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

Optimal. Leaf size=120 \[ \frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{a^3 c^3}+\frac{x}{a^2 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac{2 x}{a^2 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}-\frac{1}{2 a^3 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}+\frac{1}{2 a^3 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2} \]

[Out]

1/(2*a^3*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^2) - 1/(2*a^3*c^3*(1 + a^2*x^2)*ArcTan[a*x]^2) - (2*x)/(a^2*c^3*(1 +
a^2*x^2)^2*ArcTan[a*x]) + x/(a^2*c^3*(1 + a^2*x^2)*ArcTan[a*x]) + CosIntegral[4*ArcTan[a*x]]/(a^3*c^3)

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Rubi [A]  time = 0.596477, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4964, 4902, 4968, 4970, 3312, 3302, 4904, 4406} \[ \frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{a^3 c^3}+\frac{x}{a^2 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}-\frac{2 x}{a^2 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}-\frac{1}{2 a^3 c^3 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2}+\frac{1}{2 a^3 c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*a^3*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^2) - 1/(2*a^3*c^3*(1 + a^2*x^2)*ArcTan[a*x]^2) - (2*x)/(a^2*c^3*(1 +
a^2*x^2)^2*ArcTan[a*x]) + x/(a^2*c^3*(1 + a^2*x^2)*ArcTan[a*x]) + CosIntegral[4*ArcTan[a*x]]/(a^3*c^3)

Rule 4964

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

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])

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]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^3} \, dx &=-\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^3} \, dx}{a^2}+\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^3} \, dx}{a^2 c}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}+\frac{2 \int \frac{x}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^2} \, dx}{a}-\frac{\int \frac{x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^2} \, dx}{a c}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac{x}{a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}-6 \int \frac{x^2}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx+\frac{2 \int \frac{1}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx}{a^2}+\frac{\int \frac{x^2}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx}{c}-\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)} \, dx}{a^2 c}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac{x}{a^2 c^3 \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^3 c^3}+\frac{\operatorname{Subst}\left (\int \frac{\sin ^2(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}+\frac{2 \operatorname{Subst}\left (\int \frac{\cos ^4(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}-\frac{6 \operatorname{Subst}\left (\int \frac{\cos ^2(x) \sin ^2(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac{x}{a^2 c^3 \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^3 c^3}-\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^3 c^3}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{3}{8 x}+\frac{\cos (2 x)}{2 x}+\frac{\cos (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}-\frac{6 \operatorname{Subst}\left (\int \left (\frac{1}{8 x}-\frac{\cos (4 x)}{8 x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac{x}{a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^3 c^3}-2 \frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^3 c^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^3 c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)^2}-\frac{1}{2 a^3 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2}-\frac{2 x}{a^2 c^3 \left (1+a^2 x^2\right )^2 \tan ^{-1}(a x)}+\frac{x}{a^2 c^3 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}+\frac{\text{Ci}\left (4 \tan ^{-1}(a x)\right )}{a^3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.132692, size = 60, normalized size = 0.5 \[ \frac{\frac{a x \left (2 \left (a^2 x^2-1\right ) \tan ^{-1}(a x)-a x\right )}{\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}+2 \text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{2 a^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*x*(-(a*x) + 2*(-1 + a^2*x^2)*ArcTan[a*x]))/((1 + a^2*x^2)^2*ArcTan[a*x]^2) + 2*CosIntegral[4*ArcTan[a*x]])
/(2*a^3*c^3)

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Maple [A]  time = 0.075, size = 52, normalized size = 0.4 \begin{align*}{\frac{16\,{\it Ci} \left ( 4\,\arctan \left ( ax \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{2}-4\,\sin \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +\cos \left ( 4\,\arctan \left ( ax \right ) \right ) -1}{16\,{a}^{3}{c}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a^3/c^3*(16*Ci(4*arctan(a*x))*arctan(a*x)^2-4*sin(4*arctan(a*x))*arctan(a*x)+cos(4*arctan(a*x))-1)/arctan
(a*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{-a x^{2} + 2 \,{\left (a^{2} x^{3} - x\right )} \arctan \left (a x\right ) + \frac{2 \,{\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}{\left (a^{4} \int \frac{x^{4}}{a^{6} x^{6} \arctan \left (a x\right ) + 3 \, a^{4} x^{4} \arctan \left (a x\right ) + 3 \, a^{2} x^{2} \arctan \left (a x\right ) + \arctan \left (a x\right )}\,{d x} - 6 \, a^{2} \int \frac{x^{2}}{a^{6} x^{6} \arctan \left (a x\right ) + 3 \, a^{4} x^{4} \arctan \left (a x\right ) + 3 \, a^{2} x^{2} \arctan \left (a x\right ) + \arctan \left (a x\right )}\,{d x} + \int \frac{1}{{\left (a^{2} x^{2} + 1\right )}^{3} \arctan \left (a x\right )}\,{d x}\right )} \arctan \left (a x\right )^{2}}{a^{2} c^{3}}}{2 \,{\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )} \arctan \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)^2*integrate((a^4*x^4 - 6*a^2*x^2 + 1)/((a^8*c^3*x^6
+ 3*a^6*c^3*x^4 + 3*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)), x) - a*x^2 + 2*(a^2*x^3 - x)*arctan(a*x))/((a^6*c^3*
x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)^2)

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Fricas [C]  time = 1.7319, size = 497, normalized size = 4.14 \begin{align*} -\frac{a^{2} x^{2} -{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \logintegral \left (\frac{a^{4} x^{4} + 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) -{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} \logintegral \left (\frac{a^{4} x^{4} - 4 i \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 4 i \, a x + 1}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - 2 \,{\left (a^{3} x^{3} - a x\right )} \arctan \left (a x\right )}{2 \,{\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )} \arctan \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*x^2 - (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 + 4*I*a^3*x^3 - 6*a^2*x^2 - 4*I*
a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) - (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2*log_integral((a^4*x^4 - 4*I*a^3*
x^3 - 6*a^2*x^2 + 4*I*a*x + 1)/(a^4*x^4 + 2*a^2*x^2 + 1)) - 2*(a^3*x^3 - a*x)*arctan(a*x))/((a^7*c^3*x^4 + 2*a
^5*c^3*x^2 + a^3*c^3)*arctan(a*x)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{2}}{a^{6} x^{6} \operatorname{atan}^{3}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname{atan}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atan}^{3}{\left (a x \right )} + \operatorname{atan}^{3}{\left (a x \right )}}\, dx}{c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a**6*x**6*atan(a*x)**3 + 3*a**4*x**4*atan(a*x)**3 + 3*a**2*x**2*atan(a*x)**3 + atan(a*x)**3), x
)/c**3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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