3.100 \(\int \frac{(a+b \tan ^{-1}(c x))^2}{x^2 (d+i c d x)} \, dx\)

Optimal. Leaf size=186 \[ \frac{b c \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{i b^2 c \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{i c \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d} \]

[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])*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

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Rubi [A]  time = 0.398114, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4870, 4852, 4924, 4868, 2447, 4884, 4994, 6610} \[ \frac{b c \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{i b^2 c \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{i c \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[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 4870

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 4852

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

Rule 4924

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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 4868

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 2447

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 4884

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 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[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 6610

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*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+i c d x)} \, dx &=-\left ((i c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x (d+i c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{(2 b c) \int \frac{a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac{\left (2 i b c^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{b c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}+\frac{(2 i b c) \int \frac{a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d}-\frac{\left (b^2 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{b c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}-\frac{i b^2 c \text{Li}_3\left (-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 \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{i b^2 c \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}+\frac{b c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}-\frac{i b^2 c \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.859642, size = 265, normalized size = 1.42 \[ -\frac{2 a b c \left (\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+2 \left (-\log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+\tan ^{-1}(c x)^2+\tan ^{-1}(c x) \left (\frac{1}{c x}+i \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )\right )\right )+2 i b^2 c \left (i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{i \tan ^{-1}(c x)^2}{c x}+\tan ^{-1}(c x)^2+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+2 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )-i a^2 c \log \left (c^2 x^2+1\right )+2 i a^2 c \log (x)+2 a^2 c \tan ^{-1}(c x)+\frac{2 a^2}{x}}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((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*c*(2*(ArcTan[c*x]^2
+ ArcTan[c*x]*(1/(c*x) + I*Log[1 - E^((2*I)*ArcTan[c*x])]) - Log[(c*x)/Sqrt[1 + c^2*x^2]]) + PolyLog[2, E^((2*
I)*ArcTan[c*x])]) + (2*I)*b^2*c*((-I/24)*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*ArcTan[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))/(2*d)

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Maple [C]  time = 0.783, size = 9235, normalized size = 49.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{i \, b^{2} \log \left (-\frac{c x + i}{c x - i}\right )^{2} + 4 \, a b \log \left (-\frac{c x + i}{c x - i}\right ) - 4 i \, a^{2}}{4 \,{\left (c d x^{3} - i \, d x^{2}\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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