\(\int \frac {e^{-2 \arctan (a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\) [299]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {e^{-2 \arctan (a x)} (2-3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {6 e^{-2 \arctan (a x)} (2-a x)}{65 a c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/13*(3*a*x-2)/a/c/exp(2*arctan(a*x))/(a^2*c*x^2+c)^(3/2)-6/65*(-a*x+2)/a/c^2/exp(2*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} \[ \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {6 (2-a x) e^{-2 \arctan (a x)}}{65 a c^2 \sqrt {a^2 c x^2+c}}-\frac {(2-3 a x) e^{-2 \arctan (a x)}}{13 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

-1/13*(2 - 3*a*x)/(a*c*E^(2*ArcTan[a*x])*(c + a^2*c*x^2)^(3/2)) - (6*(2 - a*x))/(65*a*c^2*E^(2*ArcTan[a*x])*Sq
rt[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^{-2 \arctan (a x)} (2-3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {6 \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{13 c} \\ & = -\frac {e^{-2 \arctan (a x)} (2-3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {6 e^{-2 \arctan (a x)} (2-a x)}{65 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 \[ \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{-2 \arctan (a x)} \left (-22+21 a x-12 a^2 x^2+6 a^3 x^3\right )}{65 c^2 \left (a+a^3 x^2\right ) \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

(-22 + 21*a*x - 12*a^2*x^2 + 6*a^3*x^3)/(65*c^2*E^(2*ArcTan[a*x])*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{3} x^{3}-12 a^{2} x^{2}+21 a x -22\right ) {\mathrm e}^{-2 \arctan \left (a x \right )}}{65 a \left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(58\)

[In]

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

[Out]

1/65*(a^2*x^2+1)*(6*a^3*x^3-12*a^2*x^2+21*a*x-22)/a/exp(2*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 \[ \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 21 \, a x - 22\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (-2 \, \arctan \left (a x\right )\right )}}{65 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

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

[Out]

1/65*(6*a^3*x^3 - 12*a^2*x^2 + 21*a*x - 22)*sqrt(a^2*c*x^2 + c)*e^(-2*arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^
2 + a*c^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-2 \arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\mathrm {e}}^{-2\,\mathrm {atan}\left (a\,x\right )}\,\left (\frac {22}{65\,a^3\,c^2}-\frac {6\,x^3}{65\,c^2}-\frac {21\,x}{65\,a^2\,c^2}+\frac {12\,x^2}{65\,a\,c^2}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^2}+x^2\,\sqrt {c\,a^2\,x^2+c}} \]

[In]

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

[Out]

-(exp(-2*atan(a*x))*(22/(65*a^3*c^2) - (6*x^3)/(65*c^2) - (21*x)/(65*a^2*c^2) + (12*x^2)/(65*a*c^2)))/((c + a^
2*c*x^2)^(1/2)/a^2 + x^2*(c + a^2*c*x^2)^(1/2))