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

Optimal. Leaf size=16 \[ -\frac {i \log (-a x+i)}{a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5073, 31} \[ -\frac {i \log (-a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Log[I - a*x])/a

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ -\frac {i \log (-a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Log[I - a*x])/a

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fricas [A]  time = 0.39, size = 15, normalized size = 0.94 \[ -\frac {i \, \log \left (\frac {a x - i}{a}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*log((a*x - I)/a)/a

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giac [A]  time = 0.11, size = 13, normalized size = 0.81 \[ -\frac {i \log \left (a i x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-i*log(a*i*x + 1)/a

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maple [A]  time = 0.03, size = 26, normalized size = 1.62 \[ -\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*a*x),x)

[Out]

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

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maxima [A]  time = 0.32, size = 12, normalized size = 0.75 \[ -\frac {i \, \log \left (i \, a x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*log(I*a*x + 1)/a

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mupad [B]  time = 0.48, size = 15, normalized size = 0.94 \[ -\frac {\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x*1i + 1),x)

[Out]

-(log(x - 1i/a)*1i)/a

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sympy [A]  time = 0.05, size = 14, normalized size = 0.88 \[ - \frac {i \log {\left (- i a x - 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*log(-I*a*x - 1)/a

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