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

Optimal. Leaf size=86 \[ \frac{x^3}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x}{32 a^3 c^3 \left (a^2 x^2+1\right )}+\frac{x^4 \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{3 \tan ^{-1}(a x)}{32 a^4 c^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0655447, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4944, 288, 205} \[ \frac{x^3}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x}{32 a^3 c^3 \left (a^2 x^2+1\right )}+\frac{x^4 \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{3 \tan ^{-1}(a x)}{32 a^4 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 0.141649, size = 58, normalized size = 0.67 \[ \frac{a x \left (5 a^2 x^2+3\right )+\left (5 a^4 x^4-6 a^2 x^2-3\right ) \tan ^{-1}(a x)}{32 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 102, normalized size = 1.2 \begin{align*}{\frac{\arctan \left ( ax \right ) }{4\,{a}^{4}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{\arctan \left ( ax \right ) }{2\,{a}^{4}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,{x}^{3}}{32\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,x}{32\,{a}^{3}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{5\,\arctan \left ( ax \right ) }{32\,{a}^{4}{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.54169, size = 146, normalized size = 1.7 \begin{align*} \frac{1}{32} \, a{\left (\frac{5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac{5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} - \frac{{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )}{4 \,{\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*a*((5*a^2*x^3 + 3*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) + 5*arctan(a*x)/(a^5*c^3)) - 1/4*(2*a^2*x^2
+ 1)*arctan(a*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

________________________________________________________________________________________

Fricas [A]  time = 1.64915, size = 146, normalized size = 1.7 \begin{align*} \frac{5 \, a^{3} x^{3} + 3 \, a x +{\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )}{32 \,{\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(5*a^3*x^3 + 3*a*x + (5*a^4*x^4 - 6*a^2*x^2 - 3)*arctan(a*x))/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

________________________________________________________________________________________

Sympy [A]  time = 3.95857, size = 243, normalized size = 2.83 \begin{align*} \begin{cases} \frac{5 a^{4} x^{4} \operatorname{atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} + \frac{5 a^{3} x^{3}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} - \frac{6 a^{2} x^{2} \operatorname{atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} + \frac{3 a x}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} - \frac{3 \operatorname{atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} & \text{for}\: c \neq 0 \\\tilde{\infty } \left (\frac{x^{4} \operatorname{atan}{\left (a x \right )}}{4} - \frac{x^{3}}{12 a} + \frac{x}{4 a^{3}} - \frac{\operatorname{atan}{\left (a x \right )}}{4 a^{4}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*a**4*x**4*atan(a*x)/(32*a**8*c**3*x**4 + 64*a**6*c**3*x**2 + 32*a**4*c**3) + 5*a**3*x**3/(32*a**8
*c**3*x**4 + 64*a**6*c**3*x**2 + 32*a**4*c**3) - 6*a**2*x**2*atan(a*x)/(32*a**8*c**3*x**4 + 64*a**6*c**3*x**2
+ 32*a**4*c**3) + 3*a*x/(32*a**8*c**3*x**4 + 64*a**6*c**3*x**2 + 32*a**4*c**3) - 3*atan(a*x)/(32*a**8*c**3*x**
4 + 64*a**6*c**3*x**2 + 32*a**4*c**3), Ne(c, 0)), (zoo*(x**4*atan(a*x)/4 - x**3/(12*a) + x/(4*a**3) - atan(a*x
)/(4*a**4)), True))

________________________________________________________________________________________

Giac [A]  time = 1.15508, size = 104, normalized size = 1.21 \begin{align*} \frac{5 \, \arctan \left (a x\right )}{32 \, a^{4} c^{3}} + \frac{5 \, a^{2} x^{3} + 3 \, x}{32 \,{\left (a^{2} x^{2} + 1\right )}^{2} a^{3} c^{3}} - \frac{{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )}{4 \,{\left (a^{2} x^{2} + 1\right )}^{2} a^{4} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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