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

Optimal. Leaf size=62 \[ \frac {x}{4 a c^2 \left (a^2 x^2+1\right )}-\frac {\tan ^{-1}(a x)}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{4 a^2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4930, 199, 205} \[ \frac {x}{4 a c^2 \left (a^2 x^2+1\right )}-\frac {\tan ^{-1}(a x)}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{4 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*a*c^2*(1 + a^2*x^2)) + ArcTan[a*x]/(4*a^2*c^2) - ArcTan[a*x]/(2*a^2*c^2*(1 + a^2*x^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 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 {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac {\tan ^{-1}(a x)}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a}\\ &=\frac {x}{4 a c^2 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 a c}\\ &=\frac {x}{4 a c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)}{4 a^2 c^2}-\frac {\tan ^{-1}(a x)}{2 a^2 c^2 \left (1+a^2 x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 39, normalized size = 0.63 \[ \frac {\left (a^2 x^2-1\right ) \tan ^{-1}(a x)+a x}{4 a^2 c^2 \left (a^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 40, normalized size = 0.65 \[ \frac {a x + {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{4 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*x + (a^2*x^2 - 1)*arctan(a*x))/(a^4*c^2*x^2 + a^2*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(x*arctan(a*x)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.03, size = 57, normalized size = 0.92 \[ \frac {x}{4 a \,c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{4 a^{2} c^{2}}-\frac {\arctan \left (a x \right )}{2 a^{2} c^{2} \left (a^{2} x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 59, normalized size = 0.95 \[ \frac {\frac {x}{a^{2} c x^{2} + c} + \frac {\arctan \left (a x\right )}{a c}}{4 \, a c} - \frac {\arctan \left (a x\right )}{2 \, {\left (a^{2} c x^{2} + c\right )} a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 40, normalized size = 0.65 \[ \frac {a\,x-\mathrm {atan}\left (a\,x\right )+a^2\,x^2\,\mathrm {atan}\left (a\,x\right )}{4\,a^2\,c^2\,\left (a^2\,x^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x - atan(a*x) + a^2*x^2*atan(a*x))/(4*a^2*c^2*(a^2*x^2 + 1))

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sympy [A]  time = 1.42, size = 107, normalized size = 1.73 \[ \begin {cases} \frac {a^{2} x^{2} \operatorname {atan}{\left (a x \right )}}{4 a^{4} c^{2} x^{2} + 4 a^{2} c^{2}} + \frac {a x}{4 a^{4} c^{2} x^{2} + 4 a^{2} c^{2}} - \frac {\operatorname {atan}{\left (a x \right )}}{4 a^{4} c^{2} x^{2} + 4 a^{2} c^{2}} & \text {for}\: c \neq 0 \\\tilde {\infty } \left (\frac {x^{2} \operatorname {atan}{\left (a x \right )}}{2} - \frac {x}{2 a} + \frac {\operatorname {atan}{\left (a x \right )}}{2 a^{2}}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**2*atan(a*x)/(4*a**4*c**2*x**2 + 4*a**2*c**2) + a*x/(4*a**4*c**2*x**2 + 4*a**2*c**2) - atan(
a*x)/(4*a**4*c**2*x**2 + 4*a**2*c**2), Ne(c, 0)), (zoo*(x**2*atan(a*x)/2 - x/(2*a) + atan(a*x)/(2*a**2)), True
))

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