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

Optimal. Leaf size=82 \[ -i c \left (a+b \tan ^{-1}(c x)\right )^2-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b^2 c \text {Li}_2\left (\frac {2}{1-i c x}-1\right ) \]

[Out]

-I*c*(a+b*arctan(c*x))^2-(a+b*arctan(c*x))^2/x+2*b*c*(a+b*arctan(c*x))*ln(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 = 0.15, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.15, size = 102, normalized size = 1.24 \[ \frac {-a \left (a+b c x \log \left (c^2 x^2+1\right )-2 b c x \log (c x)\right )+2 b \tan ^{-1}(c x) \left (-a+b c x \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-i b^2 c x \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+b^2 (-1-i c x) \tan ^{-1}(c x)^2}{x} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{x^{2}}, x\right ) \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.02, size = 323, normalized size = 3.94 \[ -\frac {a^{2}}{x}-\frac {b^{2} \arctan \left (c x \right )^{2}}{x}+2 c \,b^{2} \ln \left (c x \right ) \arctan \left (c x \right )-c \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )-i c \,b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )-i c \,b^{2} \dilog \left (-i c x +1\right )+\frac {i c \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+i c \,b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )-\frac {i c \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+\frac {i c \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i c \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {i c \,b^{2} \ln \left (c x +i\right )^{2}}{4}-\frac {i c \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i c \,b^{2} \ln \left (c x -i\right )^{2}}{4}-\frac {i c \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+i c \,b^{2} \dilog \left (i c x +1\right )-\frac {2 a b \arctan \left (c x \right )}{x}+2 c a b \ln \left (c x \right )-c a b \ln \left (c^{2} x^{2}+1\right ) \]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/x^2,x)

[Out]

int((a + b*atan(c*x))^2/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \]

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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