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

Optimal. Leaf size=41 \[ \frac {\text {Ci}\left (2 \tan ^{-1}(a x)\right )}{a^2 c^2}-\frac {x}{a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)} \]

[Out]

-x/a/c^2/(a^2*x^2+1)/arctan(a*x)+Ci(2*arctan(a*x))/a^2/c^2

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Rubi [A]  time = 0.21, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4968, 4970, 3312, 3302, 4904} \[ \frac {\text {CosIntegral}\left (2 \tan ^{-1}(a x)\right )}{a^2 c^2}-\frac {x}{a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 36, normalized size = 0.88 \[ \frac {\text {Ci}\left (2 \tan ^{-1}(a x)\right )-\frac {a x}{\left (a^2 x^2+1\right ) \tan ^{-1}(a x)}}{a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.45, size = 115, normalized size = 2.80 \[ \frac {{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} + 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) + {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\left (-\frac {a^{2} x^{2} - 2 i \, a x - 1}{a^{2} x^{2} + 1}\right ) - 2 \, a x}{2 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} \arctan \left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.25, size = 38, normalized size = 0.93 \[ \frac {2 \Ci \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\sin \left (2 \arctan \left (a x \right )\right )}{2 a^{2} c^{2} \arctan \left (a x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^2/c^2*(2*Ci(2*arctan(a*x))*arctan(a*x)-sin(2*arctan(a*x)))/arctan(a*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (a^{3} c^{2} x^{2} + a c^{2}\right )} \mathit {sage}_{0} x \arctan \left (a x\right ) + x}{{\left (a^{3} c^{2} x^{2} + a c^{2}\right )} \arctan \left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x}{a^{4} x^{4} \operatorname {atan}^{2}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{2}{\left (a x \right )} + \operatorname {atan}^{2}{\left (a x \right )}}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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