\(\int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx\) [73]

Optimal result

Integrand size = 23, antiderivative size = 228 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=-i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+2 b c d (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4946, 5044, 4988, 2497, 4942, 5108, 5004, 5114, 6745} \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=2 i c d \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2+b c d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))-b c d \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))-i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-i b^2 c d \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )-\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right ) \]

[In]

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

[Out]

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

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 4942

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[(a + b*ArcTan[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 + I*c*x)]/(1 + c^2*x^2)), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

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 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])

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 5108

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L
og[1 + u]*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] - Dist[1/2, Int[Log[1 - u]*((a + b*ArcTan[c*x])^p/(d + e
*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - 2*(I/(I - c*x)))^
2, 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}& = \int \left (\frac {d (a+b \arctan (c x))^2}{x^2}+\frac {i c d (a+b \arctan (c x))^2}{x}\right ) \, dx \\ & = d \int \frac {(a+b \arctan (c x))^2}{x^2} \, dx+(i c d) \int \frac {(a+b \arctan (c x))^2}{x} \, dx \\ & = -\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+(2 b c d) \int \frac {a+b \arctan (c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (4 i b c^2 d\right ) \int \frac {(a+b \arctan (c x)) \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = -i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+(2 i b c d) \int \frac {a+b \arctan (c x)}{x (i+c x)} \, dx+\left (2 i b c^2 d\right ) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b c^2 d\right ) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = -i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+2 b c d (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )+b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\left (b^2 c^2 d\right ) \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (b^2 c^2 d\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b^2 c^2 d\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx \\ & = -i c d (a+b \arctan (c x))^2-\frac {d (a+b \arctan (c x))^2}{x}+2 i c d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+2 b c d (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-b c d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.27 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))^2}{x^2} \, dx=\frac {i d \left (i a^2+a^2 c x \log (x)+i a b \left (2 \arctan (c x)+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )\right )+i b^2 \left (\arctan (c x)^2-2 c x \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+i c x \left (\arctan (c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )\right )+i a b c x (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+\frac {1}{24} b^2 c x \left (-i \pi ^3+16 i \arctan (c x)^3+24 \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-24 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-12 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )\right )}{x} \]

[In]

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

[Out]

(I*d*(I*a^2 + a^2*c*x*Log[x] + I*a*b*(2*ArcTan[c*x] + c*x*(-2*Log[c*x] + Log[1 + c^2*x^2])) + I*b^2*(ArcTan[c*
x]^2 - 2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + I*c*x*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*
x])])) + I*a*b*c*x*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*c*x*((-I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 2
4*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 24*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*Arc
Tan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 12*PolyL
og[3, E^((-2*I)*ArcTan[c*x])] - 12*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/24))/x

Maple [C] (warning: unable to verify)

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

Time = 15.89 (sec) , antiderivative size = 5609, normalized size of antiderivative = 24.60

method	result	size
parts	\(\text {Expression too large to display}\)	\(5609\)
derivativedivides	\(\text {Expression too large to display}\)	\(5613\)
default	\(\text {Expression too large to display}\)	\(5613\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

I*a^2*c*d*log(x) - (c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b*d - a^2*d/x - 1/96*(24*b^2*d*arctan
(c*x)^2 + 24*I*b^2*d*arctan(c*x)*log(c^2*x^2 + 1) - 6*b^2*d*log(c^2*x^2 + 1)^2 - 24*(b^2*c*d*arctan(c*x)^3 + 1
6*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 4*b^2*c^2*d*integrate(1/16*x
^2*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x) - 16*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^4 + x^2),
x) + 16*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 32*b^2*c*d*integrate(1/16*
x*arctan(c*x)/(c^2*x^4 + x^2), x) + 48*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 4*b^2*d*integr
ate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x))*x - I*(1152*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x
^4 + x^2), x) + 3072*a*b*c^3*d*integrate(1/16*x^3*arctan(c*x)/(c^2*x^4 + x^2), x) + b^2*c*d*log(c^2*x^2 + 1)^3
+ 24*b^2*c*d*arctan(c*x)^2 - 384*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x
) + 1152*b^2*c*d*integrate(1/16*x*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 96*b^2*c*d*integrate(1/16*x*log(c^2*x^2
+ 1)^2/(c^2*x^4 + x^2), x) + 3072*a*b*c*d*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^2), x) + 384*b^2*c*d*integ
rate(1/16*x*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 384*b^2*d*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*
x^4 + x^2), x))*x)/x

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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