\(\int \frac {a+b \arctan (c x)}{d+e x} \, dx\) [137]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 138 \[ \int \frac {a+b \arctan (c x)}{d+e x} \, dx=-\frac {(a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {(a+b \arctan (c x)) \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4966, 2449, 2352, 2497} \[ \int \frac {a+b \arctan (c x)}{d+e x} \, dx=\frac {(a+b \arctan (c x)) \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e}-\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{e}-\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e} \]

[In]

Int[(a + b*ArcTan[c*x])/(d + e*x),x]

[Out]

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

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 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 4966

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \arctan (c x)}{d+e x} \, dx=\frac {2 a \log (d+e x)+i b \log (1-i c x) \log \left (\frac {c (d+e x)}{c d-i e}\right )-i b \log (1+i c x) \log \left (\frac {c (d+e x)}{c d+i e}\right )+i b \operatorname {PolyLog}\left (2,\frac {e (1-i c x)}{i c d+e}\right )-i b \operatorname {PolyLog}\left (2,-\frac {e (-i+c x)}{c d+i e}\right )}{2 e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.13

method result size
parts \(\frac {a \ln \left (e x +d \right )}{e}+\frac {b \left (\frac {c \ln \left (e c x +c d \right ) \arctan \left (c x \right )}{e}-c \left (-\frac {i \ln \left (e c x +c d \right ) \left (\ln \left (\frac {-e c x +i e}{c d +i e}\right )-\ln \left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}-\frac {i \left (\operatorname {dilog}\left (\frac {-e c x +i e}{c d +i e}\right )-\operatorname {dilog}\left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}\right )\right )}{c}\) \(156\)
derivativedivides \(\frac {\frac {a c \ln \left (e c x +c d \right )}{e}+b c \left (\frac {\ln \left (e c x +c d \right ) \arctan \left (c x \right )}{e}+\frac {i \ln \left (e c x +c d \right ) \left (\ln \left (\frac {-e c x +i e}{c d +i e}\right )-\ln \left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}+\frac {i \left (\operatorname {dilog}\left (\frac {-e c x +i e}{c d +i e}\right )-\operatorname {dilog}\left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}\right )}{c}\) \(157\)
default \(\frac {\frac {a c \ln \left (e c x +c d \right )}{e}+b c \left (\frac {\ln \left (e c x +c d \right ) \arctan \left (c x \right )}{e}+\frac {i \ln \left (e c x +c d \right ) \left (\ln \left (\frac {-e c x +i e}{c d +i e}\right )-\ln \left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}+\frac {i \left (\operatorname {dilog}\left (\frac {-e c x +i e}{c d +i e}\right )-\operatorname {dilog}\left (\frac {e c x +i e}{-c d +i e}\right )\right )}{2 e}\right )}{c}\) \(157\)
risch \(\frac {i b \operatorname {dilog}\left (\frac {-i c d +\left (-i c x +1\right ) e -e}{-i c d -e}\right )}{2 e}+\frac {i b \ln \left (-i c x +1\right ) \ln \left (\frac {-i c d +\left (-i c x +1\right ) e -e}{-i c d -e}\right )}{2 e}+\frac {a \ln \left (i c d -\left (-i c x +1\right ) e +e \right )}{e}-\frac {i b \operatorname {dilog}\left (\frac {i c d +\left (i c x +1\right ) e -e}{i c d -e}\right )}{2 e}-\frac {i b \ln \left (i c x +1\right ) \ln \left (\frac {i c d +\left (i c x +1\right ) e -e}{i c d -e}\right )}{2 e}\) \(193\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

2*b*integrate(1/2*arctan(c*x)/(e*x + d), x) + a*log(e*x + d)/e

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((a + b*atan(c*x))/(d + e*x),x)

[Out]

int((a + b*atan(c*x))/(d + e*x), x)