\(\int \frac {e^{-\arctan (a x)}}{(c+a^2 c x^2)^3} \, dx\) [279]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 89 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {24 e^{-\arctan (a x)}}{85 a c^3}-\frac {e^{-\arctan (a x)} (1-4 a x)}{17 a c^3 \left (1+a^2 x^2\right )^2}-\frac {12 e^{-\arctan (a x)} (1-2 a x)}{85 a c^3 \left (1+a^2 x^2\right )} \]

[Out]

-24/85/a/c^3/exp(arctan(a*x))+1/17*(4*a*x-1)/a/c^3/exp(arctan(a*x))/(a^2*x^2+1)^2-12/85*(-2*a*x+1)/a/c^3/exp(a
rctan(a*x))/(a^2*x^2+1)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5178, 5179} \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {(1-4 a x) e^{-\arctan (a x)}}{17 a c^3 \left (a^2 x^2+1\right )^2}-\frac {12 (1-2 a x) e^{-\arctan (a x)}}{85 a c^3 \left (a^2 x^2+1\right )}-\frac {24 e^{-\arctan (a x)}}{85 a c^3} \]

[In]

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

[Out]

-24/(85*a*c^3*E^ArcTan[a*x]) - (1 - 4*a*x)/(17*a*c^3*E^ArcTan[a*x]*(1 + a^2*x^2)^2) - (12*(1 - 2*a*x))/(85*a*c
^3*E^ArcTan[a*x]*(1 + a^2*x^2))

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]

Rule 5179

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*} \text {integral}& = -\frac {e^{-\arctan (a x)} (1-4 a x)}{17 a c^3 \left (1+a^2 x^2\right )^2}+\frac {12 \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{17 c} \\ & = -\frac {e^{-\arctan (a x)} (1-4 a x)}{17 a c^3 \left (1+a^2 x^2\right )^2}-\frac {12 e^{-\arctan (a x)} (1-2 a x)}{85 a c^3 \left (1+a^2 x^2\right )}+\frac {24 \int \frac {e^{-\arctan (a x)}}{c+a^2 c x^2} \, dx}{85 c^2} \\ & = -\frac {24 e^{-\arctan (a x)}}{85 a c^3}-\frac {e^{-\arctan (a x)} (1-4 a x)}{17 a c^3 \left (1+a^2 x^2\right )^2}-\frac {12 e^{-\arctan (a x)} (1-2 a x)}{85 a c^3 \left (1+a^2 x^2\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5 e^{-\arctan (a x)} (-1+4 a x)-12 (1-i a x)^{-\frac {i}{2}} (1+i a x)^{\frac {i}{2}} \left (1+a^2 x^2\right ) \left (3-2 a x+2 a^2 x^2\right )}{85 a c^3 \left (1+a^2 x^2\right )^2} \]

[In]

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

[Out]

((5*(-1 + 4*a*x))/E^ArcTan[a*x] - (12*(1 + I*a*x)^(I/2)*(1 + a^2*x^2)*(3 - 2*a*x + 2*a^2*x^2))/(1 - I*a*x)^(I/
2))/(85*a*c^3*(1 + a^2*x^2)^2)

Maple [A] (verified)

Time = 12.62 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.64

method result size
gosper \(-\frac {\left (24 a^{4} x^{4}-24 a^{3} x^{3}+60 a^{2} x^{2}-44 a x +41\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{85 \left (a^{2} x^{2}+1\right )^{2} c^{3} a}\) \(57\)
parallelrisch \(\frac {\left (-24 a^{4} x^{4}+24 a^{3} x^{3}-60 a^{2} x^{2}+44 a x -41\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{85 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}\) \(57\)

[In]

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

[Out]

-1/85*(24*a^4*x^4-24*a^3*x^3+60*a^2*x^2-44*a*x+41)/(a^2*x^2+1)^2/c^3/exp(arctan(a*x))/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {{\left (24 \, a^{4} x^{4} - 24 \, a^{3} x^{3} + 60 \, a^{2} x^{2} - 44 \, a x + 41\right )} e^{\left (-\arctan \left (a x\right )\right )}}{85 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

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

[Out]

-1/85*(24*a^4*x^4 - 24*a^3*x^3 + 60*a^2*x^2 - 44*a*x + 41)*e^(-arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c
^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (76) = 152\).

Time = 87.60 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.27 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\begin {cases} - \frac {24 a^{4} x^{4}}{85 a^{5} c^{3} x^{4} e^{\operatorname {atan}{\left (a x \right )}} + 170 a^{3} c^{3} x^{2} e^{\operatorname {atan}{\left (a x \right )}} + 85 a c^{3} e^{\operatorname {atan}{\left (a x \right )}}} + \frac {24 a^{3} x^{3}}{85 a^{5} c^{3} x^{4} e^{\operatorname {atan}{\left (a x \right )}} + 170 a^{3} c^{3} x^{2} e^{\operatorname {atan}{\left (a x \right )}} + 85 a c^{3} e^{\operatorname {atan}{\left (a x \right )}}} - \frac {60 a^{2} x^{2}}{85 a^{5} c^{3} x^{4} e^{\operatorname {atan}{\left (a x \right )}} + 170 a^{3} c^{3} x^{2} e^{\operatorname {atan}{\left (a x \right )}} + 85 a c^{3} e^{\operatorname {atan}{\left (a x \right )}}} + \frac {44 a x}{85 a^{5} c^{3} x^{4} e^{\operatorname {atan}{\left (a x \right )}} + 170 a^{3} c^{3} x^{2} e^{\operatorname {atan}{\left (a x \right )}} + 85 a c^{3} e^{\operatorname {atan}{\left (a x \right )}}} - \frac {41}{85 a^{5} c^{3} x^{4} e^{\operatorname {atan}{\left (a x \right )}} + 170 a^{3} c^{3} x^{2} e^{\operatorname {atan}{\left (a x \right )}} + 85 a c^{3} e^{\operatorname {atan}{\left (a x \right )}}} & \text {for}\: a \neq 0 \\\frac {x}{c^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-24*a**4*x**4/(85*a**5*c**3*x**4*exp(atan(a*x)) + 170*a**3*c**3*x**2*exp(atan(a*x)) + 85*a*c**3*exp
(atan(a*x))) + 24*a**3*x**3/(85*a**5*c**3*x**4*exp(atan(a*x)) + 170*a**3*c**3*x**2*exp(atan(a*x)) + 85*a*c**3*
exp(atan(a*x))) - 60*a**2*x**2/(85*a**5*c**3*x**4*exp(atan(a*x)) + 170*a**3*c**3*x**2*exp(atan(a*x)) + 85*a*c*
*3*exp(atan(a*x))) + 44*a*x/(85*a**5*c**3*x**4*exp(atan(a*x)) + 170*a**3*c**3*x**2*exp(atan(a*x)) + 85*a*c**3*
exp(atan(a*x))) - 41/(85*a**5*c**3*x**4*exp(atan(a*x)) + 170*a**3*c**3*x**2*exp(atan(a*x)) + 85*a*c**3*exp(ata
n(a*x))), Ne(a, 0)), (x/c**3, True))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {12\,{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (2\,a\,x-1\right )}{85\,a\,c^3\,\left (a^2\,x^2+1\right )}-\frac {24\,{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}}{85\,a\,c^3}+\frac {{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (4\,a\,x-1\right )}{17\,a\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

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

[Out]

(12*exp(-atan(a*x))*(2*a*x - 1))/(85*a*c^3*(a^2*x^2 + 1)) - (24*exp(-atan(a*x)))/(85*a*c^3) + (exp(-atan(a*x))
*(4*a*x - 1))/(17*a*c^3*(a^2*x^2 + 1)^2)