\(\int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx\) [100]

Optimal result

Integrand size = 25, antiderivative size = 186 \[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx=-\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}+\frac {2 b c (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}+\frac {b c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \]

[Out]

-I*c*(a+b*arctan(c*x))^2/d-(a+b*arctan(c*x))^2/d/x+2*b*c*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d-I*c*(a+b*arctan
(c*x))^2*ln(2-2/(1+I*c*x))/d-I*b^2*c*polylog(2,-1+2/(1-I*c*x))/d+b*c*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x
))/d-1/2*I*b^2*c*polylog(3,-1+2/(1+I*c*x))/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4990, 4946, 5044, 4988, 2497, 5004, 5114, 6745} \[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx=\frac {b c \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{d}-\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}+\frac {2 b c \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{2 d} \]

[In]

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

[Out]

((-I)*c*(a + b*ArcTan[c*x])^2)/d - (a + b*ArcTan[c*x])^2/(d*x) + (2*b*c*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c
*x)])/d - (I*c*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 + I*c*x)])/d - (I*b^2*c*PolyLog[2, -1 + 2/(1 - I*c*x)])/d +
(b*c*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d - ((I/2)*b^2*c*PolyLog[3, -1 + 2/(1 + I*c*x)])/d

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4988

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] - Dist[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]

Rule 4990

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[1/d, I
nt[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f), Int[(f*x)^(m + 1)*((a + b*ArcTan[c*x])^p/(d + e*x)),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && LtQ[m, -1]

Rule 5004

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

Rule 5044

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

Rule 5114

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

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps \begin{align*} \text {integral}& = -\left ((i c) \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx\right )+\frac {\int \frac {(a+b \arctan (c x))^2}{x^2} \, dx}{d} \\ & = -\frac {(a+b \arctan (c x))^2}{d x}-\frac {i c (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(2 b c) \int \frac {a+b \arctan (c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (2 i b c^2\right ) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}-\frac {i c (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {b c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}+\frac {(2 i b c) \int \frac {a+b \arctan (c x)}{x (i+c x)} \, dx}{d}-\frac {\left (b^2 c^2\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}+\frac {2 b c (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {b c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (2 b^2 c^2\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {i c (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{d x}+\frac {2 b c (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}+\frac {b c (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {i b^2 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx=-\frac {\frac {2 a^2}{x}+2 a^2 c \arctan (c x)+2 i a^2 c \log (x)-i a^2 c \log \left (1+c^2 x^2\right )+\frac {2 a b \left (2 c x \arctan (c x)^2+\arctan (c x) \left (2+2 i c x \log \left (1-e^{2 i \arctan (c x)}\right )\right )+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )+c x \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )}{x}+2 i b^2 c \left (-\frac {i \pi ^3}{24}+\arctan (c x)^2-\frac {i \arctan (c x)^2}{c x}+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )+2 i \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )\right )}{2 d} \]

[In]

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

[Out]

-1/2*((2*a^2)/x + 2*a^2*c*ArcTan[c*x] + (2*I)*a^2*c*Log[x] - I*a^2*c*Log[1 + c^2*x^2] + (2*a*b*(2*c*x*ArcTan[c
*x]^2 + ArcTan[c*x]*(2 + (2*I)*c*x*Log[1 - E^((2*I)*ArcTan[c*x])]) + c*x*(-2*Log[c*x] + Log[1 + c^2*x^2]) + c*
x*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/x + (2*I)*b^2*c*((-1/24*I)*Pi^3 + ArcTan[c*x]^2 - (I*ArcTan[c*x]^2)/(c*x
) + ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (2*I)*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + I*ArcTa
n[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + PolyLog[2, E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*
x])]/2))/d

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 14.05 (sec) , antiderivative size = 8468, normalized size of antiderivative = 45.53

method	result	size
parts	\(\text {Expression too large to display}\)	\(8468\)
derivativedivides	\(\text {Expression too large to display}\)	\(8470\)
default	\(\text {Expression too large to display}\)	\(8470\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

1/4*(I*b^2*c*x*log(2*c*x/(c*x - I))*log(-(c*x + I)/(c*x - I))^2 + 2*I*b^2*c*x*dilog(-2*c*x/(c*x - I) + 1)*log(
-(c*x + I)/(c*x - I)) - 2*I*b^2*c*x*polylog(3, -(c*x + I)/(c*x - I)) + b^2*log(-(c*x + I)/(c*x - I))^2 + 4*d*x
*integral((-I*a^2*c*x + a^2 + ((a*b + I*b^2)*c*x + I*a*b)*log(-(c*x + I)/(c*x - I)))/(c^2*d*x^4 + d*x^2), x))/
(d*x)

Sympy [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx=- \frac {i \left (\int \frac {a^{2}}{c x^{3} - i x^{2}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx\right )}{d} \]

[In]

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

[Out]

-I*(Integral(a**2/(c*x**3 - I*x**2), x) + Integral(b**2*atan(c*x)**2/(c*x**3 - I*x**2), x) + Integral(2*a*b*at
an(c*x)/(c*x**3 - I*x**2), x))/d

Maxima [F]

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

[In]

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

[Out]

a^2*(I*c*log(I*c*x + 1)/d - I*c*log(x)/d - 1/(d*x)) - 1/96*(24*b^2*c*x*arctan(c*x)^3 + 3*I*b^2*c*x*log(c^2*x^2
+ 1)^3 + 24*b^2*arctan(c*x)^2 - 2*I*(384*b^2*c^3*integrate(1/16*x^3*arctan(c*x)^2/(c^2*d*x^4 + d*x^2), x) + b
^2*c*log(c^2*x^2 + 1)^3/d + 12*b^2*c*arctan(c*x)^2/d - 576*b^2*c*integrate(1/16*x*arctan(c*x)^2/(c^2*d*x^4 + d
*x^2), x) - 48*b^2*c*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) - 1536*a*b*c*integrate(1/16*x
*arctan(c*x)/(c^2*d*x^4 + d*x^2), x) + 192*b^2*c*integrate(1/16*x*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x) - 1
92*b^2*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x))*d*x - 16*(b^2*c*arctan(c*x)^3/d +
12*b^2*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) - 24*b^2*c^2*integrate(1/16*x^2*log(c
^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x) - 24*b^2*c*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^
2), x) + 48*b^2*c*integrate(1/16*x*arctan(c*x)/(c^2*d*x^4 + d*x^2), x) + 72*b^2*integrate(1/16*arctan(c*x)^2/(
c^2*d*x^4 + d*x^2), x) + 6*b^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) + 192*a*b*integrate(1
/16*arctan(c*x)/(c^2*d*x^4 + d*x^2), x))*d*x + 6*(b^2*c*x*arctan(c*x) - b^2)*log(c^2*x^2 + 1)^2 + 12*(I*b^2*c*
x*arctan(c*x)^2 + 2*I*b^2*arctan(c*x))*log(c^2*x^2 + 1))/(d*x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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