\(\int \frac {1}{(c+a^2 c x^2)^2 \arctan (a x)^3} \, dx\) [630]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=-\frac {1}{2 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^2}+\frac {x}{c^2 \left (1+a^2 x^2\right ) \arctan (a x)}-\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{a c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5022, 5088, 5090, 3393, 3383, 5024} \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=\frac {x}{c^2 \left (a^2 x^2+1\right ) \arctan (a x)}-\frac {1}{2 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^2}-\frac {\operatorname {CosIntegral}(2 \arctan (a x))}{a c^2} \]

[In]

Int[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^3),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=\frac {-1+2 a x \arctan (a x)-2 \left (1+a^2 x^2\right ) \arctan (a x)^2 \operatorname {CosIntegral}(2 \arctan (a x))}{2 c^2 \left (a+a^3 x^2\right ) \arctan (a x)^2} \]

[In]

Integrate[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^3),x]

[Out]

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

Maple [A] (verified)

Time = 8.88 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80

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

[In]

int(1/(a^2*c*x^2+c)^2/arctan(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4/a/c^2*(4*Ci(2*arctan(a*x))*arctan(a*x)^2-2*sin(2*arctan(a*x))*arctan(a*x)+cos(2*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 = 122, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=-\frac {{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 \, a x \arctan \left (a x\right ) + 1}{2 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )} \arctan \left (a x\right )^{2}} \]

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^3,x, algorithm="fricas")

[Out]

-1/2*((a^2*x^2 + 1)*arctan(a*x)^2*log_integral(-(a^2*x^2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) + (a^2*x^2 + 1)*arctan(
a*x)^2*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)) - 2*a*x*arctan(a*x) + 1)/((a^3*c^2*x^2 + a*c^2)*ar
ctan(a*x)^2)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(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]

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

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^3,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(1/(a^2*c*x^2+c)^2/arctan(a*x)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^3} \, dx=\int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^3\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

[In]

int(1/(atan(a*x)^3*(c + a^2*c*x^2)^2),x)

[Out]

int(1/(atan(a*x)^3*(c + a^2*c*x^2)^2), x)