2i - 2*b*c^3*d^4*(dilog(- c*x*1i + 1) - dilog(c*x*1i + 1)) - (b*c^3*d^4* 
log(c^2*x^2 + 1))/2 - (b*c^3*d^4*log(x))/3 + (b*c^3*d^4*log(- (3*c^4)/2 - 
(3*c^6*x^2)/2))/6 - 6*b*c*d^4*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2) - (b 
*c*d^4)/(6*x^2) - (a*d^4*(c*x*6i - 18*c^2*x^2 - 3*c^4*x^4 + c^3*x^3*log(x) 
*12i + 1))/(3*x^3) - (b*d^4*atan(c*x))/(3*x^3) - (b*c*d^4*atan(c*x)*2i)/x^ 
2 + b*c^4*d^4*x*atan(c*x) + (6*b*c^2*d^4*atan(c*x))/x)
 

Reduce [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\frac {d^{4} \left (6 \mathit {atan} \left (c x \right ) b \,c^{4} x^{4}-12 \mathit {atan} \left (c x \right ) b \,c^{3} i \,x^{3}+36 \mathit {atan} \left (c x \right ) b \,c^{2} x^{2}-12 \mathit {atan} \left (c x \right ) b c i x -2 \mathit {atan} \left (c x \right ) b -24 \left (\int \frac {\mathit {atan} \left (c x \right )}{x}d x \right ) b \,c^{3} i \,x^{3}+16 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{3} x^{3}-24 \,\mathrm {log}\left (x \right ) a \,c^{3} i \,x^{3}-38 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{3}+6 a \,c^{4} x^{4}+36 a \,c^{2} x^{2}-12 a c i x -2 a -12 b \,c^{2} i \,x^{2}-b c x \right )}{6 x^{3}} \] Input:

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

Output:

(d**4*(6*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 - 2*atan(c*x)*b - 24*int(atan(c*x)/x,x 
)*b*c**3*i*x**3 + 16*log(c**2*x**2 + 1)*b*c**3*x**3 - 24*log(x)*a*c**3*i*x 
**3 - 38*log(x)*b*c**3*x**3 + 6*a*c**4*x**4 + 36*a*c**2*x**2 - 12*a*c*i*x 
- 2*a - 12*b*c**2*i*x**2 - b*c*x))/(6*x**3)