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

Optimal. Leaf size=35 \[ \frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{4 a^2 c^3}+\frac{\text{Si}\left (4 \tan ^{-1}(a x)\right )}{8 a^2 c^3} \]

[Out]

SinIntegral[2*ArcTan[a*x]]/(4*a^2*c^3) + SinIntegral[4*ArcTan[a*x]]/(8*a^2*c^3)

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Rubi [A]  time = 0.0899071, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4970, 4406, 3299} \[ \frac{\text{Si}\left (2 \tan ^{-1}(a x)\right )}{4 a^2 c^3}+\frac{\text{Si}\left (4 \tan ^{-1}(a x)\right )}{8 a^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

SinIntegral[2*ArcTan[a*x]]/(4*a^2*c^3) + SinIntegral[4*ArcTan[a*x]]/(8*a^2*c^3)

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

Mathematica [A]  time = 0.0809775, size = 27, normalized size = 0.77 \[ \frac{2 \text{Si}\left (2 \tan ^{-1}(a x)\right )+\text{Si}\left (4 \tan ^{-1}(a x)\right )}{8 a^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*SinIntegral[2*ArcTan[a*x]] + SinIntegral[4*ArcTan[a*x]])/(8*a^2*c^3)

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Maple [A]  time = 0.058, size = 32, normalized size = 0.9 \begin{align*}{\frac{{\it Si} \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{c}^{3}{a}^{2}}}+{\frac{{\it Si} \left ( 4\,\arctan \left ( ax \right ) \right ) }{8\,{c}^{3}{a}^{2}}} \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),x)

[Out]

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

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Maxima [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 )}\,{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),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 1.61181, size = 435, normalized size = 12.43 \begin{align*} \frac{i \, \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 ) - i \, \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 i \, \logintegral \left (-\frac{a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 i \, \logintegral \left (-\frac{a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right )}{16 \, a^{2} c^{3}} \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),x, algorithm="fricas")

[Out]

1/16*(I*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*log_inte
gral((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*log_integral(-(a^2*x^2
+ 2*I*a*x - 1)/(a^2*x^2 + 1)) - 2*I*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)))/(a^2*c^3)

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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}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname{atan}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atan}{\left (a x \right )} + \operatorname{atan}{\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),x)

[Out]

Integral(x/(a**6*x**6*atan(a*x) + 3*a**4*x**4*atan(a*x) + 3*a**2*x**2*atan(a*x) + atan(a*x)), 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 )}\,{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),x, algorithm="giac")

[Out]

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