3.286 \(\int \frac{e^{-\tan ^{-1}(a x)}}{(c+a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=115 \[ -\frac{3 (1-a x) e^{-\tan ^{-1}(a x)}}{13 a c^3 \sqrt{a^2 c x^2+c}}-\frac{(1-3 a x) e^{-\tan ^{-1}(a x)}}{13 a c^2 \left (a^2 c x^2+c\right )^{3/2}}-\frac{(1-5 a x) e^{-\tan ^{-1}(a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}} \]

[Out]

-(1 - 5*a*x)/(26*a*c*E^ArcTan[a*x]*(c + a^2*c*x^2)^(5/2)) - (1 - 3*a*x)/(13*a*c^2*E^ArcTan[a*x]*(c + a^2*c*x^2
)^(3/2)) - (3*(1 - a*x))/(13*a*c^3*E^ArcTan[a*x]*Sqrt[c + a^2*c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.123632, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {5070, 5069} \[ -\frac{3 (1-a x) e^{-\tan ^{-1}(a x)}}{13 a c^3 \sqrt{a^2 c x^2+c}}-\frac{(1-3 a x) e^{-\tan ^{-1}(a x)}}{13 a c^2 \left (a^2 c x^2+c\right )^{3/2}}-\frac{(1-5 a x) e^{-\tan ^{-1}(a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(7/2)),x]

[Out]

-(1 - 5*a*x)/(26*a*c*E^ArcTan[a*x]*(c + a^2*c*x^2)^(5/2)) - (1 - 3*a*x)/(13*a*c^2*E^ArcTan[a*x]*(c + a^2*c*x^2
)^(3/2)) - (3*(1 - a*x))/(13*a*c^3*E^ArcTan[a*x]*Sqrt[c + a^2*c*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 5069

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^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx &=-\frac{e^{-\tan ^{-1}(a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac{10 \int \frac{e^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{13 c}\\ &=-\frac{e^{-\tan ^{-1}(a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}-\frac{e^{-\tan ^{-1}(a x)} (1-3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac{6 \int \frac{e^{-\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{13 c^2}\\ &=-\frac{e^{-\tan ^{-1}(a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}-\frac{e^{-\tan ^{-1}(a x)} (1-3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}-\frac{3 e^{-\tan ^{-1}(a x)} (1-a x)}{13 a c^3 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.045607, size = 81, normalized size = 0.7 \[ \frac{\left (6 a^5 x^5-6 a^4 x^4+18 a^3 x^3-14 a^2 x^2+17 a x-9\right ) e^{-\tan ^{-1}(a x)}}{26 a c^3 \left (a^2 x^2+1\right )^2 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTan[a*x]*(c + a^2*c*x^2)^(7/2)),x]

[Out]

(-9 + 17*a*x - 14*a^2*x^2 + 18*a^3*x^3 - 6*a^4*x^4 + 6*a^5*x^5)/(26*a*c^3*E^ArcTan[a*x]*(1 + a^2*x^2)^2*Sqrt[c
+ a^2*c*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 72, normalized size = 0.6 \begin{align*}{\frac{ \left ({a}^{2}{x}^{2}+1 \right ) \left ( 6\,{a}^{5}{x}^{5}-6\,{a}^{4}{x}^{4}+18\,{a}^{3}{x}^{3}-14\,{a}^{2}{x}^{2}+17\,ax-9 \right ) }{26\,a{{\rm e}^{\arctan \left ( ax \right ) }}} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{7}{2}}}} \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)^(7/2),x)

[Out]

1/26*(a^2*x^2+1)*(6*a^5*x^5-6*a^4*x^4+18*a^3*x^3-14*a^2*x^2+17*a*x-9)/a/exp(arctan(a*x))/(a^2*c*x^2+c)^(7/2)

________________________________________________________________________________________

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 )}^{\frac{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^(7/2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.9506, size = 216, normalized size = 1.88 \begin{align*} \frac{{\left (6 \, a^{5} x^{5} - 6 \, a^{4} x^{4} + 18 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 17 \, a x - 9\right )} \sqrt{a^{2} c x^{2} + c} e^{\left (-\arctan \left (a x\right )\right )}}{26 \,{\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)^(7/2),x, algorithm="fricas")

[Out]

1/26*(6*a^5*x^5 - 6*a^4*x^4 + 18*a^3*x^3 - 14*a^2*x^2 + 17*a*x - 9)*sqrt(a^2*c*x^2 + c)*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)

________________________________________________________________________________________

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)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{\frac{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate(e^(-arctan(a*x))/(a^2*c*x^2 + c)^(7/2), x)