\(\int \frac {x}{(c+a^2 c x^2)^2 \arctan (a x)^3} \, dx\) [629]

Optimal result

Integrand size = 20, antiderivative size = 81 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=-\frac {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {\text {Si}(2 \arctan (a x))}{a^2 c^2} \]

[Out]

-1/2*x/a/c^2/(a^2*x^2+1)/arctan(a*x)^2+1/2*(a^2*x^2-1)/a^2/c^2/(a^2*x^2+1)/arctan(a*x)-Si(2*arctan(a*x))/a^2/c
^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5052, 5090, 4491, 12, 3380} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=-\frac {\text {Si}(2 \arctan (a x))}{a^2 c^2}-\frac {x}{2 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (a^2 x^2+1\right ) \arctan (a x)} \]

[In]

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

[Out]

-1/2*x/(a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^2) - (1 - a^2*x^2)/(2*a^2*c^2*(1 + a^2*x^2)*ArcTan[a*x]) - SinIntegral
[2*ArcTan[a*x]]/(a^2*c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3380

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

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 5052

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTan
[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2))), x] + (-Dist[4/(b^2*(p + 1)*(p + 2)), Int[x*((a + b*ArcTan[c*x])^(
p + 2)/(d + e*x^2)^2), x], x] - Simp[(1 - c^2*x^2)*((a + b*ArcTan[c*x])^(p + 2)/(b^2*e*(p + 1)*(p + 2)*(d + e*
x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]

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 {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-2 \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)} \, dx \\ & = -\frac {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {2 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\arctan (a x)\right )}{a^2 c^2} \\ & = -\frac {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\arctan (a x)\right )}{a^2 c^2} \\ & = -\frac {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\arctan (a x)\right )}{a^2 c^2} \\ & = -\frac {x}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}-\frac {1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {\text {Si}(2 \arctan (a x))}{a^2 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=\frac {-a x+\left (-1+a^2 x^2\right ) \arctan (a x)-2 \left (1+a^2 x^2\right ) \arctan (a x)^2 \text {Si}(2 \arctan (a x))}{2 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2} \]

[In]

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

[Out]

(-(a*x) + (-1 + a^2*x^2)*ArcTan[a*x] - 2*(1 + a^2*x^2)*ArcTan[a*x]^2*SinIntegral[2*ArcTan[a*x]])/(2*a^2*c^2*(1
+ a^2*x^2)*ArcTan[a*x]^2)

Maple [A] (verified)

Time = 8.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(-\frac {4 \,\operatorname {Si}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+2 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sin \left (2 \arctan \left (a x \right )\right )}{4 a^{2} c^{2} \arctan \left (a x \right )^{2}}\)	\(51\)
default	\(-\frac {4 \,\operatorname {Si}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )^{2}+2 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sin \left (2 \arctan \left (a x \right )\right )}{4 a^{2} c^{2} \arctan \left (a x \right )^{2}}\)	\(51\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.67 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=\frac {{\left (-i \, a^{2} x^{2} - i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (i \, a^{2} x^{2} + i\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - a x + {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{2 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} \arctan \left (a x\right )^{2}} \]

[In]

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

[Out]

1/2*((-I*a^2*x^2 - I)*arctan(a*x)^2*log_integral(-(a^2*x^2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) + (I*a^2*x^2 + I)*arc
tan(a*x)^2*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)) - a*x + (a^2*x^2 - 1)*arctan(a*x))/((a^4*c^2*x
^2 + a^2*c^2)*arctan(a*x)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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