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

Optimal. Leaf size=82 \[ -i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )-i c \left (a+b \tan ^{-1}(c x)\right )^2-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.146934, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4852, 4924, 4868, 2447} \[ -i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )-i c \left (a+b \tan ^{-1}(c x)\right )^2-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/x^2,x]

[Out]

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

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]]

Rubi steps

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

Mathematica [A]  time = 0.14126, size = 102, normalized size = 1.24 \[ \frac{-i b^2 c x \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )-a \left (a+b c x \log \left (c^2 x^2+1\right )-2 b c x \log (c x)\right )+2 b \tan ^{-1}(c x) \left (-a+b c x \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^2 (-1-i c x) \tan ^{-1}(c x)^2}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 323, normalized size = 3.9 \begin{align*} -{\frac{{a}^{2}}{x}}-{\frac{{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{x}}-c{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) +2\,c{b}^{2}\ln \left ( cx \right ) \arctan \left ( cx \right ) +{\frac{i}{2}}c{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) +ic{b}^{2}{\it dilog} \left ( 1+icx \right ) +{\frac{i}{2}}c{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) -ic{b}^{2}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) -{\frac{i}{2}}c{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) -{\frac{i}{2}}c{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) -{\frac{i}{4}}c{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}-{\frac{i}{2}}c{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) +ic{b}^{2}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) +{\frac{i}{2}}c{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) -ic{b}^{2}{\it dilog} \left ( 1-icx \right ) +{\frac{i}{4}}c{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}-2\,{\frac{ab\arctan \left ( cx \right ) }{x}}-cab\ln \left ({c}^{2}{x}^{2}+1 \right ) +2\,cab\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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,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{b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{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,x, algorithm="giac")

[Out]

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