\(\int \frac {x^2}{(c+a^2 c x^2)^3 \arctan (a x)^3} \, dx\) [636]

Optimal result

Integrand size = 22, antiderivative size = 120 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}+\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{a^3 c^3} \]

[Out]

1/2/a^3/c^3/(a^2*x^2+1)^2/arctan(a*x)^2-1/2/a^3/c^3/(a^2*x^2+1)/arctan(a*x)^2-2*x/a^2/c^3/(a^2*x^2+1)^2/arctan
(a*x)+x/a^2/c^3/(a^2*x^2+1)/arctan(a*x)+Ci(4*arctan(a*x))/a^3/c^3

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5084, 5022, 5088, 5090, 3393, 3383, 5024, 4491} \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{a^3 c^3}+\frac {x}{a^2 c^3 \left (a^2 x^2+1\right ) \arctan (a x)}-\frac {2 x}{a^2 c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)}-\frac {1}{2 a^3 c^3 \left (a^2 x^2+1\right ) \arctan (a x)^2}+\frac {1}{2 a^3 c^3 \left (a^2 x^2+1\right )^2 \arctan (a x)^2} \]

[In]

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

[Out]

1/(2*a^3*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^2) - 1/(2*a^3*c^3*(1 + a^2*x^2)*ArcTan[a*x]^2) - (2*x)/(a^2*c^3*(1 +
a^2*x^2)^2*ArcTan[a*x]) + x/(a^2*c^3*(1 + a^2*x^2)*ArcTan[a*x]) + CosIntegral[4*ArcTan[a*x]]/(a^3*c^3)

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5022

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

Rule 5024

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

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 5088

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

Rule 5090

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(m
+ 1), Subst[Int[(a + b*x)^p*(Sin[x]^m/Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d,
e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx}{a^2}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx}{a^2 c} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}+\frac {2 \int \frac {x}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^2} \, dx}{a}-\frac {\int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^2} \, dx}{a c} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}-6 \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx+\frac {2 \int \frac {1}{\left (c+a^2 c x^2\right )^3 \arctan (a x)} \, dx}{a^2}+\frac {\int \frac {x^2}{\left (c+a^2 c x^2\right )^2 \arctan (a x)} \, dx}{c}-\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)} \, dx}{a^2 c} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {\text {Subst}\left (\int \frac {\cos ^2(x)}{x} \, dx,x,\arctan (a x)\right )}{a^3 c^3}+\frac {\text {Subst}\left (\int \frac {\sin ^2(x)}{x} \, dx,x,\arctan (a x)\right )}{a^3 c^3}+\frac {2 \text {Subst}\left (\int \frac {\cos ^4(x)}{x} \, dx,x,\arctan (a x)\right )}{a^3 c^3}-\frac {6 \text {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{x} \, dx,x,\arctan (a x)\right )}{a^3 c^3} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}+\frac {\text {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cos (2 x)}{2 x}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3}-\frac {\text {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cos (2 x)}{2 x}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3}+\frac {2 \text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cos (2 x)}{2 x}+\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3}-\frac {6 \text {Subst}\left (\int \left (\frac {1}{8 x}-\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}+\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arctan (a x)\right )}{4 a^3 c^3}-2 \frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arctan (a x)\right )}{2 a^3 c^3}+\frac {3 \text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arctan (a x)\right )}{4 a^3 c^3}+\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arctan (a x)\right )}{a^3 c^3} \\ & = \frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}-\frac {1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \arctan (a x)}+\frac {x}{a^2 c^3 \left (1+a^2 x^2\right ) \arctan (a x)}+\frac {\operatorname {CosIntegral}(4 \arctan (a x))}{a^3 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.50 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\frac {\frac {a x \left (-a x+2 \left (-1+a^2 x^2\right ) \arctan (a x)\right )}{\left (1+a^2 x^2\right )^2 \arctan (a x)^2}+2 \operatorname {CosIntegral}(4 \arctan (a x))}{2 a^3 c^3} \]

[In]

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

[Out]

((a*x*(-(a*x) + 2*(-1 + a^2*x^2)*ArcTan[a*x]))/((1 + a^2*x^2)^2*ArcTan[a*x]^2) + 2*CosIntegral[4*ArcTan[a*x]])
/(2*a^3*c^3)

Maple [A] (verified)

Time = 10.58 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43

method	result	size
derivativedivides	\(\frac {16 \,\operatorname {Ci}\left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-4 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\cos \left (4 \arctan \left (a x \right )\right )-1}{16 a^{3} c^{3} \arctan \left (a x \right )^{2}}\)	\(52\)
default	\(\frac {16 \,\operatorname {Ci}\left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}-4 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\cos \left (4 \arctan \left (a x \right )\right )-1}{16 a^{3} c^{3} \arctan \left (a x \right )^{2}}\)	\(52\)

[In]

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

[Out]

1/16/a^3/c^3*(16*Ci(4*arctan(a*x))*arctan(a*x)^2-4*sin(4*arctan(a*x))*arctan(a*x)+cos(4*arctan(a*x))-1)/arctan
(a*x)^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.79 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=-\frac {a^{2} x^{2} - {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (\frac {a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - 2 \, {\left (a^{3} x^{3} - a x\right )} \arctan \left (a x\right )}{2 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )} \arctan \left (a x\right )^{2}} \]

[In]

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

[Out]

-1/2*(a^2*x^2 - (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 + 4*I*a^3*x^3 - 6*a^2*x^2 - 4*I*
a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) - (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 - 4*I*a^3*
x^3 - 6*a^2*x^2 + 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) - 2*(a^3*x^3 - a*x)*arctan(a*x))/((a^7*c^3*x^4 + 2*a
^5*c^3*x^2 + a^3*c^3)*arctan(a*x)^2)

Sympy [F]

\[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^3} \, dx=\frac {\int \frac {x^{2}}{a^{6} x^{6} \operatorname {atan}^{3}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{3}{\left (a x \right )} + \operatorname {atan}^{3}{\left (a x \right )}}\, dx}{c^{3}} \]

[In]

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

[Out]

Integral(x**2/(a**6*x**6*atan(a*x)**3 + 3*a**4*x**4*atan(a*x)**3 + 3*a**2*x**2*atan(a*x)**3 + atan(a*x)**3), x
)/c**3

Maxima [F]

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

[In]

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

[Out]

1/2*(2*(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)^2*integrate((a^4*x^4 - 6*a^2*x^2 + 1)/((a^8*c^3*x^6
+ 3*a^6*c^3*x^4 + 3*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)), x) - a*x^2 + 2*(a^2*x^3 - x)*arctan(a*x))/((a^6*c^3*
x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)^2)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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