3.194 \(\int \frac{x \tan ^{-1}(a x)}{(c+a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=84 \[ \frac{3 x}{32 a c^3 \left (a^2 x^2+1\right )}+\frac{x}{16 a c^3 \left (a^2 x^2+1\right )^2}-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)}{32 a^2 c^3} \]

[Out]

x/(16*a*c^3*(1 + a^2*x^2)^2) + (3*x)/(32*a*c^3*(1 + a^2*x^2)) + (3*ArcTan[a*x])/(32*a^2*c^3) - ArcTan[a*x]/(4*
a^2*c^3*(1 + a^2*x^2)^2)

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Rubi [A]  time = 0.0496819, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4930, 199, 205} \[ \frac{3 x}{32 a c^3 \left (a^2 x^2+1\right )}+\frac{x}{16 a c^3 \left (a^2 x^2+1\right )^2}-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)}{32 a^2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[a*x])/(c + a^2*c*x^2)^3,x]

[Out]

x/(16*a*c^3*(1 + a^2*x^2)^2) + (3*x)/(32*a*c^3*(1 + a^2*x^2)) + (3*ArcTan[a*x])/(32*a^2*c^3) - ArcTan[a*x]/(4*
a^2*c^3*(1 + a^2*x^2)^2)

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]

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]

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx &=-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^3} \, dx}{4 a}\\ &=\frac{x}{16 a c^3 \left (1+a^2 x^2\right )^2}-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{16 a c}\\ &=\frac{x}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x}{32 a c^3 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 \int \frac{1}{c+a^2 c x^2} \, dx}{32 a c^2}\\ &=\frac{x}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x}{32 a c^3 \left (1+a^2 x^2\right )}+\frac{3 \tan ^{-1}(a x)}{32 a^2 c^3}-\frac{\tan ^{-1}(a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0454185, size = 55, normalized size = 0.65 \[ \frac{a x \left (3 a^2 x^2+5\right )+\left (3 a^4 x^4+6 a^2 x^2-5\right ) \tan ^{-1}(a x)}{32 c^3 \left (a^3 x^2+a\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTan[a*x])/(c + a^2*c*x^2)^3,x]

[Out]

(a*x*(5 + 3*a^2*x^2) + (-5 + 6*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x])/(32*c^3*(a + a^3*x^2)^2)

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Maple [A]  time = 0.029, size = 77, normalized size = 0.9 \begin{align*}{\frac{x}{16\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,x}{32\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,\arctan \left ( ax \right ) }{32\,{a}^{2}{c}^{3}}}-{\frac{\arctan \left ( ax \right ) }{4\,{a}^{2}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x/a/c^3/(a^2*x^2+1)^2+3/32*x/a/c^3/(a^2*x^2+1)+3/32*arctan(a*x)/a^2/c^3-1/4*arctan(a*x)/a^2/c^3/(a^2*x^2+
1)^2

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Maxima [A]  time = 1.48182, size = 116, normalized size = 1.38 \begin{align*} \frac{\frac{3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}} + \frac{3 \, \arctan \left (a x\right )}{a c^{2}}}{32 \, a c} - \frac{\arctan \left (a x\right )}{4 \,{\left (a^{2} c x^{2} + c\right )}^{2} a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((3*a^2*x^3 + 5*x)/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2) + 3*arctan(a*x)/(a*c^2))/(a*c) - 1/4*arctan(a*x)/(
(a^2*c*x^2 + c)^2*a^2*c)

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Fricas [A]  time = 1.54218, size = 146, normalized size = 1.74 \begin{align*} \frac{3 \, a^{3} x^{3} + 5 \, a x +{\left (3 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 5\right )} \arctan \left (a x\right )}{32 \,{\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(3*a^3*x^3 + 5*a*x + (3*a^4*x^4 + 6*a^2*x^2 - 5)*arctan(a*x))/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)

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Sympy [A]  time = 3.81606, size = 235, normalized size = 2.8 \begin{align*} \begin{cases} \frac{3 a^{4} x^{4} \operatorname{atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac{3 a^{3} x^{3}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac{6 a^{2} x^{2} \operatorname{atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac{5 a x}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} - \frac{5 \operatorname{atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} & \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**4*x**4*atan(a*x)/(32*a**6*c**3*x**4 + 64*a**4*c**3*x**2 + 32*a**2*c**3) + 3*a**3*x**3/(32*a**6
*c**3*x**4 + 64*a**4*c**3*x**2 + 32*a**2*c**3) + 6*a**2*x**2*atan(a*x)/(32*a**6*c**3*x**4 + 64*a**4*c**3*x**2
+ 32*a**2*c**3) + 5*a*x/(32*a**6*c**3*x**4 + 64*a**4*c**3*x**2 + 32*a**2*c**3) - 5*atan(a*x)/(32*a**6*c**3*x**
4 + 64*a**4*c**3*x**2 + 32*a**2*c**3), Ne(c, 0)), (zoo*(x**2*atan(a*x)/2 - x/(2*a) + atan(a*x)/(2*a**2)), True
))

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Giac [A]  time = 1.18818, size = 92, normalized size = 1.1 \begin{align*} \frac{3 \, \arctan \left (a x\right )}{32 \, a^{2} c^{3}} - \frac{\arctan \left (a x\right )}{4 \,{\left (a^{2} c x^{2} + c\right )}^{2} a^{2} c} + \frac{3 \, a^{2} x^{3} + 5 \, x}{32 \,{\left (a^{2} x^{2} + 1\right )}^{2} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*arctan(a*x)/(a^2*c^3) - 1/4*arctan(a*x)/((a^2*c*x^2 + c)^2*a^2*c) + 1/32*(3*a^2*x^3 + 5*x)/((a^2*x^2 + 1)
^2*a*c^3)