3.1.30 \(\int e^{4 i \text {ArcTan}(a x)} \, dx\) [30]

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

[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 = {5169, 45} \begin {gather*} \frac {4}{a (a x+i)}-\frac {4 i \log (a x+i)}{a}+x \end {gather*}

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 45

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 5169

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 \begin {gather*} x+\frac {4}{a (i+a x)}-\frac {4 \text {ArcTan}(a x)}{a}-\frac {2 i \log \left (1+a^2 x^2\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.08, size = 33, normalized size = 1.06

method	result	size
default	\(x -4 a \left (-\frac {1}{a^{2} \left (a x +i\right )}+\frac {i \ln \left (a x +i\right )}{a^{2}}\right )\)	\(33\)
risch	\(x +\frac {4}{a \left (a x +i\right )}-\frac {2 i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {4 \arctan \left (a x \right )}{a}\)	\(41\)
meijerg	\(\frac {\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}}{2 \sqrt {a^{2}}}+\frac {2 i a \,x^{2}}{a^{2} x^{2}+1}-\frac {3 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{\sqrt {a^{2}}}-\frac {2 i \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{a}+\frac {\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}}{2 \sqrt {a^{2}}}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 44, normalized size = 1.42 \begin {gather*} x + \frac {4 \, {\left (a x - i\right )}}{a^{3} x^{2} + a} - \frac {4 \, \arctan \left (a x\right )}{a} - \frac {2 i \, \log \left (a^{2} x^{2} + 1\right )}{a} \end {gather*}

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

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Fricas [A]
time = 1.65, size = 43, normalized size = 1.39 \begin {gather*} \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} \end {gather*}

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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Sympy [A]
time = 0.09, size = 22, normalized size = 0.71 \begin {gather*} x + \frac {4}{a^{2} x + i a} - \frac {4 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)**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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Giac [A]
time = 0.40, size = 25, normalized size = 0.81 \begin {gather*} x - \frac {4 i \, \log \left (a x + i\right )}{a} + \frac {4}{{\left (a x + i\right )} a} \end {gather*}

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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Mupad [B]
time = 0.43, size = 32, normalized size = 1.03 \begin {gather*} 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} \end {gather*}

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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