3.30 \(\int e^{4 i \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {4}{a (a x+i)}-\frac {4 i \log (a x+i)}{a}+x \]

[Out]

x+4/a/(I+a*x)-4*I*ln(I+a*x)/a

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5061, 43} \[ \frac {4}{a (a x+i)}-\frac {4 i \log (a x+i)}{a}+x \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x]),x]

[Out]

x + 4/(a*(I + a*x)) - ((4*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 5061

Int[E^(ArcTan[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 - I*a*x)^((I*n)/2)/(1 + I*a*x)^((I*n)/2), x] /; FreeQ[{a
, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 42, normalized size = 1.35 \[ -\frac {2 i \log \left (a^2 x^2+1\right )}{a}+\frac {4}{a (a x+i)}-\frac {4 \tan ^{-1}(a x)}{a}+x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((4*I)*ArcTan[a*x]),x]

[Out]

x + 4/(a*(I + a*x)) - (4*ArcTan[a*x])/a - ((2*I)*Log[1 + a^2*x^2])/a

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fricas [A]  time = 0.52, size = 43, normalized size = 1.39 \[ \frac {a^{2} x^{2} + i \, a x - 4 \, {\left (i \, a x - 1\right )} \log \left (\frac {a x + i}{a}\right ) + 4}{a^{2} x + i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 26, normalized size = 0.84 \[ x - \frac {4 \, i \log \left (a x + i\right )}{a} + \frac {4}{{\left (a x + i\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 4*i*log(a*x + i)/a + 4/((a*x + i)*a)

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maple [A]  time = 0.05, size = 41, normalized size = 1.32 \[ x +\frac {4}{a \left (a x +i\right )}-\frac {2 i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {4 \arctan \left (a x \right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 45, normalized size = 1.45 \[ x + \frac {8 \, a x - 8 i}{2 \, {\left (a^{3} x^{2} + a\right )}} - \frac {4 \, \arctan \left (a x\right )}{a} - \frac {2 i \, \log \left (a^{2} x^{2} + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 1/2*(8*a*x - 8*I)/(a^3*x^2 + a) - 4*arctan(a*x)/a - 2*I*log(a^2*x^2 + 1)/a

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mupad [B]  time = 0.43, size = 32, normalized size = 1.03 \[ x+\frac {4}{a^2\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,4{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^4/(a^2*x^2 + 1)^2,x)

[Out]

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

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sympy [A]  time = 0.17, size = 22, normalized size = 0.71 \[ x + \frac {4}{a^{2} x + i a} - \frac {4 i \log {\left (a x + i \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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