\(\int \frac {x^3 \arctan (a x)^3}{(c+a^2 c x^2)^2} \, dx\) [396]

Optimal result

Integrand size = 22, antiderivative size = 270 \[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3 x}{8 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)}{8 a^4 c^2}-\frac {3 \arctan (a x)}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {3 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}-\frac {3 \arctan (a x) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}+\frac {3 i \operatorname {PolyLog}\left (4,1-\frac {2}{1+i a x}\right )}{4 a^4 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5084, 5040, 4964, 5004, 5114, 5118, 6745, 5050, 5012, 205, 211} \[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {3 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i a x+1}\right )}{2 a^4 c^2}-\frac {3 \arctan (a x) \operatorname {PolyLog}\left (3,1-\frac {2}{i a x+1}\right )}{2 a^4 c^2}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {3 \arctan (a x)}{8 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}+\frac {3 i \operatorname {PolyLog}\left (4,1-\frac {2}{i a x+1}\right )}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {3 \arctan (a x)}{4 a^4 c^2 \left (a^2 x^2+1\right )}-\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (a^2 x^2+1\right )}+\frac {3 x}{8 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

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 5118

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 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*} \text {integral}& = -\frac {\int \frac {x \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx}{a^2}+\frac {\int \frac {x \arctan (a x)^3}{c+a^2 c x^2} \, dx}{a^2 c} \\ & = \frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {3 \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^3}-\frac {\int \frac {\arctan (a x)^3}{i-a x} \, dx}{a^3 c^2} \\ & = -\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}+\frac {3 \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^2}+\frac {3 \int \frac {\arctan (a x)^2 \log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c^2} \\ & = -\frac {3 \arctan (a x)}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {3 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}+\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{4 a^3}+\frac {(3 i) \int \frac {\arctan (a x) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^3 c^2} \\ & = \frac {3 x}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {3 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}-\frac {3 \arctan (a x) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}+\frac {3 \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{2 a^3 c^2}+\frac {3 \int \frac {1}{c+a^2 c x^2} \, dx}{8 a^3 c} \\ & = \frac {3 x}{8 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)}{8 a^4 c^2}-\frac {3 \arctan (a x)}{4 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {3 x \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)^3}{4 a^4 c^2}+\frac {\arctan (a x)^3}{2 a^4 c^2 \left (1+a^2 x^2\right )}-\frac {i \arctan (a x)^4}{4 a^4 c^2}-\frac {\arctan (a x)^3 \log \left (\frac {2}{1+i a x}\right )}{a^4 c^2}-\frac {3 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}-\frac {3 \arctan (a x) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i a x}\right )}{2 a^4 c^2}+\frac {3 i \operatorname {PolyLog}\left (4,1-\frac {2}{1+i a x}\right )}{4 a^4 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.58 \[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {4 i \arctan (a x)^4-6 \arctan (a x) \cos (2 \arctan (a x))+4 \arctan (a x)^3 \cos (2 \arctan (a x))-16 \arctan (a x)^3 \log \left (1+e^{2 i \arctan (a x)}\right )+24 i \arctan (a x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (a x)}\right )-24 \arctan (a x) \operatorname {PolyLog}\left (3,-e^{2 i \arctan (a x)}\right )-12 i \operatorname {PolyLog}\left (4,-e^{2 i \arctan (a x)}\right )+3 \sin (2 \arctan (a x))-6 \arctan (a x)^2 \sin (2 \arctan (a x))}{16 a^4 c^2} \]

[In]

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

[Out]

((4*I)*ArcTan[a*x]^4 - 6*ArcTan[a*x]*Cos[2*ArcTan[a*x]] + 4*ArcTan[a*x]^3*Cos[2*ArcTan[a*x]] - 16*ArcTan[a*x]^
3*Log[1 + E^((2*I)*ArcTan[a*x])] + (24*I)*ArcTan[a*x]^2*PolyLog[2, -E^((2*I)*ArcTan[a*x])] - 24*ArcTan[a*x]*Po
lyLog[3, -E^((2*I)*ArcTan[a*x])] - (12*I)*PolyLog[4, -E^((2*I)*ArcTan[a*x])] + 3*Sin[2*ArcTan[a*x]] - 6*ArcTan
[a*x]^2*Sin[2*ArcTan[a*x]])/(16*a^4*c^2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 24.99 (sec) , antiderivative size = 936, normalized size of antiderivative = 3.47

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(936\)
default	\(\text {Expression too large to display}\)	\(936\)
parts	\(\text {Expression too large to display}\)	\(971\)

[In]

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

[Out]

1/a^4*(1/2/c^2*arctan(a*x)^3*ln(a^2*x^2+1)+1/2*arctan(a*x)^3/c^2/(a^2*x^2+1)-3/2/c^2*(2/3*arctan(a*x)^3*ln((1+
I*a*x)/(a^2*x^2+1)^(1/2))-1/6*I*arctan(a*x)^4-I*arctan(a*x)^2*(I+a*x)/(8*a*x-8*I)-1/8*arctan(a*x)*(I+a*x)/(a*x
-I)+I*(I+a*x)/(16*a*x-16*I)+I*arctan(a*x)^2*(a*x-I)/(8*a*x+8*I)-1/8*arctan(a*x)*(a*x-I)/(I+a*x)-I*(a*x-I)/(16*
a*x+16*I)-I*arctan(a*x)^2*polylog(2,-(1+I*a*x)^2/(a^2*x^2+1))+arctan(a*x)*polylog(3,-(1+I*a*x)^2/(a^2*x^2+1))+
1/2*I*polylog(4,-(1+I*a*x)^2/(a^2*x^2+1))+1/6*(I*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-2*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*cs
gn(I*((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))^3+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))*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))*cs
gn(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)/(a^2*x^2+1)^(1/2))^2*csgn(I*
(1+I*a*x)^2/(a^2*x^2+1))-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)+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+4*ln(2)+1)*arctan(a
*x)^3))

Fricas [F]

\[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{3} \operatorname {atan}^{3}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\int \frac {x^3\,{\mathrm {atan}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

[In]

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

[Out]

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