3.3.98 \(\int \frac {e^{-2 \text {ArcTan}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [298]

Optimal. Leaf size=38 \[ -\frac {e^{-2 \text {ArcTan}(a x)} (2-a x)}{5 a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {5177} \begin {gather*} -\frac {(2-a x) e^{-2 \text {ArcTan}(a x)}}{5 a c \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(2 - a*x)/(a*c*E^(2*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*} \int \frac {e^{-2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\frac {e^{-2 \tan ^{-1}(a x)} (2-a x)}{5 a c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 37, normalized size = 0.97 \begin {gather*} \frac {e^{-2 \text {ArcTan}(a x)} (-2+a x)}{5 a c \sqrt {c+a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 41, normalized size = 1.08

method	result	size
gosper	\(\frac {\left (a^{2} x^{2}+1\right ) \left (a x -2\right ) {\mathrm e}^{-2 \arctan \left (a x \right )}}{5 a \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.47, size = 44, normalized size = 1.16 \begin {gather*} \frac {\sqrt {a^{2} c x^{2} + c} {\left (a x - 2\right )} e^{\left (-2 \, \arctan \left (a x\right )\right )}}{5 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{- 2 \operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.18, size = 35, normalized size = 0.92 \begin {gather*} \frac {{\mathrm {e}}^{-2\,\mathrm {atan}\left (a\,x\right )}\,\left (\frac {x}{5\,c}-\frac {2}{5\,a\,c}\right )}{\sqrt {c\,a^2\,x^2+c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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