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

Optimal. Leaf size=122 \[ \frac {3 a \text {Li}_3\left (\frac {2}{1-i a x}-1\right )}{2 c}-\frac {3 i a \text {Li}_2\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)}{c}-\frac {a \tan ^{-1}(a x)^4}{4 c}-\frac {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}+\frac {3 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4918, 4852, 4924, 4868, 4884, 4992, 6610} \[ \frac {3 a \text {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 i a \tan ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{c}-\frac {a \tan ^{-1}(a x)^4}{4 c}-\frac {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}+\frac {3 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)^3}{x^2 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^3}{c+a^2 c x^2} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^3}{x^2} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^3}{c x}-\frac {a \tan ^{-1}(a x)^4}{4 c}+\frac {(3 a) \int \frac {\tan ^{-1}(a x)^2}{x \left (1+a^2 x^2\right )} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}-\frac {a \tan ^{-1}(a x)^4}{4 c}+\frac {(3 i a) \int \frac {\tan ^{-1}(a x)^2}{x (i+a x)} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}-\frac {a \tan ^{-1}(a x)^4}{4 c}+\frac {3 a \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {\left (6 a^2\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 {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}-\frac {a \tan ^{-1}(a x)^4}{4 c}+\frac {3 a \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i a \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}+\frac {\left (3 i a^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {i a \tan ^{-1}(a x)^3}{c}-\frac {\tan ^{-1}(a x)^3}{c x}-\frac {a \tan ^{-1}(a x)^4}{4 c}+\frac {3 a \tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i a \tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c}+\frac {3 a \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 108, normalized size = 0.89 \[ \frac {a \left (3 i \tan ^{-1}(a x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(a x)}\right )+\frac {3}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(a x)}\right )-\frac {1}{4} \tan ^{-1}(a x)^4-\frac {\tan ^{-1}(a x)^3}{a x}+i \tan ^{-1}(a x)^3+3 \tan ^{-1}(a x)^2 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-\frac {i \pi ^3}{8}\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{3}}{a^{2} c x^{4} + c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [C]  time = 0.34, size = 1829, normalized size = 14.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*I*a/c*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3*arctan(a*x)^2+3/2*I*a/c*Pi*csgn(I
*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3*arctan(a*x)^2+3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I*((
1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3-3/4*I*a/c*arctan(a
*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-3/2*I*a/c*Pi*csgn(((1+I*a*x)^2/(a^2*x
^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2*arctan(a*x)^2+3*a/c*arctan(a*x)^2*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))-ar
ctan(a*x)^3/c/x-1/4*a*arctan(a*x)^4/c+3*a/c*arctan(a*x)^2*ln(2)-3/2*a/c*arctan(a*x)^2*ln(a^2*x^2+1)+3*a/c*arct
an(a*x)^2*ln(a*x)-3*a/c*arctan(a*x)^2*ln((1+I*a*x)^2/(a^2*x^2+1)-1)+3*a/c*arctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+
1)^(1/2))+3*a/c*arctan(a*x)^2*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+3/2*I*a/c*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1
)/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*arctan(a*x)^2+3/2
*I*a/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2-3/2*I*a/c*Pi*csg
n(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^
2*x^2+1)+1))^2*arctan(a*x)^2-3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))^2*csgn(I*(1+I*a*x)
^2/(a^2*x^2+1))+3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1
)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2+3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))^2*csgn(I*((1
+I*a*x)^2/(a^2*x^2+1)+1)^2)+3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2
*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-3/2*I*a/c*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1))*csgn(I*((1+I*a*x)^2/
(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2*arctan(a*x)^2-3/2*I*a/c*arctan(a*x)^2*Pi*csgn(I*((1+I*a*x)^2/(a^
2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-3/2*I*a/c*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*(
(1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2*arctan(a*x)^2-6*I*a/c*arctan(a*x)*polylog(2,-(1+I*a*
x)/(a^2*x^2+1)^(1/2))-6*I*a/c*arctan(a*x)*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))+3/2*I*a/c*Pi*arctan(a*x)^2+3/
2*I*a/c*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1))*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^
2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*arctan(a*x)^2-3/4*I*a/c*arctan(a*x)^2*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+
1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)+6*a/c*poly
log(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*a/c*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*a*arctan(a*x)^3/c

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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)^3/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 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x^2\,\left (c\,a^2\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{a^{2} x^{4} + x^{2}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**3/(a**2*x**4 + x**2), x)/c

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