3.4.5 \(\int \frac {e^{i \text {ArcTan}(a x)}}{\sqrt {1+a^2 x^2}} \, dx\) [305]

Optimal. Leaf size=15 \[ \frac {i \log (i+a x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 15, 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 = {5181, 31} \begin {gather*} \frac {i \log (a x+i)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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 5181

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 = 15, normalized size = 1.00 \begin {gather*} \frac {i \log (i+a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Log[I + a*x])/a

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Maple [A]
time = 0.06, size = 26, normalized size = 1.73

method	result	size
default	\(\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\)	\(26\)
meijerg	\(\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\)	\(26\)
risch	\(\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)/(a^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (11) = 22\).
time = 0.46, size = 24, normalized size = 1.60 \begin {gather*} \frac {\arctan \left (a x\right )}{a} + \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.08, size = 15, normalized size = 1.00 \begin {gather*} \frac {i \, \log \left (\frac {a x + i}{a}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*log((a*x + I)/a)/a

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Sympy [A]
time = 0.01, size = 8, normalized size = 0.53 \begin {gather*} \frac {i \log {\left (a x + i \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*log(a*x + I)/a

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Giac [A]
time = 0.45, size = 12, normalized size = 0.80 \begin {gather*} \frac {i \, \log \left (-i \, a x + 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*log(-I*a*x + 1)/a

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Mupad [B]
time = 0.47, size = 15, normalized size = 1.00 \begin {gather*} \frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,1{}\mathrm {i}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)/(a^2*x^2 + 1),x)

[Out]

(log(x + 1i/a)*1i)/a

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