Integrand size = 20, antiderivative size = 49 \[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\frac {\arctan (a x) \log \left (2-\frac {2}{1+i a x}\right )}{c}+\frac {i \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i a x}\right )}{2 c} \] Output:
arctan(a*x)*ln(2-2/(1+I*a*x))/c+1/2*I*polylog(2,-1+2/(1+I*a*x))/c
Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.80 \[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\frac {\arctan (a x) \log \left (\frac {2 i}{i-a x}\right )}{c}+\frac {i \operatorname {PolyLog}(2,-i a x)}{2 c}-\frac {i \operatorname {PolyLog}(2,i a x)}{2 c}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i+a x}{i-a x}\right )}{2 c} \] Input:
Integrate[ArcTan[a*x]/(c*x + I*a*c*x^2),x]
Output:
(ArcTan[a*x]*Log[(2*I)/(I - a*x)])/c + ((I/2)*PolyLog[2, (-I)*a*x])/c - (( I/2)*PolyLog[2, I*a*x])/c + ((I/2)*PolyLog[2, -((I + a*x)/(I - a*x))])/c
Time = 0.29 (sec) , antiderivative size = 49, 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 = {2026, 5403, 2897}
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.
\(\displaystyle \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\arctan (a x)}{x (c+i a c x)}dx\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle \frac {\arctan (a x) \log \left (2-\frac {2}{1+i a x}\right )}{c}-\frac {a \int \frac {\log \left (2-\frac {2}{i a x+1}\right )}{a^2 x^2+1}dx}{c}\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {\arctan (a x) \log \left (2-\frac {2}{1+i a x}\right )}{c}+\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i a x+1}-1\right )}{2 c}\) |
Input:
Int[ArcTan[a*x]/(c*x + I*a*c*x^2),x]
Output:
(ArcTan[a*x]*Log[2 - 2/(1 + I*a*x)])/c + ((I/2)*PolyLog[2, -1 + 2/(1 + I*a *x)])/c
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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] - Si mp[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (44 ) = 88\).
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.12
method | result | size |
risch | \(\frac {i \ln \left (i a x +1\right )^{2}}{4 c}+\frac {i \operatorname {dilog}\left (i a x +1\right )}{2 c}-\frac {i \ln \left (-i a x +1\right ) \ln \left (\frac {1}{2}+\frac {i a x}{2}\right )}{2 c}+\frac {i \ln \left (\frac {1}{2}-\frac {i a x}{2}\right ) \ln \left (\frac {1}{2}+\frac {i a x}{2}\right )}{2 c}-\frac {i \operatorname {dilog}\left (-i a x +1\right )}{2 c}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i a x}{2}\right )}{2 c}\) | \(104\) |
derivativedivides | \(\frac {\frac {a \arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {a \arctan \left (a x \right ) \ln \left (a x -i\right )}{c}-\frac {a \left (-\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2}+\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i a x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i a x +1\right )}{2}+\frac {i \ln \left (a x -i\right )^{2}}{4}-\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )+\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )\right )}{2}\right )}{c}}{a}\) | \(139\) |
default | \(\frac {\frac {a \arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {a \arctan \left (a x \right ) \ln \left (a x -i\right )}{c}-\frac {a \left (-\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2}+\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i a x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i a x +1\right )}{2}+\frac {i \ln \left (a x -i\right )^{2}}{4}-\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )+\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )\right )}{2}\right )}{c}}{a}\) | \(139\) |
parts | \(-\frac {\arctan \left (a x \right ) \ln \left (-a x +i\right )}{c}+\frac {\arctan \left (a x \right ) \ln \left (x \right )}{c}-\frac {a \left (-\frac {i \ln \left (x \right ) \left (-\ln \left (-i a x +1\right )+\ln \left (i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}+\frac {-\frac {i \left (\left (\ln \left (-a x +i\right )-\ln \left (-\frac {i \left (-a x +i\right )}{2}\right )\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\operatorname {dilog}\left (-\frac {i \left (-a x +i\right )}{2}\right )\right )}{2}+\frac {i \ln \left (-a x +i\right )^{2}}{4}}{a}\right )}{c}\) | \(156\) |
Input:
int(arctan(a*x)/(c*x+I*a*c*x^2),x,method=_RETURNVERBOSE)
Output:
1/4*I/c*ln(1+I*a*x)^2+1/2*I/c*dilog(1+I*a*x)-1/2*I/c*ln(1-I*a*x)*ln(1/2+1/ 2*I*a*x)+1/2*I/c*ln(1/2-1/2*I*a*x)*ln(1/2+1/2*I*a*x)-1/2*I/c*dilog(1-I*a*x )+1/2*I/c*dilog(1/2-1/2*I*a*x)
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.43 \[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=-\frac {i \, {\rm Li}_2\left (\frac {a x + i}{a x - i} + 1\right )}{2 \, c} \] Input:
integrate(arctan(a*x)/(c*x+I*a*c*x^2),x, algorithm="fricas")
Output:
-1/2*I*dilog((a*x + I)/(a*x - I) + 1)/c
\[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=- \frac {i \int \frac {\operatorname {atan}{\left (a x \right )}}{a x^{2} - i x}\, dx}{c} \] Input:
integrate(atan(a*x)/(c*x+I*a*c*x**2),x)
Output:
-I*Integral(atan(a*x)/(a*x**2 - I*x), x)/c
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (40) = 80\).
Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.57 \[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\frac {1}{4} \, a {\left (-\frac {i \, \log \left (i \, a x + 1\right )^{2}}{a c} + \frac {2 i \, {\left (\log \left (i \, a x + 1\right ) \log \left (-\frac {1}{2} i \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, a x + \frac {1}{2}\right )\right )}}{a c} + \frac {2 i \, {\left (\log \left (i \, a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-i \, a x\right )\right )}}{a c} - \frac {2 i \, {\left (\log \left (-i \, a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (i \, a x\right )\right )}}{a c}\right )} - {\left (\frac {\log \left (i \, a x + 1\right )}{c} - \frac {\log \left (x\right )}{c}\right )} \arctan \left (a x\right ) \] Input:
integrate(arctan(a*x)/(c*x+I*a*c*x^2),x, algorithm="maxima")
Output:
1/4*a*(-I*log(I*a*x + 1)^2/(a*c) + 2*I*(log(I*a*x + 1)*log(-1/2*I*a*x + 1/ 2) + dilog(1/2*I*a*x + 1/2))/(a*c) + 2*I*(log(I*a*x + 1)*log(x) + dilog(-I *a*x))/(a*c) - 2*I*(log(-I*a*x + 1)*log(x) + dilog(I*a*x))/(a*c)) - (log(I *a*x + 1)/c - log(x)/c)*arctan(a*x)
\[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\int { \frac {\arctan \left (a x\right )}{i \, a c x^{2} + c x} \,d x } \] Input:
integrate(arctan(a*x)/(c*x+I*a*c*x^2),x, algorithm="giac")
Output:
integrate(arctan(a*x)/(I*a*c*x^2 + c*x), x)
Timed out. \[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{1{}\mathrm {i}\,a\,c\,x^2+c\,x} \,d x \] Input:
int(atan(a*x)/(c*x + a*c*x^2*1i),x)
Output:
int(atan(a*x)/(c*x + a*c*x^2*1i), x)
\[ \int \frac {\arctan (a x)}{c x+i a c x^2} \, dx=\frac {-\mathit {atan} \left (a x \right )^{2} i -2 \left (\int \frac {\mathit {atan} \left (a x \right )}{a^{3} x^{4}-a^{2} i \,x^{3}+a \,x^{2}-i x}d x \right ) i +2 \left (\int \frac {\mathit {atan} \left (a x \right )}{a^{3} x^{3}-a^{2} i \,x^{2}+a x -i}d x \right ) a}{2 c} \] Input:
int(atan(a*x)/(c*x+I*a*c*x^2),x)
Output:
( - atan(a*x)**2*i - 2*int(atan(a*x)/(a**3*x**4 - a**2*i*x**3 + a*x**2 - i *x),x)*i + 2*int(atan(a*x)/(a**3*x**3 - a**2*i*x**2 + a*x - i),x)*a)/(2*c)