3.3.91 \(\int \frac {x^3 \text {ArcTan}(a x)^2}{(c+a^2 c x^2)^2} \, dx\) [291]

Optimal. Leaf size=192 \[ -\frac {1}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {x \text {ArcTan}(a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\text {ArcTan}(a x)^2}{4 a^4 c^2}+\frac {\text {ArcTan}(a x)^2}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \text {ArcTan}(a x)^3}{3 a^4 c^2}-\frac {\text {ArcTan}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {i \text {ArcTan}(a x) \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {\text {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.22, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {5084, 5040, 4964, 5004, 5114, 6745, 5050, 5012, 267} \begin {gather*} -\frac {i \text {ArcTan}(a x) \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{a^4 c^2}-\frac {i \text {ArcTan}(a x)^3}{3 a^4 c^2}-\frac {\text {ArcTan}(a x)^2}{4 a^4 c^2}-\frac {\text {ArcTan}(a x)^2 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {\text {Li}_3\left (1-\frac {2}{i a x+1}\right )}{2 a^4 c^2}+\frac {\text {ArcTan}(a x)^2}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {1}{4 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {x \text {ArcTan}(a x)}{2 a^3 c^2 \left (a^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 5004

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 5012

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

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5050

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

Rule 5084

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rule 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 6745

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

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 117, normalized size = 0.61 \begin {gather*} \frac {\frac {1}{3} i \text {ArcTan}(a x)^3+\frac {1}{8} \left (-1+2 \text {ArcTan}(a x)^2\right ) \cos (2 \text {ArcTan}(a x))-\text {ArcTan}(a x)^2 \log \left (1+e^{2 i \text {ArcTan}(a x)}\right )+i \text {ArcTan}(a x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(a x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(a x)}\right )-\frac {1}{4} \text {ArcTan}(a x) \sin (2 \text {ArcTan}(a x))}{a^4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.00, size = 855, normalized size = 4.45 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/2/c^2*arctan(a*x)^2*ln(a^2*x^2+1)+1/2*arctan(a*x)^2/c^2/(a^2*x^2+1)-1/c^2*(arctan(a*x)^2*ln((1+I*a*x)
/(a^2*x^2+1)^(1/2))-1/3*I*arctan(a*x)^3-I*arctan(a*x)*(I+a*x)/(8*a*x-8*I)-1/16*(I+a*x)/(a*x-I)+I*arctan(a*x)*(
a*x-I)/(8*a*x+8*I)-1/16*(a*x-I)/(I+a*x)+1/4*arctan(a*x)^2*(-2*I*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+I*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)-I*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))+2*I*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+I*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-I*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)-I*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+I*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+I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-I*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3+1+4*ln
(2))-I*arctan(a*x)*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))+1/2*polylog(3,-(1+I*a*x)^2/(a^2*x^2+1))))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arctan(a*x)^2/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {x^{3} \operatorname {atan}^{2}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*atan(a*x)**2/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________