\(\int \frac {x \arctan (a x)}{(c+a^2 c x^2)^3} \, dx\) [194]

Optimal result

Integrand size = 18, antiderivative size = 84 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {x}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x}{32 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)}{32 a^2 c^3}-\frac {\arctan (a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2} \]

[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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5050, 205, 211} \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {\arctan (a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \arctan (a x)}{32 a^2 c^3}+\frac {3 x}{32 a c^3 \left (a^2 x^2+1\right )}+\frac {x}{16 a c^3 \left (a^2 x^2+1\right )^2} \]

[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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 5050

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {a x \left (5+3 a^2 x^2\right )+\left (-5+6 a^2 x^2+3 a^4 x^4\right ) \arctan (a x)}{32 c^3 \left (a+a^3 x^2\right )^2} \]

[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)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {3 \arctan \left (a x \right ) a^{4} x^{4}+3 a^{3} x^{3}+6 a^{2} \arctan \left (a x \right ) x^{2}+5 a x -5 \arctan \left (a x \right )}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{2}}\)	\(63\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {a x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )}{8}}{4 c^{3}}}{a^{2}}\)	\(68\)
default	\(\frac {-\frac {\arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {a x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )}{8}}{4 c^{3}}}{a^{2}}\)	\(68\)
parts	\(-\frac {\arctan \left (a x \right )}{4 a^{2} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )}{8 a}}{4 a \,c^{3}}\)	\(71\)
risch	\(\frac {i \ln \left (i a x +1\right )}{8 a^{2} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {i \left (8 \ln \left (-i a x +1\right )+3 \ln \left (-a x +i\right ) a^{4} x^{4}+6 \ln \left (-a x +i\right ) a^{2} x^{2}+3 \ln \left (-a x +i\right )-3 \ln \left (a x +i\right ) a^{4} x^{4}-6 \ln \left (a x +i\right ) a^{2} x^{2}-3 \ln \left (a x +i\right )+6 i a^{3} x^{3}+10 i a x \right )}{64 a^{2} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}\)	\(161\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{3} x^{3} + 5 \, a x + {\left (3 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 5\right )} \arctan \left (a x\right )}{32 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (75) = 150\).

Time = 0.59 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.49 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\begin {cases} \frac {3 a^{4} x^{4} \operatorname {atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac {3 a^{3} x^{3}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac {6 a^{2} x^{2} \operatorname {atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} + \frac {5 a x}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} - \frac {5 \operatorname {atan}{\left (a x \right )}}{32 a^{6} c^{3} x^{4} + 64 a^{4} c^{3} x^{2} + 32 a^{2} c^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[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(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}} + \frac {3 \, \arctan \left (a x\right )}{a c^{2}}}{32 \, a c} - \frac {\arctan \left (a x\right )}{4 \, {\left (a^{2} c x^{2} + c\right )}^{2} a^{2} c} \]

[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)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.23 \[ \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\frac {5\,x}{32\,a}+\frac {a\,x^3}{4}-\frac {\mathrm {atan}\left (a\,x\right )}{4\,a^2}-\frac {x^2\,\mathrm {atan}\left (a\,x\right )}{4}+\frac {3\,a^3\,x^5}{32}}{a^6\,c^3\,x^6+3\,a^4\,c^3\,x^4+3\,a^2\,c^3\,x^2+c^3}+\frac {3\,\mathrm {atan}\left (\frac {a^2\,x}{\sqrt {a^2}}\right )}{32\,a\,c^3\,\sqrt {a^2}} \]

[In]

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

[Out]

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