\(\int \frac {e^{-\arctan (a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\) [285]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {e^{-\arctan (a x)} (1-3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 e^{-\arctan (a x)} (1-a x)}{10 a c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/10*(3*a*x-1)/a/c/exp(arctan(a*x))/(a^2*c*x^2+c)^(3/2)-3/10*(-a*x+1)/a/c^2/exp(arctan(a*x))/(a^2*c*x^2+c)^(1/
2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {5178, 5177} \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 (1-a x) e^{-\arctan (a x)}}{10 a c^2 \sqrt {a^2 c x^2+c}}-\frac {(1-3 a x) e^{-\arctan (a x)}}{10 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

Int[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(5/2)),x]

[Out]

-1/10*(1 - 3*a*x)/(a*c*E^ArcTan[a*x]*(c + a^2*c*x^2)^(3/2)) - (3*(1 - a*x))/(10*a*c^2*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]

Rule 5178

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n - 2*a*(p + 1)*x)*(c + d*x^2)
^(p + 1)*(E^(n*ArcTan[a*x])/(a*c*(n^2 + 4*(p + 1)^2))), x] + Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 + 4*(p + 1)^2))
), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&&  !IntegerQ[I*n] && NeQ[n^2 + 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {e^{-\arctan (a x)} (1-3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {3 \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{5 c} \\ & = -\frac {e^{-\arctan (a x)} (1-3 a x)}{10 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 e^{-\arctan (a x)} (1-a x)}{10 a c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{-\arctan (a x)} \left (-4+6 a x-3 a^2 x^2+3 a^3 x^3\right )}{10 c^2 \left (a+a^3 x^2\right ) \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(5/2)),x]

[Out]

(-4 + 6*a*x - 3*a^2*x^2 + 3*a^3*x^3)/(10*c^2*E^ArcTan[a*x]*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (3 a^{3} x^{3}-3 a^{2} x^{2}+6 a x -4\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{10 a \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(56\)

[In]

int(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/10*(a^2*x^2+1)*(3*a^3*x^3-3*a^2*x^2+6*a*x-4)/a/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, a^{3} x^{3} - 3 \, a^{2} x^{2} + 6 \, a x - 4\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (-\arctan \left (a x\right )\right )}}{10 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/10*(3*a^3*x^3 - 3*a^2*x^2 + 6*a*x - 4)*sqrt(a^2*c*x^2 + c)*e^(-arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a
*c^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/exp(atan(a*x))/(a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)

Giac [F]

\[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (\frac {2}{5\,a^3\,c^2}-\frac {3\,x^3}{10\,c^2}-\frac {3\,x}{5\,a^2\,c^2}+\frac {3\,x^2}{10\,a\,c^2}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^2}+x^2\,\sqrt {c\,a^2\,x^2+c}} \]

[In]

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

[Out]

-(exp(-atan(a*x))*(2/(5*a^3*c^2) - (3*x^3)/(10*c^2) - (3*x)/(5*a^2*c^2) + (3*x^2)/(10*a*c^2)))/((c + a^2*c*x^2
)^(1/2)/a^2 + x^2*(c + a^2*c*x^2)^(1/2))