3.278 \(\int \frac{e^{-\tan ^{-1}(a x)}}{(c+a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=54 \[ -\frac{(1-2 a x) e^{-\tan ^{-1}(a x)}}{5 a c^2 \left (a^2 x^2+1\right )}-\frac{2 e^{-\tan ^{-1}(a x)}}{5 a c^2} \]

[Out]

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

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Rubi [A]  time = 0.0539602, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {5070, 5071} \[ -\frac{(1-2 a x) e^{-\tan ^{-1}(a x)}}{5 a c^2 \left (a^2 x^2+1\right )}-\frac{2 e^{-\tan ^{-1}(a x)}}{5 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5070

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]

Rule 5071

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcTan[a*x])/(a*c*n), x] /; Fre
eQ[{a, c, d, n}, x] && EqQ[d, a^2*c]

Rubi steps

\begin{align*} \int \frac{e^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac{e^{-\tan ^{-1}(a x)} (1-2 a x)}{5 a c^2 \left (1+a^2 x^2\right )}+\frac{2 \int \frac{e^{-\tan ^{-1}(a x)}}{c+a^2 c x^2} \, dx}{5 c}\\ &=-\frac{2 e^{-\tan ^{-1}(a x)}}{5 a c^2}-\frac{e^{-\tan ^{-1}(a x)} (1-2 a x)}{5 a c^2 \left (1+a^2 x^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.0222546, size = 60, normalized size = 1.11 \[ -\frac{(1-i a x)^{-\frac{i}{2}} (1+i a x)^{\frac{i}{2}} \left (2 a^2 x^2-2 a x+3\right )}{5 c^2 \left (a^3 x^2+a\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((1 + I*a*x)^(I/2)*(3 - 2*a*x + 2*a^2*x^2))/(5*c^2*(1 - I*a*x)^(I/2)*(a + a^3*x^2))

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Maple [A]  time = 0.036, size = 41, normalized size = 0.8 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}-2\,ax+3}{ \left ( 5\,{a}^{2}{x}^{2}+5 \right ){c}^{2}{{\rm e}^{\arctan \left ( ax \right ) }}a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.79043, size = 93, normalized size = 1.72 \begin{align*} -\frac{{\left (2 \, a^{2} x^{2} - 2 \, a x + 3\right )} e^{\left (-\arctan \left (a x\right )\right )}}{5 \,{\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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