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

Optimal. Leaf size=92 \[ -\frac {i a \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{c}-\frac {a \tan ^{-1}(a x)^3}{3 c}-\frac {i a \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{c x}+\frac {2 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4918, 4852, 4924, 4868, 2447, 4884} \[ -\frac {i a \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{c}-\frac {a \tan ^{-1}(a x)^3}{3 c}-\frac {i a \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{c x}+\frac {2 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4918

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

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^2}{c x}-\frac {a \tan ^{-1}(a x)^3}{3 c}+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{c x}-\frac {a \tan ^{-1}(a x)^3}{3 c}+\frac {(2 i a) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{c x}-\frac {a \tan ^{-1}(a x)^3}{3 c}+\frac {2 a \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {\left (2 a^2\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{c x}-\frac {a \tan ^{-1}(a x)^3}{3 c}+\frac {2 a \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {i a \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 73, normalized size = 0.79 \[ \frac {a \left (-i \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )-\frac {1}{3} \tan ^{-1}(a x) \left (\left (\tan ^{-1}(a x)+3 i\right ) \tan ^{-1}(a x)+\frac {3 \tan ^{-1}(a x)}{a x}-6 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )\right )\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(-1/3*(ArcTan[a*x]*((3*ArcTan[a*x])/(a*x) + ArcTan[a*x]*(3*I + ArcTan[a*x]) - 6*Log[1 - E^((2*I)*ArcTan[a*x
])])) - I*PolyLog[2, E^((2*I)*ArcTan[a*x])]))/c

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{2}}{a^{2} c x^{4} + c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)^2/(a^2*c*x^4 + c*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.12, size = 292, normalized size = 3.17 \[ -\frac {\arctan \left (a x \right )^{2}}{c x}-\frac {a \arctan \left (a x \right )^{3}}{3 c}+\frac {2 a \arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {a \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{c}+\frac {i a \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c}-\frac {i a \ln \left (a x +i\right )^{2}}{4 c}-\frac {i a \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{2 c}-\frac {i a \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{2 c}-\frac {i a \dilog \left (-i a x +1\right )}{c}+\frac {i a \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c}+\frac {i a \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}+\frac {i a \ln \left (a x \right ) \ln \left (i a x +1\right )}{c}+\frac {i a \dilog \left (i a x +1\right )}{c}-\frac {i a \ln \left (a x \right ) \ln \left (-i a x +1\right )}{c}+\frac {i a \ln \left (a x -i\right )^{2}}{4 c}-\frac {i a \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(a*x)^2/c/x-1/3*a*arctan(a*x)^3/c+2*a/c*arctan(a*x)*ln(a*x)-a/c*arctan(a*x)*ln(a^2*x^2+1)+1/2*I*a/c*ln(
a*x-I)*ln(-1/2*I*(I+a*x))-1/4*I*a/c*ln(I+a*x)^2-1/2*I*a/c*ln(I+a*x)*ln(1/2*I*(a*x-I))+1/2*I*a/c*dilog(-1/2*I*(
I+a*x))-1/2*I*a/c*dilog(1/2*I*(a*x-I))+1/2*I*a/c*ln(I+a*x)*ln(a^2*x^2+1)+I*a/c*ln(a*x)*ln(1+I*a*x)+I*a/c*dilog
(1+I*a*x)-I*a/c*dilog(1-I*a*x)-I*a/c*ln(a*x)*ln(1-I*a*x)+1/4*I*a/c*ln(a*x-I)^2-1/2*I*a/c*ln(a*x-I)*ln(a^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(arctan(a*x)^2/x^2/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{2} x^{4} + x^{2}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**2/(a**2*x**4 + x**2), x)/c

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