3.392 \(\int \frac {\tan ^{-1}(a x)^3}{x (c+a^2 c x^2)} \, dx\)
Optimal. Leaf size=124 \[ \frac {3 i \text {Li}_4\left (\frac {2}{1-i a x}-1\right )}{4 c}-\frac {3 i \text {Li}_2\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)^2}{2 c}+\frac {3 \text {Li}_3\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)}{2 c}-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c} \]
[Out]
-1/4*I*arctan(a*x)^4/c+arctan(a*x)^3*ln(2-2/(1-I*a*x))/c-3/2*I*arctan(a*x)^2*polylog(2,-1+2/(1-I*a*x))/c+3/2*a
rctan(a*x)*polylog(3,-1+2/(1-I*a*x))/c+3/4*I*polylog(4,-1+2/(1-I*a*x))/c
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Rubi [A] time = 0.23, antiderivative size = 124, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{4924, 4868, 4884, 4992, 4996, 6610} \[ \frac {3 i \text {PolyLog}\left (4,-1+\frac {2}{1-i a x}\right )}{4 c}-\frac {3 i \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {3 \tan ^{-1}(a x) \text {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c} \]
Antiderivative was successfully verified.
[In]
Int[ArcTan[a*x]^3/(x*(c + a^2*c*x^2)),x]
[Out]
((-I/4)*ArcTan[a*x]^4)/c + (ArcTan[a*x]^3*Log[2 - 2/(1 - I*a*x)])/c - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -1 +
2/(1 - I*a*x)])/c + (3*ArcTan[a*x]*PolyLog[3, -1 + 2/(1 - I*a*x)])/(2*c) + (((3*I)/4)*PolyLog[4, -1 + 2/(1 -
I*a*x)])/c
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 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 4996
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[
k + 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[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 \left (c+a^2 c x^2\right )} \, dx &=-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {i \int \frac {\tan ^{-1}(a x)^3}{x (i+a x)} \, dx}{c}\\ &=-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {(3 a) \int \frac {\tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {(3 i a) \int \frac {\tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {(3 a) \int \frac {\text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{2 c}\\ &=-\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {3 \tan ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {3 i \text {Li}_4\left (-1+\frac {2}{1-i a x}\right )}{4 c}\\ \end {align*}
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Mathematica [B] time = 0.06, size = 354, normalized size = 2.85 \[ -\frac {3 i \text {Li}_4\left (\frac {-a x-i}{a x-i}\right )}{4 c}-\frac {3 i \text {Li}_4\left (-\frac {a x+i}{i-a x}\right )}{4 c}+\frac {3 i \text {Li}_4\left (\frac {a x+i}{a x-i}\right )}{4 c}+\frac {3 i \text {Li}_2\left (\frac {-a x-i}{a x-i}\right ) \tan ^{-1}(a x)^2}{2 c}+\frac {3 i \text {Li}_2\left (-\frac {a x+i}{i-a x}\right ) \tan ^{-1}(a x)^2}{2 c}-\frac {3 i \text {Li}_2\left (\frac {a x+i}{a x-i}\right ) \tan ^{-1}(a x)^2}{2 c}+\frac {3 \text {Li}_3\left (\frac {-a x-i}{a x-i}\right ) \tan ^{-1}(a x)}{2 c}+\frac {3 \text {Li}_3\left (-\frac {a x+i}{i-a x}\right ) \tan ^{-1}(a x)}{2 c}-\frac {3 \text {Li}_3\left (\frac {a x+i}{a x-i}\right ) \tan ^{-1}(a x)}{2 c}+\frac {i \tan ^{-1}(a x)^4}{4 c}+\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)^3}{c}+\frac {2 \tan ^{-1}(a x)^3 \tanh ^{-1}\left (1-\frac {2 i}{-a x+i}\right )}{c} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTan[a*x]^3/(x*(c + a^2*c*x^2)),x]
[Out]
((I/4)*ArcTan[a*x]^4)/c + (2*ArcTan[a*x]^3*ArcTanh[1 - (2*I)/(I - a*x)])/c + (ArcTan[a*x]^3*Log[(2*I)/(I - a*x
)])/c + (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, (-I - a*x)/(-I + a*x)])/c + (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -(
(I + a*x)/(I - a*x))])/c - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, (I + a*x)/(-I + a*x)])/c + (3*ArcTan[a*x]*PolyL
og[3, (-I - a*x)/(-I + a*x)])/(2*c) + (3*ArcTan[a*x]*PolyLog[3, -((I + a*x)/(I - a*x))])/(2*c) - (3*ArcTan[a*x
]*PolyLog[3, (I + a*x)/(-I + a*x)])/(2*c) - (((3*I)/4)*PolyLog[4, (-I - a*x)/(-I + a*x)])/c - (((3*I)/4)*PolyL
og[4, -((I + a*x)/(I - a*x))])/c + (((3*I)/4)*PolyLog[4, (I + a*x)/(-I + a*x)])/c
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{3}}{a^{2} c x^{3} + c x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctan(a*x)^3/x/(a^2*c*x^2+c),x, algorithm="fricas")
[Out]
integral(arctan(a*x)^3/(a^2*c*x^3 + c*x), 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/(a^2*c*x^2+c),x, algorithm="giac")
[Out]
sage0*x
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maple [C] time = 0.54, size = 1834, normalized size = 14.79 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctan(a*x)^3/x/(a^2*c*x^2+c),x)
[Out]
-1/4*I*arctan(a*x)^4/c-1/4*I/c*arctan(a*x)^3*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)+1/c*arctan(a*x)^3*ln(a*x)+1/2*I/c*arctan
(a*x)^3*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))+1/2*I/c*arctan(a*x)^3*Pi-3*I/c*arctan(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*x^
2+1)^(1/2))-3*I/c*arctan(a*x)^2*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-1/4*I/c*arctan(a*x)^3*Pi*csgn(I*(1+I*a*
x)^2/(a^2*x^2+1))^3-1/c*arctan(a*x)^3*ln((1+I*a*x)^2/(a^2*x^2+1)-1)+1/4*I/c*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)
^2/(a^2*x^2+1)+1)^2)^3+1/2*I/c*arctan(a*x)^3*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+1/2*I/c*arctan(a*x)^3*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3-1/2*I/c*arctan(a*
x)^3*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2-1/4*I/c*arctan(a*x)^3*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+6*I/c*polylog(4,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I/c*polylog
(4,(1+I*a*x)/(a^2*x^2+1)^(1/2))+6/c*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/c*arctan(a*x)^3*ln((1
+I*a*x)/(a^2*x^2+1)^(1/2))+1/c*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+6/c*arctan(a*x)*polylog(3,-(1+I
*a*x)/(a^2*x^2+1)^(1/2))-1/2/c*ln(a^2*x^2+1)*arctan(a*x)^3+1/c*arctan(a*x)^3*ln(2)+1/c*arctan(a*x)^3*ln(1-(1+I
*a*x)/(a^2*x^2+1)^(1/2))+1/2*I/c*arctan(a*x)^3*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))-1/2*I/c*arctan(a*x)^3*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+1/4*I/c*arctan(a*x)^3*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)+1/2*I/c*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*c
sgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2-1/2*I/c*arctan(a*x)^3*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+1/4*I/c*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csg
n(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-1/2*I/c*arctan(a*x)^3*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))^2-1/2*I/c*arctan(a*x)^3*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))^2+1/4*I/c*arctan(a*x)^3*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)^2-1/4*I/c*arctan(a*x)^3*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))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctan(a*x)^3/x/(a^2*c*x^2+c),x, algorithm="maxima")
[Out]
integrate(arctan(a*x)^3/((a^2*c*x^2 + c)*x), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(atan(a*x)^3/(x*(c + a^2*c*x^2)),x)
[Out]
int(atan(a*x)^3/(x*(c + a^2*c*x^2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{a^{2} x^{3} + x}\, dx}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atan(a*x)**3/x/(a**2*c*x**2+c),x)
[Out]
Integral(atan(a*x)**3/(a**2*x**3 + x), x)/c
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