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

Optimal. Leaf size=65 \[ \frac{x}{c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac{1}{2 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}-\frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{a c^2} \]

[Out]

-1/(2*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)

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Rubi [A]  time = 0.245925, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {4902, 4968, 4970, 3312, 3302, 4904} \[ \frac{x}{c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac{1}{2 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}-\frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*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 4902

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 4968

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 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 3312

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 3302

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 4904

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])

Rubi steps

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

Mathematica [A]  time = 0.0653508, size = 58, normalized size = 0.89 \[ \frac{-2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2 \text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )+2 a x \tan ^{-1}(a x)-1}{2 c^2 \left (a^3 x^2+a\right ) \tan ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.071, size = 52, normalized size = 0.8 \begin{align*} -{\frac{4\,{\it Ci} \left ( 2\,\arctan \left ( ax \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{2}-2\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +\cos \left ( 2\,\arctan \left ( ax \right ) \right ) +1}{4\,a{c}^{2} \left ( \arctan \left ( ax \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, a x \arctan \left (a x\right ) + \frac{2 \,{\left (a^{3} c^{2} x^{2} + a c^{2}\right )}{\left (a^{2} \int \frac{x^{2}}{a^{4} x^{4} \arctan \left (a x\right ) + 2 \, a^{2} x^{2} \arctan \left (a x\right ) + \arctan \left (a x\right )}\,{d x} - \int \frac{1}{a^{4} x^{4} \arctan \left (a x\right ) + 2 \, a^{2} x^{2} \arctan \left (a x\right ) + \arctan \left (a x\right )}\,{d x}\right )} \arctan \left (a x\right )^{2}}{c^{2}} - 1}{2 \,{\left (a^{3} c^{2} x^{2} + a c^{2}\right )} \arctan \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [C]  time = 1.74219, size = 316, normalized size = 4.86 \begin{align*} -\frac{{\left (a^{2} x^{2} + 1\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 (a^{2} x^{2} + 1\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 ) - 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a^2*c*x^2+c)^2/arctan(a*x)^3,x, algorithm="giac")

[Out]

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