\(\int \frac {x}{(c+a^2 c x^2)^2 \arctan (a x)^2} \, dx\) [553]

Optimal result

Integrand size = 20, antiderivative size = 41 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^2} \, dx=-\frac {x}{a c^2 \left (1+a^2 x^2\right ) \arctan (a x)}+\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{a^2 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5088, 5090, 3393, 3383, 5024} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^2} \, dx=\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{a^2 c^2}-\frac {x}{a c^2 \left (a^2 x^2+1\right ) \arctan (a x)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^2} \, dx=\frac {-\frac {a x}{\left (1+a^2 x^2\right ) \arctan (a x)}+\operatorname {CosIntegral}(2 \arctan (a x))}{a^2 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {2 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\sin \left (2 \arctan \left (a x \right )\right )}{2 a^{2} c^{2} \arctan \left (a x \right )}\)	\(38\)
default	\(\frac {2 \,\operatorname {Ci}\left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\sin \left (2 \arctan \left (a x \right )\right )}{2 a^{2} c^{2} \arctan \left (a x \right )}\)	\(38\)

[In]

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

[Out]

1/2/a^2/c^2*(2*Ci(2*arctan(a*x))*arctan(a*x)-sin(2*arctan(a*x)))/arctan(a*x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.80 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^2} \, dx=\frac {{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 \, a x}{2 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} \arctan \left (a x\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-((a^3*c^2*x^2 + a*c^2)*arctan(a*x)*integrate((a^2*x^2 - 1)/((a^5*c^2*x^4 + 2*a^3*c^2*x^2 + a*c^2)*arctan(a*x)
), x) + x)/((a^3*c^2*x^2 + a*c^2)*arctan(a*x))

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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