\(\int \frac {e^{-\arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [284]

Optimal result

Integrand size = 23, antiderivative size = 38 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {e^{-\arctan (a x)} (1-a x)}{2 a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {5177} \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {(1-a x) e^{-\arctan (a x)}}{2 a c \sqrt {a^2 c x^2+c}} \]

[In]

Int[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(3/2)),x]

[Out]

-1/2*(1 - a*x)/(a*c*E^ArcTan[a*x]*Sqrt[c + a^2*c*x^2])

Rule 5177

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(n + a*x)*(E^(n*ArcTan[a*x])/(
a*c*(n^2 + 1)*Sqrt[c + d*x^2])), x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] &&  !IntegerQ[I*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {e^{-\arctan (a x)} (1-a x)}{2 a c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {e^{-\arctan (a x)} (-1+a x)}{2 a c \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(3/2)),x]

[Out]

(-1 + a*x)/(2*a*c*E^ArcTan[a*x]*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (a x -1\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{2 a \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(39\)

[In]

int(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(a^2*x^2+1)*(a*x-1)/a/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} {\left (a x - 1\right )} e^{\left (-\arctan \left (a x\right )\right )}}{2 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \]

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

1/2*sqrt(a^2*c*x^2 + c)*(a*x - 1)*e^(-arctan(a*x))/(a^3*c^2*x^2 + a*c^2)

Sympy [F]

\[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {e^{- \operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/exp(atan(a*x))/(a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(exp(-atan(a*x))/(c*(a**2*x**2 + 1))**(3/2), x)

Maxima [F]

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

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^(3/2), x)

Giac [F]

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

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (\frac {x}{2\,c}-\frac {1}{2\,a\,c}\right )}{\sqrt {c\,a^2\,x^2+c}} \]

[In]

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

[Out]

(exp(-atan(a*x))*(x/(2*c) - 1/(2*a*c)))/(c + a^2*c*x^2)^(1/2)