3.1.25 \(\int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx\) [25]

3.1.25.1 Optimal result
3.1.25.2 Mathematica [A] (verified)
3.1.25.3 Rubi [A] (verified)
3.1.25.4 Maple [A] (verified)
3.1.25.5 Fricas [F]
3.1.25.6 Sympy [F]
3.1.25.7 Maxima [A] (verification not implemented)
3.1.25.8 Giac [F]
3.1.25.9 Mupad [B] (verification not implemented)

3.1.25.1 Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=-3 a c^2 d^3 x+\frac {1}{2} i b c^2 d^3 x-\frac {1}{2} i b c d^3 \arctan (c x)-3 b c^2 d^3 x \arctan (c x)-\frac {d^3 (a+b \arctan (c x))}{x}-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))+3 i a c d^3 \log (x)+b c d^3 \log (x)+b c d^3 \log \left (1+c^2 x^2\right )-\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,-i c x)+\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,i c x) \]

output
-3*a*c^2*d^3*x+1/2*I*b*c^2*d^3*x-1/2*I*b*c*d^3*arctan(c*x)-3*b*c^2*d^3*x*a 
rctan(c*x)-d^3*(a+b*arctan(c*x))/x-1/2*I*c^3*d^3*x^2*(a+b*arctan(c*x))+3*I 
*a*c*d^3*ln(x)+b*c*d^3*ln(x)+b*c*d^3*ln(c^2*x^2+1)-3/2*b*c*d^3*polylog(2,- 
I*c*x)+3/2*b*c*d^3*polylog(2,I*c*x)
 
3.1.25.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=\frac {d^3 \left (-2 a-6 a c^2 x^2+i b c^2 x^2-i a c^3 x^3-2 b \arctan (c x)-i b c x \arctan (c x)-6 b c^2 x^2 \arctan (c x)-i b c^3 x^3 \arctan (c x)+6 i a c x \log (x)+2 b c x \log (c x)+2 b c x \log \left (1+c^2 x^2\right )-3 b c x \operatorname {PolyLog}(2,-i c x)+3 b c x \operatorname {PolyLog}(2,i c x)\right )}{2 x} \]

input
Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^2,x]
 
output
(d^3*(-2*a - 6*a*c^2*x^2 + I*b*c^2*x^2 - I*a*c^3*x^3 - 2*b*ArcTan[c*x] - I 
*b*c*x*ArcTan[c*x] - 6*b*c^2*x^2*ArcTan[c*x] - I*b*c^3*x^3*ArcTan[c*x] + ( 
6*I)*a*c*x*Log[x] + 2*b*c*x*Log[c*x] + 2*b*c*x*Log[1 + c^2*x^2] - 3*b*c*x* 
PolyLog[2, (-I)*c*x] + 3*b*c*x*PolyLog[2, I*c*x]))/(2*x)
 
3.1.25.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 162, 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.

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

\(\Big \downarrow \) 5411

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

\(\Big \downarrow \) 2009

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

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

3.1.25.3.1 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])
 
3.1.25.4 Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98

method result size
parts \(a \,d^{3} \left (-\frac {i c^{3} x^{2}}{2}-3 c^{2} x +3 i c \ln \left (x \right )-\frac {1}{x}\right )+b \,d^{3} c \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\) \(159\)
derivativedivides \(c \left (a \,d^{3} \left (-3 c x -\frac {i c^{2} x^{2}}{2}+3 i \ln \left (c x \right )-\frac {1}{c x}\right )+b \,d^{3} \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\right )\) \(162\)
default \(c \left (a \,d^{3} \left (-3 c x -\frac {i c^{2} x^{2}}{2}+3 i \ln \left (c x \right )-\frac {1}{c x}\right )+b \,d^{3} \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\right )\) \(162\)
risch \(-\frac {b \,c^{3} d^{3} \ln \left (i c x +1\right ) x^{2}}{4}-\frac {7 i d^{3} c a}{2}+\frac {3 b c \,d^{3} \ln \left (i c x +1\right )}{4}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{2 x}-3 b c \,d^{3}-\frac {3 b c \,d^{3} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {b c \,d^{3} \ln \left (i c x \right )}{2}-\frac {3 i c^{2} d^{3} b x \ln \left (-i c x +1\right )}{2}+\frac {3 i b \,c^{2} d^{3} \ln \left (i c x +1\right ) x}{2}-3 c^{2} x \,d^{3} a +\frac {i b \,c^{2} d^{3} x}{2}+3 i c \,d^{3} \ln \left (-i c x \right ) a -\frac {d^{3} a}{x}+\frac {c^{3} d^{3} x^{2} b \ln \left (-i c x +1\right )}{4}-\frac {i c^{3} d^{3} a \,x^{2}}{2}+\frac {5 c \,d^{3} \ln \left (-i c x +1\right ) b}{4}-\frac {i d^{3} b \ln \left (-i c x +1\right )}{2 x}+\frac {3 c \,d^{3} \operatorname {dilog}\left (-i c x +1\right ) b}{2}+\frac {c \,d^{3} b \ln \left (-i c x \right )}{2}\) \(274\)

input
int((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^2,x,method=_RETURNVERBOSE)
 
output
a*d^3*(-1/2*I*c^3*x^2-3*c^2*x+3*I*c*ln(x)-1/x)+b*d^3*c*(-3*c*x*arctan(c*x) 
-1/2*I*arctan(c*x)*c^2*x^2+3*I*arctan(c*x)*ln(c*x)-1/c/x*arctan(c*x)-3/2*l 
n(c*x)*ln(1+I*c*x)+3/2*ln(c*x)*ln(1-I*c*x)-3/2*dilog(1+I*c*x)+3/2*dilog(1- 
I*c*x)+1/2*I*c*x+ln(c*x)+ln(c^2*x^2+1)-1/2*I*arctan(c*x))
 
3.1.25.5 Fricas [F]

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

input
integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^2,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^2, x)
 
3.1.25.6 Sympy [F]

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

input
integrate((d+I*c*d*x)**3*(a+b*atan(c*x))/x**2,x)
 
output
-I*d**3*(Integral(-3*I*a*c**2, x) + Integral(I*a/x**2, x) + Integral(-3*a* 
c/x, x) + Integral(a*c**3*x, x) + Integral(-3*I*b*c**2*atan(c*x), x) + Int 
egral(I*b*atan(c*x)/x**2, x) + Integral(-3*b*c*atan(c*x)/x, x) + Integral( 
b*c**3*x*atan(c*x), x))
 
3.1.25.7 Maxima [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.24 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=-\frac {1}{2} i \, a c^{3} d^{3} x^{2} - 3 \, a c^{2} d^{3} x + \frac {1}{2} i \, b c^{2} d^{3} x - \frac {3}{4} i \, \pi b c d^{3} \log \left (c^{2} x^{2} + 1\right ) + 3 i \, b c d^{3} \arctan \left (c x\right ) \log \left (c x\right ) - \frac {3}{2} \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c d^{3} + \frac {3}{2} \, b c d^{3} {\rm Li}_2\left (i \, c x + 1\right ) - \frac {3}{2} \, b c d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + 3 i \, a c d^{3} \log \left (x\right ) - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{3} - \frac {a d^{3}}{x} + \frac {1}{2} \, {\left (-i \, b c^{3} d^{3} x^{2} - i \, b c d^{3}\right )} \arctan \left (c x\right ) \]

input
integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^2,x, algorithm="maxima")
 
output
-1/2*I*a*c^3*d^3*x^2 - 3*a*c^2*d^3*x + 1/2*I*b*c^2*d^3*x - 3/4*I*pi*b*c*d^ 
3*log(c^2*x^2 + 1) + 3*I*b*c*d^3*arctan(c*x)*log(c*x) - 3/2*(2*c*x*arctan( 
c*x) - log(c^2*x^2 + 1))*b*c*d^3 + 3/2*b*c*d^3*dilog(I*c*x + 1) - 3/2*b*c* 
d^3*dilog(-I*c*x + 1) + 3*I*a*c*d^3*log(x) - 1/2*(c*(log(c^2*x^2 + 1) - lo 
g(x^2)) + 2*arctan(c*x)/x)*b*d^3 - a*d^3/x + 1/2*(-I*b*c^3*d^3*x^2 - I*b*c 
*d^3)*arctan(c*x)
 
3.1.25.8 Giac [F]

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

input
integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^2,x, algorithm="giac")
 
output
sage0*x
 
3.1.25.9 Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^3}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d^3\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}-\frac {a\,c^3\,d^3\,x^2\,1{}\mathrm {i}}{2}-\frac {a\,d^3}{x}+\frac {3\,b\,c\,d^3\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}+\frac {3\,b\,c\,d^3\,\ln \left (c^2\,x^2+1\right )}{2}-3\,a\,c^2\,d^3\,x+\frac {b\,c^2\,d^3\,x\,1{}\mathrm {i}}{2}+a\,c\,d^3\,\ln \left (x\right )\,3{}\mathrm {i}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{x}-3\,b\,c^2\,d^3\,x\,\mathrm {atan}\left (c\,x\right )-b\,c^3\,d^3\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\,1{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

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