\(\int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx\) [55]

Optimal result

Integrand size = 23, antiderivative size = 150 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=\frac {b}{2 d^2 (i-c x)}-\frac {b \arctan (c x)}{2 d^2}+\frac {i (a+b \arctan (c x))}{d^2 (i-c x)}+\frac {a \log (x)}{d^2}+\frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {i b \operatorname {PolyLog}(2,-i c x)}{2 d^2}-\frac {i b \operatorname {PolyLog}(2,i c x)}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4940, 2438, 4972, 641, 46, 209, 4964, 2449, 2352} \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=\frac {i (a+b \arctan (c x))}{d^2 (-c x+i)}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{d^2}+\frac {a \log (x)}{d^2}-\frac {b \arctan (c x)}{2 d^2}+\frac {i b \operatorname {PolyLog}(2,-i c x)}{2 d^2}-\frac {i b \operatorname {PolyLog}(2,i c x)}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 d^2}+\frac {b}{2 d^2 (-c x+i)} \]

[In]

Int[(a + b*ArcTan[c*x])/(x*(d + I*c*d*x)^2),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 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 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 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 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]

Rule 4972

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

Rule 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=\frac {b \left (\frac {1}{i-c x}-\arctan (c x)\right )-\frac {2 i (a+b \arctan (c x))}{-i+c x}+2 a \log (x)+2 (a+b \arctan (c x)) \log \left (\frac {2 i}{i-c x}\right )+i b \operatorname {PolyLog}(2,-i c x)-i b \operatorname {PolyLog}(2,i c x)+i b \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )}{2 d^2} \]

[In]

Integrate[(a + b*ArcTan[c*x])/(x*(d + I*c*d*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.38

method	result	size
parts	\(\frac {a \ln \left (x \right )}{d^{2}}+\frac {i a}{d^{2} \left (-c x +i\right )}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}-\frac {i a \arctan \left (c x \right )}{d^{2}}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\frac {i \arctan \left (c x \right )}{c x -i}-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\arctan \left (c x \right )}{2}-\frac {1}{2 \left (c x -i\right )}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d^{2}}\)	\(207\)
derivativedivides	\(\frac {a \ln \left (c x \right )}{d^{2}}-\frac {i a}{d^{2} \left (c x -i\right )}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}-\frac {i a \arctan \left (c x \right )}{d^{2}}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\frac {i \arctan \left (c x \right )}{c x -i}-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\arctan \left (c x \right )}{2}-\frac {1}{2 \left (c x -i\right )}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d^{2}}\)	\(208\)
default	\(\frac {a \ln \left (c x \right )}{d^{2}}-\frac {i a}{d^{2} \left (c x -i\right )}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}-\frac {i a \arctan \left (c x \right )}{d^{2}}+\frac {b \left (\arctan \left (c x \right ) \ln \left (c x \right )-\frac {i \arctan \left (c x \right )}{c x -i}-\arctan \left (c x \right ) \ln \left (c x -i\right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {\arctan \left (c x \right )}{2}-\frac {1}{2 \left (c x -i\right )}+\frac {i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}-\frac {i \ln \left (c x -i\right )^{2}}{4}\right )}{d^{2}}\)	\(208\)
risch	\(\frac {i b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{2}}+\frac {i b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{2}}-\frac {i b \operatorname {dilog}\left (-i c x +1\right )}{2 d^{2}}-\frac {i b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d^{2}}-\frac {i b \ln \left (-i c x +1\right )}{4 d^{2} \left (-i c x -1\right )}-\frac {b \arctan \left (c x \right )}{4 d^{2}}-\frac {b \ln \left (-i c x +1\right ) c x}{4 d^{2} \left (-i c x -1\right )}-\frac {i a \arctan \left (c x \right )}{d^{2}}+\frac {a \ln \left (-i c x \right )}{d^{2}}-\frac {a}{d^{2} \left (-i c x -1\right )}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {i b \ln \left (c^{2} x^{2}+1\right )}{8 d^{2}}-\frac {i b \ln \left (i c x +1\right )}{2 d^{2} \left (i c x +1\right )}-\frac {i b}{2 d^{2} \left (i c x +1\right )}+\frac {i b \operatorname {dilog}\left (i c x +1\right )}{2 d^{2}}+\frac {i b \ln \left (i c x +1\right )^{2}}{4 d^{2}}\)	\(279\)

[In]

int((a+b*arctan(c*x))/x/(d+I*c*d*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=-\frac {2 \, {\left (i \, b c x + b\right )} {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) - 4 \, {\left (a c x - i \, a\right )} \log \left (x\right ) - 2 \, b \log \left (-\frac {c x + i}{c x - i}\right ) - {\left (-i \, b c x - b\right )} \log \left (\frac {c x + i}{c}\right ) + {\left ({\left (4 \, a - i \, b\right )} c x - 4 i \, a - b\right )} \log \left (\frac {c x - i}{c}\right ) + 4 i \, a + 2 \, b}{4 \, {\left (c d^{2} x - i \, d^{2}\right )}} \]

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x)^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x)^2,x, algorithm="maxima")

[Out]

(-2*I*c*integrate(arctan(c*x)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), x) - integrate((c^2*x^2 - 1)*arctan(c*x)/(c
^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x))*b + a*(-I/(c*d^2*x - I*d^2) - log(c*x - I)/d^2 + log(x)/d^2)

Giac [F]

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

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x)^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c x)}{x (d+i c d x)^2} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

int((a + b*atan(c*x))/(x*(d + c*d*x*1i)^2),x)

[Out]

int((a + b*atan(c*x))/(x*(d + c*d*x*1i)^2), x)