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]

1/26*(5*a*x-1)/a/c/exp(arctan(a*x))/(a^2*c*x^2+c)^(5/2)+1/13*(3*a*x-1)/a/c^2/exp(arctan(a*x))/(a^2*c*x^2+c)^(3
/2)-3/13*(-a*x+1)/a/c^3/exp(arctan(a*x))/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.05, size = 81, normalized size = 0.70 \[ \frac {\left (6 a^5 x^5-6 a^4 x^4+18 a^3 x^3-14 a^2 x^2+17 a x-9\right ) e^{-\tan ^{-1}(a x)}}{26 a c^3 \left (a^2 x^2+1\right )^2 \sqrt {a^2 c x^2+c}} \]

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])

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fricas [A]  time = 0.42, size = 99, normalized size = 0.86 \[ \frac {{\left (6 \, a^{5} x^{5} - 6 \, a^{4} x^{4} + 18 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 17 \, a x - 9\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (-\arctan \left (a x\right )\right )}}{26 \, {\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\right )}} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.04, size = 72, normalized size = 0.63 \[ \frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{5} x^{5}-6 a^{4} x^{4}+18 a^{3} x^{3}-14 a^{2} x^{2}+17 a x -9\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{26 a \left (a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}} \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{2}}}\,{d x} \]

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)

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mupad [B]  time = 0.67, size = 123, normalized size = 1.07 \[ -\frac {{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (\frac {9}{26\,a^5\,c^3}-\frac {3\,x^5}{13\,c^3}-\frac {17\,x}{26\,a^4\,c^3}+\frac {3\,x^4}{13\,a\,c^3}-\frac {9\,x^3}{13\,a^2\,c^3}+\frac {7\,x^2}{13\,a^3\,c^3}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^4}+x^4\,\sqrt {c\,a^2\,x^2+c}+\frac {2\,x^2\,\sqrt {c\,a^2\,x^2+c}}{a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-atan(a*x))/(c + a^2*c*x^2)^(7/2),x)

[Out]

-(exp(-atan(a*x))*(9/(26*a^5*c^3) - (3*x^5)/(13*c^3) - (17*x)/(26*a^4*c^3) + (3*x^4)/(13*a*c^3) - (9*x^3)/(13*
a^2*c^3) + (7*x^2)/(13*a^3*c^3)))/((c + a^2*c*x^2)^(1/2)/a^4 + x^4*(c + a^2*c*x^2)^(1/2) + (2*x^2*(c + a^2*c*x
^2)^(1/2))/a^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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