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

Optimal. Leaf size=102 \[ -\frac {\text {Li}_3\left (1-\frac {2}{i a x+1}\right )}{2 a^2 c}-\frac {i \text {Li}_2\left (1-\frac {2}{i a x+1}\right ) \tan ^{-1}(a x)}{a^2 c}-\frac {i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4920, 4854, 4884, 4994, 6610} \[ -\frac {\text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^2 c}-\frac {i \tan ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{a^2 c}-\frac {i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[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 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 4994

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

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Mathematica [A]  time = 0.01, size = 110, normalized size = 1.08 \[ -\frac {\text {Li}_3\left (\frac {a x+i}{a x-i}\right )}{2 a^2 c}-\frac {i \text {Li}_2\left (\frac {a x+i}{a x-i}\right ) \tan ^{-1}(a x)}{a^2 c}-\frac {i \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)^2}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \arctan \left (a x\right )^{2}}{a^{2} c x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [C]  time = 0.58, size = 897, normalized size = 8.79 \[ \frac {\arctan \left (a x \right )^{2} \ln \left (a^{2} x^{2}+1\right )}{2 a^{2} c}-\frac {\arctan \left (a x \right )^{2} \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{a^{2} c}+\frac {i \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{\left (a^{2} x^{2}+1\right ) \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right )^{3} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}-\frac {i \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )^{2} \mathrm {csgn}\left (\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )^{2} \pi }{2 a^{2} c}-\frac {i \mathrm {csgn}\left (i \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )\right )^{2} \mathrm {csgn}\left (i \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}\right ) \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}-\frac {i \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right ) \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{\left (a^{2} x^{2}+1\right ) \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right )^{2} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}+\frac {i \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right ) \mathrm {csgn}\left (\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )^{2} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}+\frac {i \arctan \left (a x \right ) \polylog \left (2, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{a^{2} c}-\frac {i \mathrm {csgn}\left (\frac {i}{\left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{\left (a^{2} x^{2}+1\right ) \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right )^{2} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}+\frac {i \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )^{3} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}-\frac {i \mathrm {csgn}\left (i \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}\right )^{3} \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}+\frac {i \arctan \left (a x \right )^{3}}{3 a^{2} c}-\frac {\ln \relax (2) \arctan \left (a x \right )^{2}}{a^{2} c}+\frac {i \mathrm {csgn}\left (\frac {i}{\left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right ) \mathrm {csgn}\left (\frac {i \left (i a x +1\right )^{2}}{\left (a^{2} x^{2}+1\right ) \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}}\right ) \arctan \left (a x \right )^{2} \pi }{4 a^{2} c}-\frac {\polylog \left (3, -\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}\right )}{2 a^{2} c}+\frac {i \mathrm {csgn}\left (i \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )\right ) \mathrm {csgn}\left (i \left (\frac {\left (i a x +1\right )^{2}}{a^{2} x^{2}+1}+1\right )^{2}\right )^{2} \arctan \left (a x \right )^{2} \pi }{2 a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x \operatorname {atan}^{2}{\left (a x \right )}}{a^{2} x^{2} + 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atan(a*x)**2/(a**2*x**2 + 1), x)/c

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