3.290 \(\int \frac {\tan ^{-1}(a x)^2}{x^4 (c+a^2 c x^2)} \, dx\)

Optimal. Leaf size=166 \[ \frac {4 i a^3 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{3 c}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}+\frac {4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac {a^3 \tan ^{-1}(a x)}{3 c}-\frac {8 a^3 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{3 c}-\frac {a^2}{3 c x}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}-\frac {\tan ^{-1}(a x)^2}{3 c x^3}-\frac {a \tan ^{-1}(a x)}{3 c x^2} \]

[Out]

-1/3*a^2/c/x-1/3*a^3*arctan(a*x)/c-1/3*a*arctan(a*x)/c/x^2+4/3*I*a^3*arctan(a*x)^2/c-1/3*arctan(a*x)^2/c/x^3+a
^2*arctan(a*x)^2/c/x+1/3*a^3*arctan(a*x)^3/c-8/3*a^3*arctan(a*x)*ln(2-2/(1-I*a*x))/c+4/3*I*a^3*polylog(2,-1+2/
(1-I*a*x))/c

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Rubi [A]  time = 0.44, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4918, 4852, 325, 203, 4924, 4868, 2447, 4884} \[ \frac {4 i a^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{3 c}-\frac {a^2}{3 c x}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}+\frac {4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac {a^3 \tan ^{-1}(a x)}{3 c}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}-\frac {8 a^3 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{3 c}-\frac {a \tan ^{-1}(a x)}{3 c x^2}-\frac {\tan ^{-1}(a x)^2}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)),x]

[Out]

-a^2/(3*c*x) - (a^3*ArcTan[a*x])/(3*c) - (a*ArcTan[a*x])/(3*c*x^2) + (((4*I)/3)*a^3*ArcTan[a*x]^2)/c - ArcTan[
a*x]^2/(3*c*x^3) + (a^2*ArcTan[a*x]^2)/(c*x) + (a^3*ArcTan[a*x]^3)/(3*c) - (8*a^3*ArcTan[a*x]*Log[2 - 2/(1 - I
*a*x)])/(3*c) + (((4*I)/3)*a^3*PolyLog[2, -1 + 2/(1 - I*a*x)])/c

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, 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 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^4 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^4} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^2}{3 c x^3}+a^4 \int \frac {\tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x^3 \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac {a^2 \int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^2}{3 c x^3}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x^3} \, dx}{3 c}-\frac {\left (2 a^3\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac {\left (2 a^3\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{c}\\ &=-\frac {a \tan ^{-1}(a x)}{3 c x^2}+\frac {4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac {\tan ^{-1}(a x)^2}{3 c x^3}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}+\frac {a^2 \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx}{3 c}-\frac {\left (2 i a^3\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{3 c}-\frac {\left (2 i a^3\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c}\\ &=-\frac {a^2}{3 c x}-\frac {a \tan ^{-1}(a x)}{3 c x^2}+\frac {4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac {\tan ^{-1}(a x)^2}{3 c x^3}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}-\frac {8 a^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{3 c}-\frac {a^4 \int \frac {1}{1+a^2 x^2} \, dx}{3 c}+\frac {\left (2 a^4\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{3 c}+\frac {\left (2 a^4\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {a^2}{3 c x}-\frac {a^3 \tan ^{-1}(a x)}{3 c}-\frac {a \tan ^{-1}(a x)}{3 c x^2}+\frac {4 i a^3 \tan ^{-1}(a x)^2}{3 c}-\frac {\tan ^{-1}(a x)^2}{3 c x^3}+\frac {a^2 \tan ^{-1}(a x)^2}{c x}+\frac {a^3 \tan ^{-1}(a x)^3}{3 c}-\frac {8 a^3 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{3 c}+\frac {4 i a^3 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{3 c}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 120, normalized size = 0.72 \[ \frac {a^3 \left (-\frac {\frac {\left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}{a^2 x^2}-4 \tan ^{-1}(a x)^2+1}{a x}+\tan ^{-1}(a x) \left (-\frac {a^2 x^2+1}{a^2 x^2}+\tan ^{-1}(a x) \left (\tan ^{-1}(a x)+4 i\right )-8 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )\right )+4 i \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )\right )}{3 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTan[a*x]^2/(x^4*(c + a^2*c*x^2)),x]

[Out]

(a^3*(-((1 - 4*ArcTan[a*x]^2 + ((1 + a^2*x^2)*ArcTan[a*x]^2)/(a^2*x^2))/(a*x)) + ArcTan[a*x]*(-((1 + a^2*x^2)/
(a^2*x^2)) + ArcTan[a*x]*(4*I + ArcTan[a*x]) - 8*Log[1 - E^((2*I)*ArcTan[a*x])]) + (4*I)*PolyLog[2, E^((2*I)*A
rcTan[a*x])]))/(3*c)

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{2}}{a^{2} c x^{6} + c x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^2/x^4/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(arctan(a*x)^2/(a^2*c*x^6 + c*x^4), 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^4/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.12, size = 374, normalized size = 2.25 \[ -\frac {\arctan \left (a x \right )^{2}}{3 c \,x^{3}}+\frac {a^{2} \arctan \left (a x \right )^{2}}{c x}+\frac {a^{3} \arctan \left (a x \right )^{3}}{3 c}-\frac {a \arctan \left (a x \right )}{3 c \,x^{2}}-\frac {8 a^{3} \arctan \left (a x \right ) \ln \left (a x \right )}{3 c}+\frac {4 a^{3} \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3 c}-\frac {a^{2}}{3 c x}-\frac {a^{3} \arctan \left (a x \right )}{3 c}-\frac {i a^{3} \ln \left (a x -i\right )^{2}}{3 c}+\frac {4 i a^{3} \ln \left (a x \right ) \ln \left (-i a x +1\right )}{3 c}+\frac {4 i a^{3} \dilog \left (-i a x +1\right )}{3 c}-\frac {2 i a^{3} \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{3 c}+\frac {2 i a^{3} \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{3 c}-\frac {2 i a^{3} \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{3 c}+\frac {2 i a^{3} \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{3 c}+\frac {2 i a^{3} \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{3 c}-\frac {4 i a^{3} \ln \left (a x \right ) \ln \left (i a x +1\right )}{3 c}+\frac {i a^{3} \ln \left (a x +i\right )^{2}}{3 c}-\frac {2 i a^{3} \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{3 c}-\frac {4 i a^{3} \dilog \left (i a x +1\right )}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arctan(a*x)^2/c/x^3+a^2*arctan(a*x)^2/c/x+1/3*a^3*arctan(a*x)^3/c-1/3*a*arctan(a*x)/c/x^2-8/3*a^3/c*arcta
n(a*x)*ln(a*x)+4/3*a^3/c*arctan(a*x)*ln(a^2*x^2+1)-1/3*a^2/c/x-1/3*a^3*arctan(a*x)/c-2/3*I*a^3/c*dilog(-1/2*I*
(I+a*x))-1/3*I*a^3/c*ln(a*x-I)^2+4/3*I*a^3/c*ln(a*x)*ln(1-I*a*x)-2/3*I*a^3/c*ln(I+a*x)*ln(a^2*x^2+1)-4/3*I*a^3
/c*dilog(1+I*a*x)+2/3*I*a^3/c*ln(a*x-I)*ln(a^2*x^2+1)+2/3*I*a^3/c*ln(I+a*x)*ln(1/2*I*(a*x-I))+4/3*I*a^3/c*dilo
g(1-I*a*x)-4/3*I*a^3/c*ln(a*x)*ln(1+I*a*x)+1/3*I*a^3/c*ln(I+a*x)^2-2/3*I*a^3/c*ln(a*x-I)*ln(-1/2*I*(I+a*x))+2/
3*I*a^3/c*dilog(1/2*I*(a*x-I))

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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^4/(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 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^4\,\left (c\,a^2\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{2} x^{6} + x^{4}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**2/(a**2*x**6 + x**4), x)/c

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