\(\int \frac {x^3 \arctan (a x)}{(c+a^2 c x^2)^3} \, dx\) [192]

Optimal result

Integrand size = 20, antiderivative size = 86 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {x^3}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x}{32 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)}{32 a^4 c^3}+\frac {x^4 \arctan (a x)}{4 c^3 \left (1+a^2 x^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5064, 294, 211} \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {3 \arctan (a x)}{32 a^4 c^3}+\frac {x^4 \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {x^3}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x}{32 a^3 c^3 \left (a^2 x^2+1\right )} \]

[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 211

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

Rule 294

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 5064

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^4 \arctan (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 \arctan (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 \arctan (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 \arctan (a x)}{32 a^4 c^3}+\frac {x^4 \arctan (a x)}{4 c^3 \left (1+a^2 x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {a x \left (3+5 a^2 x^2\right )+\left (-3-6 a^2 x^2+5 a^4 x^4\right ) \arctan (a x)}{32 a^4 c^3 \left (1+a^2 x^2\right )^2} \]

[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] (verified)

Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {5 \arctan \left (a x \right ) a^{4} x^{4}+5 a^{3} x^{3}-6 a^{2} \arctan \left (a x \right ) x^{2}+3 a x -3 \arctan \left (a x \right )}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{4}}\)	\(63\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )}{2 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {\frac {5}{8} a^{3} x^{3}+\frac {3}{8} a x}{\left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )}{8}}{4 c^{3}}}{a^{4}}\)	\(84\)
default	\(\frac {-\frac {\arctan \left (a x \right )}{2 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {\frac {5}{8} a^{3} x^{3}+\frac {3}{8} a x}{\left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )}{8}}{4 c^{3}}}{a^{4}}\)	\(84\)
parts	\(-\frac {\arctan \left (a x \right )}{2 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{4 c^{3} a^{4} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {\frac {5}{8} a^{2} x^{3}+\frac {3}{8} x}{\left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )}{8 a}}{4 c^{3} a^{3}}\)	\(91\)
risch	\(\frac {i \left (2 a^{2} x^{2}+1\right ) \ln \left (i a x +1\right )}{8 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {i \left (16 a^{2} x^{2} \ln \left (-i a x +1\right )+8 \ln \left (-i a x +1\right )-5 \ln \left (a x +i\right ) a^{4} x^{4}-10 \ln \left (a x +i\right ) a^{2} x^{2}-5 \ln \left (a x +i\right )+5 \ln \left (-a x +i\right ) a^{4} x^{4}+10 \ln \left (-a x +i\right ) a^{2} x^{2}+5 \ln \left (-a x +i\right )+10 i a^{3} x^{3}+6 i a x \right )}{64 a^{4} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}\)	\(187\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5 \, a^{3} x^{3} + 3 \, a x + {\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )}{32 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (78) = 156\).

Time = 0.60 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.43 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\begin {cases} \frac {5 a^{4} x^{4} \operatorname {atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} + \frac {5 a^{3} x^{3}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} - \frac {6 a^{2} x^{2} \operatorname {atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} + \frac {3 a x}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} - \frac {3 \operatorname {atan}{\left (a x \right )}}{32 a^{8} c^{3} x^{4} + 64 a^{6} c^{3} x^{2} + 32 a^{4} c^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[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(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{32} \, a {\left (\frac {5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac {5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} - \frac {{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[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)

Giac [F]

\[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3\,a\,x-3\,\mathrm {atan}\left (a\,x\right )+5\,a^3\,x^3-6\,a^2\,x^2\,\mathrm {atan}\left (a\,x\right )+5\,a^4\,x^4\,\mathrm {atan}\left (a\,x\right )}{32\,a^4\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

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

[Out]

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