\(\int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx\) [47]

Optimal result

Integrand size = 23, antiderivative size = 54 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx=\frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {i b \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d} \]

[Out]

(a+b*arctan(c*x))*ln(2-2/(1+I*c*x))/d+1/2*I*b*polylog(2,-1+2/(1+I*c*x))/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, 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 = {4988, 2497} \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx=\frac {\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{d}+\frac {i b \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right )}{2 d} \]

[In]

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

[Out]

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

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))
]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.89 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx=\frac {a \log (x)}{d}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 i}{i-c x}\right )}{d}+\frac {i b \operatorname {PolyLog}(2,-i c x)}{2 d}-\frac {i b \operatorname {PolyLog}(2,i c x)}{2 d}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {i+c x}{i-c x}\right )}{2 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (49 ) = 98\).

Time = 0.85 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.74

method	result	size
risch	\(\frac {i b \ln \left (i c x +1\right )^{2}}{4 d}+\frac {i b \operatorname {dilog}\left (i c x +1\right )}{2 d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {\ln \left (-i c x \right ) a}{d}-\frac {i \ln \left (-i c x +1\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) b}{2 d}+\frac {i \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) b}{2 d}-\frac {i a \arctan \left (c x \right )}{d}+\frac {i b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d}-\frac {i \operatorname {dilog}\left (-i c x +1\right ) b}{2 d}\)	\(148\)
parts	\(\frac {a \ln \left (x \right )}{d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {i a \arctan \left (c x \right )}{d}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d}\)	\(160\)
derivativedivides	\(\frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {i a \arctan \left (c x \right )}{d}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d}\)	\(162\)
default	\(\frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {i a \arctan \left (c x \right )}{d}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d}\)	\(162\)

[In]

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

[Out]

1/4*I/d*b*ln(1+I*c*x)^2+1/2*I/d*b*dilog(1+I*c*x)-1/2/d*a*ln(c^2*x^2+1)+1/d*ln(-I*c*x)*a-1/2*I/d*ln(1-I*c*x)*ln
(1/2+1/2*I*c*x)*b+1/2*I/d*ln(1/2-1/2*I*c*x)*ln(1/2+1/2*I*c*x)*b-I/d*a*arctan(c*x)+1/2*I/d*b*dilog(1/2-1/2*I*c*
x)-1/2*I/d*dilog(1-I*c*x)*b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx=\frac {-i \, b {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) + 2 \, a \log \left (x\right ) - 2 \, a \log \left (\frac {c x - i}{c}\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(-I*b*dilog((c*x + I)/(c*x - I) + 1) + 2*a*log(x) - 2*a*log((c*x - I)/c))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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