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

Optimal. Leaf size=30 \[ \frac {2}{a (a x+i)}-\frac {i \log (a x+i)}{a} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 30, 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} \[ \frac {2}{a (a x+i)}-\frac {i \log (a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

Int[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 = 30, normalized size = 1.00 \[ \frac {2}{a (a x+i)}-\frac {i \log (a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[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 = 0.41, size = 31, normalized size = 1.03 \[ \frac {{\left (-i \, a x + 1\right )} \log \left (\frac {a x + i}{a}\right ) + 2}{a^{2} x + i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^2,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 = 0.12, size = 25, normalized size = 0.83 \[ -\frac {i \log \left (a x + i\right )}{a} + \frac {2}{{\left (a x + i\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-i*log(a*x + i)/a + 2/((a*x + i)*a)

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maple [A]  time = 0.06, size = 40, normalized size = 1.33 \[ \frac {2}{a \left (a x +i\right )}-\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}-\frac {\arctan \left (a x \right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 44, normalized size = 1.47 \[ \frac {4 \, a x - 4 i}{2 \, {\left (a^{3} x^{2} + a\right )}} - \frac {\arctan \left (a x\right )}{a} - \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.49, size = 28, normalized size = 0.93 \[ \frac {2}{x\,a^2+a\,1{}\mathrm {i}}-\frac {\ln \left (a\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^3/(a^2*x^2 + 1)^2,x)

[Out]

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

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sympy [A]  time = 0.16, size = 19, normalized size = 0.63 \[ \frac {2}{a^{2} x + i a} - \frac {i \log {\left (a x + i \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a**2*x + I*a) - I*log(a*x + I)/a

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