\(\int \frac {a+b \arctan (c x)}{d+i c d x} \, dx\) [46]

Optimal result

Integrand size = 20, antiderivative size = 59 \[ \int \frac {a+b \arctan (c x)}{d+i c d x} \, dx=\frac {i (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4964, 2449, 2352} \[ \int \frac {a+b \arctan (c x)}{d+i c d x} \, dx=\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c d} \]

[In]

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

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 {i (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {(i b) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = \frac {i (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c d} \\ & = \frac {i (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \arctan (c x)}{d+i c d x} \, dx=\frac {2 i (a+b \arctan (c x)) \log \left (\frac {2 d}{d+i c d x}\right )-b \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )}{2 c d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.76

method	result	size
derivativedivides	\(\frac {-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a \arctan \left (c x \right )}{d}+\frac {b \left (-i \ln \left (i c x +1\right ) \arctan \left (c x \right )-\frac {\left (\ln \left (i c x +1\right )-\ln \left (\frac {1}{2}+\frac {i c x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {1}{2}+\frac {i c x}{2}\right )}{2}+\frac {\ln \left (i c x +1\right )^{2}}{4}\right )}{d}}{c}\)	\(104\)
default	\(\frac {-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a \arctan \left (c x \right )}{d}+\frac {b \left (-i \ln \left (i c x +1\right ) \arctan \left (c x \right )-\frac {\left (\ln \left (i c x +1\right )-\ln \left (\frac {1}{2}+\frac {i c x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {1}{2}+\frac {i c x}{2}\right )}{2}+\frac {\ln \left (i c x +1\right )^{2}}{4}\right )}{d}}{c}\)	\(104\)
parts	\(-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 c d}+\frac {a \arctan \left (c x \right )}{c d}+\frac {b \left (-i \ln \left (i c x +1\right ) \arctan \left (c x \right )-\frac {\left (\ln \left (i c x +1\right )-\ln \left (\frac {1}{2}+\frac {i c x}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {1}{2}+\frac {i c x}{2}\right )}{2}+\frac {\ln \left (i c x +1\right )^{2}}{4}\right )}{d c}\)	\(109\)
risch	\(-\frac {b \ln \left (i c x +1\right )^{2}}{4 d c}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 c d}-\frac {b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 c d}+\frac {b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 c d}+\frac {a \arctan \left (c x \right )}{c d}-\frac {b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 c d}\)	\(120\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {a+b \arctan (c x)}{d+i c d x} \, dx=\frac {b \log {\left (- i c x + 1 \right )} \log {\left (i c x + 1 \right )}}{2 c d} - \frac {i \left (\int \frac {i a}{c^{2} x^{2} + 1}\, dx + \int \frac {a c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {i b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{d} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-1/8*(8*I*c^2*d*integrate(x*arctan(c*x)/(c^2*d*x^2 + d), x) + 4*c^2*d*integrate(x*log(c^2*x^2 + 1)/(c^2*d*x^2
+ d), x) - 4*arctan(c*x)^2 - log(c^2*x^2 + 1)^2)*b/(c*d) - I*a*log(I*c*d*x + d)/(c*d)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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