\(\int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx\) [6]

Optimal result

Integrand size = 21, antiderivative size = 77 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=-\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1+c^2 x^2\right )-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x) \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {4996, 4946, 272, 36, 29, 31, 4940, 2438} \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=-\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)-\frac {1}{2} b c d \log \left (c^2 x^2+1\right )-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x)+b c d \log (x) \]

[In]

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

[Out]

-((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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(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])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d (a+b \arctan (c x))}{x^2}+\frac {i c d (a+b \arctan (c x))}{x}\right ) \, dx \\ & = d \int \frac {a+b \arctan (c x)}{x^2} \, dx+(i c d) \int \frac {a+b \arctan (c x)}{x} \, dx \\ & = -\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)-\frac {1}{2} (b c d) \int \frac {\log (1-i c x)}{x} \, dx+\frac {1}{2} (b c d) \int \frac {\log (1+i c x)}{x} \, dx+(b c d) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x)+\frac {1}{2} (b c d) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {d (a+b \arctan (c x))}{x}+i a c d \log (x)-\frac {1}{2} b c d \operatorname {PolyLog}(2,-i c x)+\frac {1}{2} b c d \operatorname {PolyLog}(2,i c x)+\frac {1}{2} (b c d) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b c^3 d\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\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) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\frac {d \left (-2 a-2 b \arctan (c x)+2 i a c x \log (x)+2 b c x \log (x)-b c x \log \left (1+c^2 x^2\right )-b c x \operatorname {PolyLog}(2,-i c x)+b c x \operatorname {PolyLog}(2,i c x)\right )}{2 x} \]

[In]

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

[Out]

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

Maple [A] (verified)

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	\(-\frac {b c d \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {b c d \ln \left (i c x \right )}{2}-\frac {b c d \ln \left (i c x +1\right )}{2}+\frac {i b d \ln \left (i c x +1\right )}{2 x}+i c d \ln \left (-i c x \right ) a -\frac {a d}{x}+\frac {c d \operatorname {dilog}\left (-i c x +1\right ) b}{2}+\frac {c d b \ln \left (-i c x \right )}{2}-\frac {\ln \left (-i c x +1\right ) b c d}{2}-\frac {i d b \ln \left (-i c x +1\right )}{2 x}\)	\(127\)

[In]

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

[Out]

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))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+\frac {b\,c\,d\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}+\frac {a\,d\,\left (-1+c\,x\,\ln \left (x\right )\,1{}\mathrm {i}\right )}{x}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

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