Integrand size = 21, antiderivative size = 77 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=-\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1+c^2 x^2\right )-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x) \]
-d*(a+b*arctan(c*x))/x+I*a*c*d*ln(x)+b*c*d*ln(x)-1/2*b*c*d*ln(c^2*x^2+1)-1 /2*b*c*d*polylog(2,-I*c*x)+1/2*b*c*d*polylog(2,I*c*x)
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\frac {d \left (-2 a-2 b \arctan (c x)+2 i a c x \log (x)+2 b c x \log (x)-b c x \log \left (1+c^2 x^2\right )-b c x \operatorname {PolyLog}(2,-i c x)+b c x \operatorname {PolyLog}(2,i c x)\right )}{2 x} \]
(d*(-2*a - 2*b*ArcTan[c*x] + (2*I)*a*c*x*Log[x] + 2*b*c*x*Log[x] - b*c*x*L og[1 + c^2*x^2] - b*c*x*PolyLog[2, (-I)*c*x] + b*c*x*PolyLog[2, I*c*x]))/( 2*x)
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx\) |
\(\Big \downarrow \) 5411 |
\(\displaystyle \int \left (\frac {d (a+b \arctan (c x))}{x^2}+\frac {i c d (a+b \arctan (c x))}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)-\frac {1}{2} b c d \log \left (c^2 x^2+1\right )-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x)+b c d \log (x)\) |
-((d*(a + b*ArcTan[c*x]))/x) + I*a*c*d*Log[x] + b*c*d*Log[x] - (b*c*d*Log[ 1 + c^2*x^2])/2 - (b*c*d*PolyLog[2, (-I)*c*x])/2 + (b*c*d*PolyLog[2, I*c*x ])/2
3.1.6.3.1 Defintions of rubi rules used
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])
Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.42
method | result | size |
parts | \(a d \left (i c \ln \left (x \right )-\frac {1}{x}\right )+b d c \left (i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )\) | \(109\) |
derivativedivides | \(c \left (a d \left (i \ln \left (c x \right )-\frac {1}{c x}\right )+b d \left (i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )\right )\) | \(114\) |
default | \(c \left (a d \left (i \ln \left (c x \right )-\frac {1}{c x}\right )+b d \left (i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {\ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {\operatorname {dilog}\left (i c x +1\right )}{2}+\frac {\operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\ln \left (c^{2} x^{2}+1\right )}{2}+\ln \left (c x \right )\right )\right )\) | \(114\) |
risch | \(-\frac {b c d \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {b c d \ln \left (i c x \right )}{2}-\frac {b c d \ln \left (i c x +1\right )}{2}+\frac {i b d \ln \left (i c x +1\right )}{2 x}+i c d \ln \left (-i c x \right ) a -\frac {a d}{x}+\frac {c d \operatorname {dilog}\left (-i c x +1\right ) b}{2}+\frac {c d b \ln \left (-i c x \right )}{2}-\frac {\ln \left (-i c x +1\right ) b c d}{2}-\frac {i d b \ln \left (-i c x +1\right )}{2 x}\) | \(127\) |
a*d*(I*c*ln(x)-1/x)+b*d*c*(I*arctan(c*x)*ln(c*x)-1/c/x*arctan(c*x)-1/2*ln( c*x)*ln(1+I*c*x)+1/2*ln(c*x)*ln(1-I*c*x)-1/2*dilog(1+I*c*x)+1/2*dilog(1-I* c*x)-1/2*ln(c^2*x^2+1)+ln(c*x))
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=i d \left (\int \left (- \frac {i a}{x^{2}}\right )\, dx + \int \frac {a c}{x}\, dx + \int \left (- \frac {i b \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \frac {b c \operatorname {atan}{\left (c x \right )}}{x}\, dx\right ) \]
I*d*(Integral(-I*a/x**2, x) + Integral(a*c/x, x) + Integral(-I*b*atan(c*x) /x**2, x) + Integral(b*c*atan(c*x)/x, x))
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]
I*b*c*d*integrate(arctan(c*x)/x, x) + I*a*c*d*log(x) - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d - a*d/x
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]
Time = 0.93 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+\frac {b\,c\,d\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}+\frac {a\,d\,\left (-1+c\,x\,\ln \left (x\right )\,1{}\mathrm {i}\right )}{x}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]