3.489 \(\int \frac{x^4}{(c+a^2 c x^2)^3 \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=50 \[ -\frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{2 a^5 c^3}+\frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{8 a^5 c^3}+\frac{3 \log \left (\tan ^{-1}(a x)\right )}{8 a^5 c^3} \]

[Out]

-CosIntegral[2*ArcTan[a*x]]/(2*a^5*c^3) + CosIntegral[4*ArcTan[a*x]]/(8*a^5*c^3) + (3*Log[ArcTan[a*x]])/(8*a^5
*c^3)

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Rubi [A]  time = 0.127434, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4970, 3312, 3302} \[ -\frac{\text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{2 a^5 c^3}+\frac{\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )}{8 a^5 c^3}+\frac{3 \log \left (\tan ^{-1}(a x)\right )}{8 a^5 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((c + a^2*c*x^2)^3*ArcTan[a*x]),x]

[Out]

-CosIntegral[2*ArcTan[a*x]]/(2*a^5*c^3) + CosIntegral[4*ArcTan[a*x]]/(8*a^5*c^3) + (3*Log[ArcTan[a*x]])/(8*a^5
*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 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]

Rubi steps

\begin{align*} \int \frac{x^4}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^4(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^5 c^3}\\ &=\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^5 c^3}\\ &=\frac{3 \log \left (\tan ^{-1}(a x)\right )}{8 a^5 c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^5 c^3}-\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^5 c^3}\\ &=-\frac{\text{Ci}\left (2 \tan ^{-1}(a x)\right )}{2 a^5 c^3}+\frac{\text{Ci}\left (4 \tan ^{-1}(a x)\right )}{8 a^5 c^3}+\frac{3 \log \left (\tan ^{-1}(a x)\right )}{8 a^5 c^3}\\ \end{align*}

Mathematica [A]  time = 0.144777, size = 34, normalized size = 0.68 \[ \frac{-4 \text{CosIntegral}\left (2 \tan ^{-1}(a x)\right )+\text{CosIntegral}\left (4 \tan ^{-1}(a x)\right )+3 \log \left (\tan ^{-1}(a x)\right )}{8 a^5 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((c + a^2*c*x^2)^3*ArcTan[a*x]),x]

[Out]

(-4*CosIntegral[2*ArcTan[a*x]] + CosIntegral[4*ArcTan[a*x]] + 3*Log[ArcTan[a*x]])/(8*a^5*c^3)

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Maple [A]  time = 0.059, size = 45, normalized size = 0.9 \begin{align*} -{\frac{{\it Ci} \left ( 2\,\arctan \left ( ax \right ) \right ) }{2\,{c}^{3}{a}^{5}}}+{\frac{{\it Ci} \left ( 4\,\arctan \left ( ax \right ) \right ) }{8\,{c}^{3}{a}^{5}}}+{\frac{3\,\ln \left ( \arctan \left ( ax \right ) \right ) }{8\,{c}^{3}{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x)

[Out]

-1/2*Ci(2*arctan(a*x))/a^5/c^3+1/8*Ci(4*arctan(a*x))/a^5/c^3+3/8*ln(arctan(a*x))/a^5/c^3

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

[Out]

integrate(x^4/((a^2*c*x^2 + c)^3*arctan(a*x)), x)

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Fricas [C]  time = 1.67151, size = 452, normalized size = 9.04 \begin{align*} \frac{6 \, \log \left (\arctan \left (a x\right )\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 ) + \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 ) - 4 \, \logintegral \left (-\frac{a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 4 \, \logintegral \left (-\frac{a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right )}{16 \, a^{5} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x),x, algorithm="fricas")

[Out]

1/16*(6*log(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)) + 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)) - 4*log_int
egral(-(a^2*x^2 + 2*I*a*x - 1)/(a^2*x^2 + 1)) - 4*log_integral(-(a^2*x^2 - 2*I*a*x - 1)/(a^2*x^2 + 1)))/(a^5*c
^3)

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

[Out]

Integral(x**4/(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^{4}}{{\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^4/(a^2*c*x^2+c)^3/arctan(a*x),x, algorithm="giac")

[Out]

integrate(x^4/((a^2*c*x^2 + c)^3*arctan(a*x)), x)