3.289 \(\int \frac {\tan ^{-1}(a x)^2}{x^3 (c+a^2 c x^2)} \, dx\)

Optimal. Leaf size=178 \[ -\frac {a^2 \text {Li}_3\left (\frac {2}{1-i a x}-1\right )}{2 c}+\frac {i a^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)}{c}-\frac {a^2 \log \left (a^2 x^2+1\right )}{2 c}+\frac {a^2 \log (x)}{c}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}-\frac {a \tan ^{-1}(a x)}{c x} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4918, 4852, 266, 36, 29, 31, 4884, 4924, 4868, 4992, 6610} \[ -\frac {a^2 \text {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {i a^2 \tan ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \log \left (a^2 x^2+1\right )}{2 c}+\frac {a^2 \log (x)}{c}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}-\frac {a \tan ^{-1}(a x)}{c x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]^2/(x^3*(c + a^2*c*x^2)),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rule 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 {\tan ^{-1}(a x)^2}{x^3 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{x \left (c+a^2 c x^2\right )} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^3} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}+\frac {a \int \frac {\tan ^{-1}(a x)}{x^2 \left (1+a^2 x^2\right )} \, dx}{c}-\frac {\left (i a^2\right ) \int \frac {\tan ^{-1}(a x)^2}{x (i+a x)} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {a \int \frac {\tan ^{-1}(a x)}{x^2} \, dx}{c}-\frac {a^3 \int \frac {\tan ^{-1}(a x)}{1+a^2 x^2} \, dx}{c}+\frac {\left (2 a^3\right ) \int \frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {a \tan ^{-1}(a x)}{c x}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {i a^2 \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}+\frac {a^2 \int \frac {1}{x \left (1+a^2 x^2\right )} \, dx}{c}-\frac {\left (i a^3\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {a \tan ^{-1}(a x)}{c x}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {i a^2 \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {a \tan ^{-1}(a x)}{c x}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}-\frac {a^2 \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {i a^2 \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 c}-\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {a \tan ^{-1}(a x)}{c x}-\frac {a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {\tan ^{-1}(a x)^2}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^3}{3 c}+\frac {a^2 \log (x)}{c}-\frac {a^2 \log \left (1+a^2 x^2\right )}{2 c}-\frac {a^2 \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {i a^2 \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 142, normalized size = 0.80 \[ \frac {a^2 \left (\log \left (\frac {a x}{\sqrt {a^2 x^2+1}}\right )-\frac {\left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}{2 a^2 x^2}-i \tan ^{-1}(a x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(a x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(a x)}\right )-\frac {1}{3} i \tan ^{-1}(a x)^3-\frac {\tan ^{-1}(a x)}{a x}-\tan ^{-1}(a x)^2 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )+\frac {i \pi ^3}{24}\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]^2/(x^3*(c + a^2*c*x^2)),x]

[Out]

(a^2*((I/24)*Pi^3 - ArcTan[a*x]/(a*x) - ((1 + a^2*x^2)*ArcTan[a*x]^2)/(2*a^2*x^2) - (I/3)*ArcTan[a*x]^3 - ArcT
an[a*x]^2*Log[1 - E^((-2*I)*ArcTan[a*x])] + Log[(a*x)/Sqrt[1 + a^2*x^2]] - I*ArcTan[a*x]*PolyLog[2, E^((-2*I)*
ArcTan[a*x])] - PolyLog[3, E^((-2*I)*ArcTan[a*x])]/2))/c

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{2}}{a^{2} c x^{5} + c x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^3/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(arctan(a*x)^2/(a^2*c*x^5 + c*x^3), 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^3/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 1.04, size = 5491, normalized size = 30.85 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^2/x^3/(a^2*c*x^2+c),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^3/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

integrate(arctan(a*x)^2/((a^2*c*x^2 + c)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^3\,\left (c\,a^2\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^2/(x^3*(c + a^2*c*x^2)),x)

[Out]

int(atan(a*x)^2/(x^3*(c + a^2*c*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{2} x^{5} + x^{3}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**2/x**3/(a**2*c*x**2+c),x)

[Out]

Integral(atan(a*x)**2/(a**2*x**5 + x**3), x)/c

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