3.629 \(\int \frac{x}{(c+a^2 c x^2)^2 \tan ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{a^2 c^2}-\frac{x}{2 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)} \]

[Out]

-x/(2*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)

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Rubi [A]  time = 0.119267, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4932, 4970, 4406, 12, 3299} \[ -\frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{a^2 c^2}-\frac{x}{2 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*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 4932

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 4970

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])

Rule 4406

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 12

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

Rule 3299

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]

Rubi steps

\begin{align*} \int \frac{x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^3} \, dx &=-\frac{x}{2 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-2 \int \frac{x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx\\ &=-\frac{x}{2 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^2}\\ &=-\frac{x}{2 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^2}\\ &=-\frac{x}{2 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac{\operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^2}\\ &=-\frac{x}{2 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{1-a^2 x^2}{2 a^2 c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-\frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{a^2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0524829, size = 70, normalized size = 0.86 \[ \frac{-2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2 \text{Si}\left (2 \tan ^{-1}(a x)\right )+\left (a^2 x^2-1\right ) \tan ^{-1}(a x)-a x}{2 a^2 c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.065, size = 51, normalized size = 0.6 \begin{align*} -{\frac{4\,{\it Si} \left ( 2\,\arctan \left ( ax \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{2}+2\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{a}^{2}{c}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \,{\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} \arctan \left (a x\right )^{2} \int \frac{x}{{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )}\,{d x} + 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [C]  time = 1.77001, size = 338, normalized size = 4.17 \begin{align*} \frac{{\left (-i \, a^{2} x^{2} - i\right )} \arctan \left (a x\right )^{2} \logintegral \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} \logintegral \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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