3.6.61 \(\int \frac {x}{(c+a^2 c x^2)^3 \text {ArcTan}(a x)^2} \, dx\) [561]

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

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5088, 5090, 4491, 3383, 5024, 3393} \begin {gather*} \frac {\text {CosIntegral}(2 \text {ArcTan}(a x))}{2 a^2 c^3}+\frac {\text {CosIntegral}(4 \text {ArcTan}(a x))}{2 a^2 c^3}-\frac {x}{a c^3 \left (a^2 x^2+1\right )^2 \text {ArcTan}(a x)} \end {gather*}

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 3383

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 3393

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 4491

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 5024

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 5088

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 5090

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 )^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 {\text {Subst}\left (\int \frac {\cos ^4(x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^3}-\frac {3 \text {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 {\text {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 \text {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 {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^3}+\frac {3 \text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^3}+\frac {\text {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*}

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Mathematica [A]
time = 0.05, size = 75, normalized size = 1.23 \begin {gather*} \frac {-2 a x+\left (1+a^2 x^2\right )^2 \text {ArcTan}(a x) \text {CosIntegral}(2 \text {ArcTan}(a x))+\left (1+a^2 x^2\right )^2 \text {ArcTan}(a x) \text {CosIntegral}(4 \text {ArcTan}(a x))}{2 c^3 \left (a+a^3 x^2\right )^2 \text {ArcTan}(a x)} \end {gather*}

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 = 0.24, size = 60, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {4 \cosineIntegral \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \cosineIntegral \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\sin \left (4 \arctan \left (a x \right )\right )-2 \sin \left (2 \arctan \left (a x \right )\right )}{8 a^{2} c^{3} \arctan \left (a x \right )}\)	\(60\)
default	\(\frac {4 \cosineIntegral \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \cosineIntegral \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\sin \left (4 \arctan \left (a x \right )\right )-2 \sin \left (2 \arctan \left (a x \right )\right )}{8 a^{2} c^{3} \arctan \left (a x \right )}\)	\(60\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 1.13, size = 286, normalized size = 4.69 \begin {gather*} \frac {{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right ) \operatorname {log\_integral}\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 ) \operatorname {log\_integral}\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 ) \operatorname {log\_integral}\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 ) \operatorname {log\_integral}\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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