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

Optimal. Leaf size=58 \[ -\frac{1}{a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}-\frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{a c^3}-\frac{\text{Si}\left (4 \tan ^{-1}(a x)\right )}{2 a c^3} \]

[Out]

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

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Rubi [A]  time = 0.114183, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4902, 4970, 4406, 3299} \[ -\frac{1}{a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}-\frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{a c^3}-\frac{\text{Si}\left (4 \tan ^{-1}(a x)\right )}{2 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0943938, size = 45, normalized size = 0.78 \[ -\frac{\frac{1}{\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)}+\text{Si}\left (2 \tan ^{-1}(a x)\right )+\frac{1}{2} \text{Si}\left (4 \tan ^{-1}(a x)\right )}{a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 59, normalized size = 1. \begin{align*} -{\frac{8\,{\it Si} \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +4\,{\it Si} \left ( 4\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +4\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) +\cos \left ( 4\,\arctan \left ( ax \right ) \right ) +3}{8\,a{c}^{3}\arctan \left ( ax \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-(8*(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)*arctan(a*x)*integrate(1/2*x/((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*
c^3*x^2 + c^3)*arctan(a*x)), x) + 1)/((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 = 2.01078, size = 720, normalized size = 12.41 \begin{align*} \frac{{\left (-i \, a^{4} x^{4} - 2 i \, a^{2} x^{2} - i\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 (i \, a^{4} x^{4} + 2 i \, a^{2} x^{2} + i\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 (-2 i \, a^{4} x^{4} - 4 i \, a^{2} x^{2} - 2 i\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 (2 i \, a^{4} x^{4} + 4 i \, a^{2} x^{2} + 2 i\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}{4 \,{\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(1/(a^2*c*x^2+c)^3/arctan(a*x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{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(1/(a**2*c*x**2+c)**3/atan(a*x)**2,x)

[Out]

Integral(1/(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{1}{{\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(1/(a^2*c*x^2+c)^3/arctan(a*x)^2,x, algorithm="giac")

[Out]

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