3.1.29 \(\int e^{4 i \text {ArcTan}(a x)} x \, dx\) [29]

Optimal. Leaf size=45 \[ -\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2} \]

[Out]

-4*I*x/a+1/2*x^2-4*I/a^2/(I+a*x)-8*ln(I+a*x)/a^2

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Rubi [A]
time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5170, 78} \begin {gather*} -\frac {4 i}{a^2 (a x+i)}-\frac {8 \log (a x+i)}{a^2}-\frac {4 i x}{a}+\frac {x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*x)/a + x^2/2 - (4*I)/(a^2*(I + a*x)) - (8*Log[I + a*x])/a^2

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5170

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

Rubi steps

\begin {align*} \int e^{4 i \tan ^{-1}(a x)} x \, dx &=\int \frac {x (1+i a x)^2}{(1-i a x)^2} \, dx\\ &=\int \left (-\frac {4 i}{a}+x+\frac {4 i}{a (i+a x)^2}-\frac {8}{a (i+a x)}\right ) \, dx\\ &=-\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 45, normalized size = 1.00 \begin {gather*} -\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*x)/a + x^2/2 - (4*I)/(a^2*(I + a*x)) - (8*Log[I + a*x])/a^2

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Maple [A]
time = 0.07, size = 50, normalized size = 1.11

method	result	size
default	\(-\frac {-\frac {1}{2} a \,x^{2}+4 i x}{a}+\frac {-\frac {4 i}{a \left (a x +i\right )}-\frac {8 \ln \left (a x +i\right )}{a}}{a}\)	\(50\)
risch	\(\frac {x^{2}}{2}-\frac {4 i x}{a}-\frac {4 i}{a^{2} \left (a x +i\right )}-\frac {4 \ln \left (a^{2} x^{2}+1\right )}{a^{2}}+\frac {8 i \arctan \left (a x \right )}{a^{2}}\)	\(53\)
meijerg	\(\frac {x^{2}}{2 a^{2} x^{2}+2}+\frac {2 i \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{a \sqrt {a^{2}}}-\frac {3 \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{a^{2}}-\frac {2 i \left (\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a \sqrt {a^{2}}}+\frac {\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{3 a^{2} x^{2}+3}-2 \ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 60, normalized size = 1.33 \begin {gather*} -\frac {4 \, {\left (i \, a x + 1\right )}}{a^{4} x^{2} + a^{2}} + \frac {a x^{2} - 8 i \, x}{2 \, a} + \frac {8 i \, \arctan \left (a x\right )}{a^{2}} - \frac {4 \, \log \left (a^{2} x^{2} + 1\right )}{a^{2}} \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,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.62, size = 53, normalized size = 1.18 \begin {gather*} \frac {a^{3} x^{3} - 7 i \, a^{2} x^{2} + 8 \, a x - 16 \, {\left (a x + i\right )} \log \left (\frac {a x + i}{a}\right ) - 8 i}{2 \, {\left (a^{3} x + i \, a^{2}\right )}} \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,x, algorithm="fricas")

[Out]

1/2*(a^3*x^3 - 7*I*a^2*x^2 + 8*a*x - 16*(a*x + I)*log((a*x + I)/a) - 8*I)/(a^3*x + I*a^2)

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Sympy [A]
time = 0.09, size = 36, normalized size = 0.80 \begin {gather*} \frac {x^{2}}{2} - \frac {4 i}{a^{3} x + i a^{2}} - \frac {4 i x}{a} - \frac {8 \log {\left (a x + i \right )}}{a^{2}} \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,x)

[Out]

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

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Giac [A]
time = 0.40, size = 43, normalized size = 0.96 \begin {gather*} -\frac {8 \, \log \left (a x + i\right )}{a^{2}} + \frac {a^{4} x^{2} - 8 i \, a^{3} x}{2 \, a^{4}} - \frac {4 i}{{\left (a x + i\right )} a^{2}} \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,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 43, normalized size = 0.96 \begin {gather*} \frac {x^2}{2}-\frac {8\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {4{}\mathrm {i}}{a^3\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {x\,4{}\mathrm {i}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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