3.301 \(\int \frac {e^{5 i \tan ^{-1}(a x)}}{\sqrt {1+a^2 x^2}} \, dx\)

Optimal. Leaf size=50 \[ \frac {4 i}{a (1-i a x)}-\frac {2 i}{a (1-i a x)^2}+\frac {i \log (a x+i)}{a} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 50, 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} \[ \frac {4 i}{a (1-i a x)}-\frac {2 i}{a (1-i a x)^2}+\frac {i \log (a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*I)*ArcTan[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

(-2*I)/(a*(1 - I*a*x)^2) + (4*I)/(a*(1 - 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^{5 i \tan ^{-1}(a x)}}{\sqrt {1+a^2 x^2}} \, dx &=\int \frac {(1+i a x)^2}{(1-i a x)^3} \, dx\\ &=\int \left (\frac {4}{(1-i a x)^3}-\frac {4}{(1-i a x)^2}+\frac {1}{1-i a x}\right ) \, dx\\ &=-\frac {2 i}{a (1-i a x)^2}+\frac {4 i}{a (1-i a x)}+\frac {i \log (i+a x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.84 \[ \frac {i \left (4 i a x+(a x+i)^2 \log (a x+i)-2\right )}{a (a x+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((5*I)*ArcTan[a*x])/Sqrt[1 + a^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.51, size = 53, normalized size = 1.06 \[ -\frac {4 \, a x - {\left (i \, a^{2} x^{2} - 2 \, a x - i\right )} \log \left (\frac {a x + i}{a}\right ) + 2 i}{a^{3} x^{2} + 2 i \, a^{2} x - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 30, normalized size = 0.60 \[ \frac {i \log \left (a x + i\right )}{a} - \frac {2 \, {\left (2 \, a x + i\right )}}{{\left (a x + i\right )}^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 45, normalized size = 0.90 \[ \frac {-4 x -\frac {2 i}{a}}{\left (a x +i\right )^{2}}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^5/(a^2*x^2+1)^3,x)

[Out]

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

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maxima [A]  time = 0.43, size = 63, normalized size = 1.26 \[ -\frac {32 \, a^{3} x^{3} - 48 i \, a^{2} x^{2} - 16 i}{8 \, {\left (a^{5} x^{4} + 2 \, a^{3} x^{2} + a\right )}} + \frac {\arctan \left (a x\right )}{a} + \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

-1/8*(32*a^3*x^3 - 48*I*a^2*x^2 - 16*I)/(a^5*x^4 + 2*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 = 0.13, size = 49, normalized size = 0.98 \[ \frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,1{}\mathrm {i}}{a}-\frac {\frac {4\,x}{a^2}+\frac {2{}\mathrm {i}}{a^3}}{x^2-\frac {1}{a^2}+\frac {x\,2{}\mathrm {i}}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^5/(a^2*x^2 + 1)^3,x)

[Out]

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

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sympy [A]  time = 0.31, size = 37, normalized size = 0.74 \[ - \frac {- 4 a x - 2 i}{- a^{3} x^{2} - 2 i a^{2} x + a} + \frac {i \log {\left (i a x - 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**5/(a**2*x**2+1)**3,x)

[Out]

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

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