3.319 \(\int \frac {e^{5 i \tan ^{-1}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=35 \[ -\frac {i}{2 a (a x+i)^2}-\frac {2}{3 a (a x+i)^3} \]

[Out]

-2/3/a/(I+a*x)^3-1/2*I/a/(I+a*x)^2

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5073, 43} \[ -\frac {i}{2 a (a x+i)^2}-\frac {2}{3 a (a x+i)^3} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

-2/(3*a*(I + a*x)^3) - (I/2)/(a*(I + a*x)^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5073

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + (I*n
)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {e^{5 i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\int \frac {1+i a x}{(1-i a x)^4} \, dx\\ &=\int \left (\frac {2}{(i+a x)^4}+\frac {i}{(i+a x)^3}\right ) \, dx\\ &=-\frac {2}{3 a (i+a x)^3}-\frac {i}{2 a (i+a x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 0.69 \[ -\frac {1+3 i a x}{6 a (a x+i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((5*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

-1/6*(1 + (3*I)*a*x)/(a*(I + a*x)^3)

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fricas [A]  time = 0.42, size = 35, normalized size = 1.00 \[ \frac {-3 i \, a x - 1}{6 \, a^{4} x^{3} + 18 i \, a^{3} x^{2} - 18 \, a^{2} x - 6 i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="fricas")

[Out]

(-3*I*a*x - 1)/(6*a^4*x^3 + 18*I*a^3*x^2 - 18*a^2*x - 6*I*a)

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giac [A]  time = 0.12, size = 19, normalized size = 0.54 \[ -\frac {3 \, a i x + 1}{6 \, {\left (a x + i\right )}^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="giac")

[Out]

-1/6*(3*a*i*x + 1)/((a*x + i)^3*a)

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maple [A]  time = 0.06, size = 20, normalized size = 0.57 \[ \frac {-\frac {i x}{2}-\frac {1}{6 a}}{\left (a x +i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^5/(a^2*x^2+1)^4,x)

[Out]

(-1/2*I*x-1/6/a)/(I+a*x)^3

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maxima [B]  time = 0.43, size = 59, normalized size = 1.69 \[ \frac {-24 i \, a^{4} x^{4} - 80 \, a^{3} x^{3} + 96 i \, a^{2} x^{2} + 48 \, a x - 8 i}{48 \, {\left (a^{7} x^{6} + 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="maxima")

[Out]

1/48*(-24*I*a^4*x^4 - 80*a^3*x^3 + 96*I*a^2*x^2 + 48*a*x - 8*I)/(a^7*x^6 + 3*a^5*x^4 + 3*a^3*x^2 + a)

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mupad [B]  time = 0.10, size = 21, normalized size = 0.60 \[ -\frac {3\,a\,x-\mathrm {i}}{6\,a\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^5/(a^2*x^2 + 1)^4,x)

[Out]

-(3*a*x - 1i)/(6*a*(a*x*1i - 1)^3)

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sympy [A]  time = 0.31, size = 37, normalized size = 1.06 \[ \frac {3 i a x + 1}{- 6 a^{4} x^{3} - 18 i a^{3} x^{2} + 18 a^{2} x + 6 i a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**5/(a**2*x**2+1)**4,x)

[Out]

(3*I*a*x + 1)/(-6*a**4*x**3 - 18*I*a**3*x**2 + 18*a**2*x + 6*I*a)

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