3.280 \(\int \frac{e^{-\tan ^{-1}(a x)}}{(c+a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=124 \[ -\frac{(1-6 a x) e^{-\tan ^{-1}(a x)}}{37 a c^4 \left (a^2 x^2+1\right )^3}-\frac{72 (1-2 a x) e^{-\tan ^{-1}(a x)}}{629 a c^4 \left (a^2 x^2+1\right )}-\frac{30 (1-4 a x) e^{-\tan ^{-1}(a x)}}{629 a c^4 \left (a^2 x^2+1\right )^2}-\frac{144 e^{-\tan ^{-1}(a x)}}{629 a c^4} \]

[Out]

-144/(629*a*c^4*E^ArcTan[a*x]) - (1 - 6*a*x)/(37*a*c^4*E^ArcTan[a*x]*(1 + a^2*x^2)^3) - (30*(1 - 4*a*x))/(629*
a*c^4*E^ArcTan[a*x]*(1 + a^2*x^2)^2) - (72*(1 - 2*a*x))/(629*a*c^4*E^ArcTan[a*x]*(1 + a^2*x^2))

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Rubi [A]  time = 0.119277, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {5070, 5071} \[ -\frac{(1-6 a x) e^{-\tan ^{-1}(a x)}}{37 a c^4 \left (a^2 x^2+1\right )^3}-\frac{72 (1-2 a x) e^{-\tan ^{-1}(a x)}}{629 a c^4 \left (a^2 x^2+1\right )}-\frac{30 (1-4 a x) e^{-\tan ^{-1}(a x)}}{629 a c^4 \left (a^2 x^2+1\right )^2}-\frac{144 e^{-\tan ^{-1}(a x)}}{629 a c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^4),x]

[Out]

-144/(629*a*c^4*E^ArcTan[a*x]) - (1 - 6*a*x)/(37*a*c^4*E^ArcTan[a*x]*(1 + a^2*x^2)^3) - (30*(1 - 4*a*x))/(629*
a*c^4*E^ArcTan[a*x]*(1 + a^2*x^2)^2) - (72*(1 - 2*a*x))/(629*a*c^4*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 )^4} \, dx &=-\frac{e^{-\tan ^{-1}(a x)} (1-6 a x)}{37 a c^4 \left (1+a^2 x^2\right )^3}+\frac{30 \int \frac{e^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^3} \, dx}{37 c}\\ &=-\frac{e^{-\tan ^{-1}(a x)} (1-6 a x)}{37 a c^4 \left (1+a^2 x^2\right )^3}-\frac{30 e^{-\tan ^{-1}(a x)} (1-4 a x)}{629 a c^4 \left (1+a^2 x^2\right )^2}+\frac{360 \int \frac{e^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{629 c^2}\\ &=-\frac{e^{-\tan ^{-1}(a x)} (1-6 a x)}{37 a c^4 \left (1+a^2 x^2\right )^3}-\frac{30 e^{-\tan ^{-1}(a x)} (1-4 a x)}{629 a c^4 \left (1+a^2 x^2\right )^2}-\frac{72 e^{-\tan ^{-1}(a x)} (1-2 a x)}{629 a c^4 \left (1+a^2 x^2\right )}+\frac{144 \int \frac{e^{-\tan ^{-1}(a x)}}{c+a^2 c x^2} \, dx}{629 c^3}\\ &=-\frac{144 e^{-\tan ^{-1}(a x)}}{629 a c^4}-\frac{e^{-\tan ^{-1}(a x)} (1-6 a x)}{37 a c^4 \left (1+a^2 x^2\right )^3}-\frac{30 e^{-\tan ^{-1}(a x)} (1-4 a x)}{629 a c^4 \left (1+a^2 x^2\right )^2}-\frac{72 e^{-\tan ^{-1}(a x)} (1-2 a x)}{629 a c^4 \left (1+a^2 x^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.346536, size = 127, normalized size = 1.02 \[ \frac{17 c (6 a x-1) e^{-\tan ^{-1}(a x)}-6 \left (a^2 c x^2+c\right ) \left (5 (1-4 a x) e^{-\tan ^{-1}(a x)}+12 (1-i a x)^{-\frac{i}{2}} (1+i a x)^{\frac{i}{2}} (a x-i) (a x+i) \left (2 a^2 x^2-2 a x+3\right )\right )}{629 a c^2 \left (a^2 c x^2+c\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^4),x]

[Out]

((17*c*(-1 + 6*a*x))/E^ArcTan[a*x] - 6*(c + a^2*c*x^2)*((5*(1 - 4*a*x))/E^ArcTan[a*x] + (12*(1 + I*a*x)^(I/2)*
(-I + a*x)*(I + a*x)*(3 - 2*a*x + 2*a^2*x^2))/(1 - I*a*x)^(I/2)))/(629*a*c^2*(c + a^2*c*x^2)^3)

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Maple [A]  time = 0.039, size = 73, normalized size = 0.6 \begin{align*} -{\frac{144\,{a}^{6}{x}^{6}-144\,{a}^{5}{x}^{5}+504\,{a}^{4}{x}^{4}-408\,{a}^{3}{x}^{3}+606\,{a}^{2}{x}^{2}-366\,ax+263}{629\,{c}^{4} \left ({a}^{2}{x}^{2}+1 \right ) ^{3}{{\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)^4,x)

[Out]

-1/629*(144*a^6*x^6-144*a^5*x^5+504*a^4*x^4-408*a^3*x^3+606*a^2*x^2-366*a*x+263)/(a^2*x^2+1)^3/c^4/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 )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^4, x)

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Fricas [A]  time = 1.98611, size = 223, normalized size = 1.8 \begin{align*} -\frac{{\left (144 \, a^{6} x^{6} - 144 \, a^{5} x^{5} + 504 \, a^{4} x^{4} - 408 \, a^{3} x^{3} + 606 \, a^{2} x^{2} - 366 \, a x + 263\right )} e^{\left (-\arctan \left (a x\right )\right )}}{629 \,{\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\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)^4,x, algorithm="fricas")

[Out]

-1/629*(144*a^6*x^6 - 144*a^5*x^5 + 504*a^4*x^4 - 408*a^3*x^3 + 606*a^2*x^2 - 366*a*x + 263)*e^(-arctan(a*x))/
(a^7*c^4*x^6 + 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 + a*c^4)

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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)**4,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 )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^4, x)