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

Optimal. Leaf size=100 \[ -\frac {x}{4 c^2 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{2 a c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}-\frac {\tan ^{-1}(a x)}{4 a c^2} \]

[Out]

-1/4*x/c^2/(a^2*x^2+1)-1/4*arctan(a*x)/a/c^2+1/2*arctan(a*x)/a/c^2/(a^2*x^2+1)+1/2*x*arctan(a*x)^2/c^2/(a^2*x^
2+1)+1/6*arctan(a*x)^3/a/c^2

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4892, 4930, 199, 205} \[ -\frac {x}{4 c^2 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{2 a c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}-\frac {\tan ^{-1}(a x)}{4 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(4*a*c^2) + ArcTan[a*x]/(2*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^2)/(2*
c^2*(1 + a^2*x^2)) + ArcTan[a*x]^3/(6*a*c^2)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 4892

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTan[c*x])
^p)/(2*d*(d + e*x^2)), x] + (-Dist[(b*c*p)/2, Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^2, x], x] + Simp
[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p,
0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx &=\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}-a \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=\frac {\tan ^{-1}(a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}-\frac {1}{2} \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=-\frac {x}{4 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}-\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 c}\\ &=-\frac {x}{4 c^2 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)}{4 a c^2}+\frac {\tan ^{-1}(a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{6 a c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.65 \[ \frac {2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3+\left (3-3 a^2 x^2\right ) \tan ^{-1}(a x)-3 a x+6 a x \tan ^{-1}(a x)^2}{12 c^2 \left (a^3 x^2+a\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x + (3 - 3*a^2*x^2)*ArcTan[a*x] + 6*a*x*ArcTan[a*x]^2 + 2*(1 + a^2*x^2)*ArcTan[a*x]^3)/(12*c^2*(a + a^3*
x^2))

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fricas [A]  time = 0.55, size = 67, normalized size = 0.67 \[ \frac {6 \, a x \arctan \left (a x\right )^{2} + 2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \, {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{12 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a*x*arctan(a*x)^2 + 2*(a^2*x^2 + 1)*arctan(a*x)^3 - 3*a*x - 3*(a^2*x^2 - 1)*arctan(a*x))/(a^3*c^2*x^2
+ a*c^2)

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

[Out]

sage0*x

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maple [A]  time = 0.04, size = 91, normalized size = 0.91 \[ -\frac {x}{4 c^{2} \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{4 a \,c^{2}}+\frac {\arctan \left (a x \right )}{2 a \,c^{2} \left (a^{2} x^{2}+1\right )}+\frac {x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{6 a \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x/c^2/(a^2*x^2+1)-1/4*arctan(a*x)/a/c^2+1/2*arctan(a*x)/a/c^2/(a^2*x^2+1)+1/2*x*arctan(a*x)^2/c^2/(a^2*x^
2+1)+1/6*arctan(a*x)^3/a/c^2

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maxima [A]  time = 0.45, size = 146, normalized size = 1.46 \[ \frac {1}{2} \, {\left (\frac {x}{a^{2} c^{2} x^{2} + c^{2}} + \frac {\arctan \left (a x\right )}{a c^{2}}\right )} \arctan \left (a x\right )^{2} + \frac {{\left (2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{12 \, {\left (a^{5} c^{2} x^{2} + a^{3} c^{2}\right )}} - \frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1\right )} a \arctan \left (a x\right )}{2 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x/(a^2*c^2*x^2 + c^2) + arctan(a*x)/(a*c^2))*arctan(a*x)^2 + 1/12*(2*(a^2*x^2 + 1)*arctan(a*x)^3 - 3*a*x
- 3*(a^2*x^2 + 1)*arctan(a*x))*a^2/(a^5*c^2*x^2 + a^3*c^2) - 1/2*((a^2*x^2 + 1)*arctan(a*x)^2 - 1)*a*arctan(a*
x)/(a^4*c^2*x^2 + a^2*c^2)

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mupad [B]  time = 0.52, size = 101, normalized size = 1.01 \[ \frac {\mathrm {atan}\left (a\,x\right )}{2\,\left (a^3\,c^2\,x^2+a\,c^2\right )}-\frac {x}{2\,\left (2\,a^2\,c^2\,x^2+2\,c^2\right )}+\frac {x\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,\left (a^2\,c^2\,x^2+c^2\right )}-\frac {\mathrm {atan}\left (a\,x\right )}{4\,a\,c^2}+\frac {{\mathrm {atan}\left (a\,x\right )}^3}{6\,a\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^2/(c + a^2*c*x^2)^2,x)

[Out]

atan(a*x)/(2*(a*c^2 + a^3*c^2*x^2)) - x/(2*(2*c^2 + 2*a^2*c^2*x^2)) + (x*atan(a*x)^2)/(2*(c^2 + a^2*c^2*x^2))
- atan(a*x)/(4*a*c^2) + atan(a*x)^3/(6*a*c^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**2/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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