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

Optimal. Leaf size=32 \[ \frac {i \log (-a x+i)}{a}-\frac {2}{a (-a x+i)} \]

[Out]

-2/a/(I-a*x)+I*ln(I-a*x)/a

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Rubi [A]  time = 0.04, antiderivative size = 32, 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} \[ \frac {i \log (-a x+i)}{a}-\frac {2}{a (-a x+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(a*(I - a*x)) + (I*Log[I - a*x])/a

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^{-3 i \tan ^{-1}(a x)}}{\sqrt {1+a^2 x^2}} \, dx &=\int \frac {1-i a x}{(1+i a x)^2} \, dx\\ &=\int \left (-\frac {2}{(-i+a x)^2}+\frac {i}{-i+a x}\right ) \, dx\\ &=-\frac {2}{a (i-a x)}+\frac {i \log (i-a x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 1.00 \[ \frac {i \log (-a x+i)}{a}-\frac {2}{a (-a x+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(a*(I - a*x)) + (I*Log[I - a*x])/a

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fricas [A]  time = 0.51, size = 31, normalized size = 0.97 \[ \frac {{\left (i \, a x + 1\right )} \log \left (\frac {a x - i}{a}\right ) + 2}{a^{2} x - i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 28, normalized size = 0.88 \[ \frac {i \log \left (a x - i\right )}{a} + \frac {2}{{\left (a x - i\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

i*log(a*x - i)/a + 2/((a*x - i)*a)

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maple [A]  time = 0.05, size = 41, normalized size = 1.28 \[ -\frac {2}{a \left (-a x +i\right )}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}-\frac {\arctan \left (a x \right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 41, normalized size = 1.28 \[ -\frac {4 \, {\left (-i \, a x - 1\right )}}{2 i \, a^{3} x^{2} + 4 \, a^{2} x - 2 i \, a} + \frac {i \, \log \left (i \, a x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.49, size = 29, normalized size = 0.91 \[ -\frac {2}{-a^2\,x+a\,1{}\mathrm {i}}+\frac {\ln \left (a\,x-\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)/(a*x*1i + 1)^3,x)

[Out]

(log(a*x - 1i)*1i)/a - 2/(a*1i - a^2*x)

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sympy [A]  time = 0.16, size = 19, normalized size = 0.59 \[ \frac {2}{a^{2} x - i a} + \frac {i \log {\left (a x - i \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)**3*(a**2*x**2+1),x)

[Out]

2/(a**2*x - I*a) + I*log(a*x - I)/a

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