\(\int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx\) [177]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {\arctan (a x)^2}{2 a c} \]

[Out]

1/2*arctan(a*x)^2/a/c

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {5004} \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {\arctan (a x)^2}{2 a c} \]

[In]

Int[ArcTan[a*x]/(c + a^2*c*x^2),x]

[Out]

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

Rule 5004

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

Rubi steps \begin{align*} \text {integral}& = \frac {\arctan (a x)^2}{2 a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {\arctan (a x)^2}{2 a c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\arctan \left (a x \right )^{2}}{2 a c}\)	\(15\)
default	\(\frac {\arctan \left (a x \right )^{2}}{2 a c}\)	\(15\)
parallelrisch	\(\frac {\arctan \left (a x \right )^{2}}{2 a c}\)	\(15\)
parts	\(\frac {\arctan \left (a x \right )^{2}}{2 a c}\)	\(15\)
risch	\(-\frac {\ln \left (i a x +1\right )^{2}}{8 c a}+\frac {\ln \left (-i a x +1\right ) \ln \left (i a x +1\right )}{4 c a}-\frac {\ln \left (-i a x +1\right )^{2}}{8 c a}\)	\(62\)

[In]

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

[Out]

1/2*arctan(a*x)^2/a/c

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {\arctan \left (a x\right )^{2}}{2 \, a c} \]

[In]

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

[Out]

1/2*arctan(a*x)^2/(a*c)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (10) = 20\).

Time = 0.70 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.25 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\begin {cases} 0 & \text {for}\: a = 0 \\\tilde {\infty } \left (\begin {cases} 0 & \text {for}\: a = 0 \\\frac {a x \operatorname {atan}{\left (a x \right )} - \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{2}}{a} & \text {otherwise} \end {cases}\right ) & \text {for}\: c = 0 \\\frac {\operatorname {atan}^{2}{\left (a x \right )}}{2 a c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((0, Eq(a, 0)), (zoo*Piecewise((0, Eq(a, 0)), ((a*x*atan(a*x) - log(a**2*x**2 + 1)/2)/a, True)), Eq(c
, 0)), (atan(a*x)**2/(2*a*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {\arctan \left (a x\right )^{2}}{2 \, a c} \]

[In]

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

[Out]

1/2*arctan(a*x)^2/(a*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=-\frac {2 \, \pi \arctan \left (a x\right ) \left \lfloor \frac {\arctan \left (a x\right )}{\pi } + \frac {1}{2} \right \rfloor - \arctan \left (a x\right )^{2}}{2 \, a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx=\frac {{\mathrm {atan}\left (a\,x\right )}^2}{2\,a\,c} \]

[In]

int(atan(a*x)/(c + a^2*c*x^2),x)

[Out]

atan(a*x)^2/(2*a*c)