\(\int \frac {\arctan (a+b x)}{\frac {a d}{b}+d x} \, dx\) [22]

Optimal result

Integrand size = 19, antiderivative size = 41 \[ \int \frac {\arctan (a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \operatorname {PolyLog}(2,-i (a+b x))}{2 d}-\frac {i \operatorname {PolyLog}(2,i (a+b x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5151, 12, 4940, 2438} \[ \int \frac {\arctan (a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \operatorname {PolyLog}(2,-i (a+b x))}{2 d}-\frac {i \operatorname {PolyLog}(2,i (a+b x))}{2 d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5151

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {\arctan (a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i (\operatorname {PolyLog}(2,-i (a+b x))-\operatorname {PolyLog}(2,i (a+b x)))}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {i \operatorname {dilog}\left (-i b x -i a +1\right )}{2 d}+\frac {i \operatorname {dilog}\left (i b x +i a +1\right )}{2 d}\)	\(38\)
parts	\(\frac {\ln \left (b x +a \right ) \arctan \left (b x +a \right )}{d}-\frac {-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}}{d}\)	\(92\)
derivativedivides	\(\frac {\frac {b \ln \left (b x +a \right ) \arctan \left (b x +a \right )}{d}-\frac {b \left (-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}\right )}{d}}{b}\)	\(98\)
default	\(\frac {\frac {b \ln \left (b x +a \right ) \arctan \left (b x +a \right )}{d}-\frac {b \left (-\frac {i \ln \left (b x +a \right ) \ln \left (1+i \left (b x +a \right )\right )}{2}+\frac {i \ln \left (b x +a \right ) \ln \left (1-i \left (b x +a \right )\right )}{2}-\frac {i \operatorname {dilog}\left (1+i \left (b x +a \right )\right )}{2}+\frac {i \operatorname {dilog}\left (1-i \left (b x +a \right )\right )}{2}\right )}{d}}{b}\)	\(98\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(b*arctan(b*x + a)/(b*d*x + a*d), x)

Sympy [F]

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

[In]

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

[Out]

b*Integral(atan(a + b*x)/(a + b*x), x)/d

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (29) = 58\).

Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 3.00 \[ \int \frac {\arctan (a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\arctan \left (b x + a\right ) \log \left (d x + \frac {a d}{b}\right )}{d} - \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + \frac {a d}{b}\right )}{d} - \frac {\arctan \left (b x + a, 0\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, \arctan \left (b x + a\right ) \log \left ({\left | b x + a \right |}\right ) + i \, {\rm Li}_2\left (i \, b x + i \, a + 1\right ) - i \, {\rm Li}_2\left (-i \, b x - i \, a + 1\right )}{2 \, d} \]

[In]

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

[Out]

arctan(b*x + a)*log(d*x + a*d/b)/d - arctan((b^2*x + a*b)/b)*log(d*x + a*d/b)/d - 1/2*(arctan2(b*x + a, 0)*log
(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*arctan(b*x + a)*log(abs(b*x + a)) + I*dilog(I*b*x + I*a + 1) - I*dilog(-I*b*
x - I*a + 1))/d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(atan(a + b*x)/(d*x + (a*d)/b),x)

[Out]

int(atan(a + b*x)/(d*x + (a*d)/b), x)