\(\int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx\) [859]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 18 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2 \arctan (a x)^{7/2}}{7 a c} \]

[Out]

2/7*arctan(a*x)^(7/2)/a/c

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {5004} \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2 \arctan (a x)^{7/2}}{7 a c} \]

[In]

Int[ArcTan[a*x]^(5/2)/(c + a^2*c*x^2),x]

[Out]

(2*ArcTan[a*x]^(7/2))/(7*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 {2 \arctan (a x)^{7/2}}{7 a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2 \arctan (a x)^{7/2}}{7 a c} \]

[In]

Integrate[ArcTan[a*x]^(5/2)/(c + a^2*c*x^2),x]

[Out]

(2*ArcTan[a*x]^(7/2))/(7*a*c)

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method result size
default \(\frac {2 \arctan \left (a x \right )^{\frac {7}{2}}}{7 a c}\) \(15\)

[In]

int(arctan(a*x)^(5/2)/(a^2*c*x^2+c),x,method=_RETURNVERBOSE)

[Out]

2/7*arctan(a*x)^(7/2)/a/c

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2 \, \arctan \left (a x\right )^{\frac {7}{2}}}{7 \, a c} \]

[In]

integrate(arctan(a*x)^(5/2)/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

2/7*arctan(a*x)^(7/2)/(a*c)

Sympy [A] (verification not implemented)

Time = 7.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\begin {cases} \frac {2 \operatorname {atan}^{\frac {7}{2}}{\left (a x \right )}}{7 a c} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(atan(a*x)**(5/2)/(a**2*c*x**2+c),x)

[Out]

Piecewise((2*atan(a*x)**(7/2)/(7*a*c), Ne(a, 0)), (0, True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(arctan(a*x)^(5/2)/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2 \, \arctan \left (a x\right )^{\frac {7}{2}}}{7 \, a c} \]

[In]

integrate(arctan(a*x)^(5/2)/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

2/7*arctan(a*x)^(7/2)/(a*c)

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {\arctan (a x)^{5/2}}{c+a^2 c x^2} \, dx=\frac {2\,{\mathrm {atan}\left (a\,x\right )}^{7/2}}{7\,a\,c} \]

[In]

int(atan(a*x)^(5/2)/(c + a^2*c*x^2),x)

[Out]

(2*atan(a*x)^(7/2))/(7*a*c)