3.6.62 \(\int \frac {1}{(c+a^2 c x^2)^3 \text {ArcTan}(a x)^2} \, dx\) [562]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5022, 5090, 4491, 3380} \begin {gather*} -\frac {1}{a c^3 \left (a^2 x^2+1\right )^2 \text {ArcTan}(a x)}-\frac {\text {Si}(2 \text {ArcTan}(a x))}{a c^3}-\frac {\text {Si}(4 \text {ArcTan}(a x))}{2 a c^3} \end {gather*}

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 3380

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]

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 5022

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 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 {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 \text {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 \text {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 {\text {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\tan ^{-1}(a x)\right )}{2 a c^3}-\frac {\text {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 = 0.08, size = 45, normalized size = 0.78 \begin {gather*} -\frac {\frac {1}{\left (1+a^2 x^2\right )^2 \text {ArcTan}(a x)}+\text {Si}(2 \text {ArcTan}(a x))+\frac {1}{2} \text {Si}(4 \text {ArcTan}(a x))}{a c^3} \end {gather*}

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))

________________________________________________________________________________________

Maple [A]
time = 0.22, size = 59, normalized size = 1.02

method	result	size
derivativedivides	\(-\frac {8 \sinIntegral \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \sinIntegral \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\cos \left (4 \arctan \left (a x \right )\right )+4 \cos \left (2 \arctan \left (a x \right )\right )+3}{8 a \,c^{3} \arctan \left (a x \right )}\)	\(59\)
default	\(-\frac {8 \sinIntegral \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+4 \sinIntegral \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\cos \left (4 \arctan \left (a x \right )\right )+4 \cos \left (2 \arctan \left (a x \right )\right )+3}{8 a \,c^{3} \arctan \left (a x \right )}\)	\(59\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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(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))

________________________________________________________________________________________

Fricas [C] Result contains complex when optimal does not.
time = 2.04, size = 287, normalized size = 4.95 \begin {gather*} \frac {{\left (-i \, a^{4} x^{4} - 2 i \, a^{2} x^{2} - i\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 (i \, a^{4} x^{4} + 2 i \, a^{2} x^{2} + i\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 ) - 2 \, {\left (i \, a^{4} x^{4} + 2 i \, a^{2} x^{2} + i\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 \, {\left (-i \, a^{4} x^{4} - 2 i \, a^{2} x^{2} - i\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}{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 {gather*}

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

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

[Out]

sage0*x

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________