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

Optimal. Leaf size=61 \[ \frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}+\frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac{x}{a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.24642, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4968, 4970, 4406, 3302, 4904, 3312} \[ \frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{2 a^2 c^3}+\frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{2 a^2 c^3}-\frac{x}{a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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])
)

Rubi steps

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

Mathematica [A]  time = 0.066865, size = 75, normalized size = 1.23 \[ \frac{\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )+\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x) \text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )-2 a x}{2 c^3 \left (a^3 x^2+a\right )^2 \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 60, normalized size = 1. \begin{align*}{\frac{4\,{\it Ci} \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +4\,{\it Ci} \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) -2\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) -\sin \left ( 4\,\arctan \left ( ax \right ) \right ) }{8\,{a}^{2}{c}^{3}\arctan \left ( ax \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a^2/c^3*(4*Ci(2*arctan(a*x))*arctan(a*x)+4*Ci(4*arctan(a*x))*arctan(a*x)-2*sin(2*arctan(a*x))-sin(4*arctan
(a*x)))/arctan(a*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{x + \frac{{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}{\left (3 \, 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}{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}\right )} \arctan \left (a x\right )}{a c^{3}}}{{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )} \arctan \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.0782, size = 693, normalized size = 11.36 \begin{align*} \frac{{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \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 ) \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 ) \logintegral \left (-\frac{a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) +{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \logintegral \left (-\frac{a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 4 \, a x}{4 \,{\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )} \arctan \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{3} \arctan \left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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