3.320 \(\int \frac{e^{4 i \tan ^{-1}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ -\frac{i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}}-\frac{i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}} \]

[Out]

((-I/5)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(5/2)) - ((I/15)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0385951, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5073, 45, 37} \[ -\frac{i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}}-\frac{i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

((-I/5)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(5/2)) - ((I/15)*(1 + I*a*x)^(3/2))/(a*(1 - I*a*x)^(3/2))

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])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{e^{4 i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\int \frac{\sqrt{1+i a x}}{(1-i a x)^{7/2}} \, dx\\ &=-\frac{i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}}+\frac{1}{5} \int \frac{\sqrt{1+i a x}}{(1-i a x)^{5/2}} \, dx\\ &=-\frac{i (1+i a x)^{3/2}}{5 a (1-i a x)^{5/2}}-\frac{i (1+i a x)^{3/2}}{15 a (1-i a x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0153257, size = 47, normalized size = 0.7 \[ \frac{(1+i a x)^{3/2} (a x+4 i)}{15 a \sqrt{1-i a x} (a x+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

((1 + I*a*x)^(3/2)*(4*I + a*x))/(15*a*Sqrt[1 - I*a*x]*(I + a*x)^2)

________________________________________________________________________________________

Maple [B]  time = 0.067, size = 269, normalized size = 4. \begin{align*}{\frac{x}{5} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,x}{15} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,x}{15}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{a}^{4} \left ( -{\frac{{x}^{3}}{2\,{a}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{\frac{3}{2\,{a}^{2}} \left ( -{\frac{x}{4\,{a}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{4\,{a}^{2}} \left ({\frac{x}{5} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,x}{15} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,x}{15}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) } \right ) } \right ) -4\,i{a}^{3} \left ( -{\frac{{x}^{2}}{3\,{a}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}}-{\frac{2}{15\,{a}^{4}} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}} \right ) -6\,{a}^{2} \left ( -1/4\,{\frac{x}{{a}^{2} \left ({a}^{2}{x}^{2}+1 \right ) ^{5/2}}}+1/4\,{\frac{1}{{a}^{2}} \left ( 1/5\,{\frac{x}{ \left ({a}^{2}{x}^{2}+1 \right ) ^{5/2}}}+{\frac{4\,x}{15\, \left ({a}^{2}{x}^{2}+1 \right ) ^{3/2}}}+{\frac{8\,x}{15\,\sqrt{{a}^{2}{x}^{2}+1}}} \right ) } \right ) -{\frac{{\frac{4\,i}{5}}}{a} \left ({a}^{2}{x}^{2}+1 \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x)

[Out]

1/5*x/(a^2*x^2+1)^(5/2)+4/15*x/(a^2*x^2+1)^(3/2)+8/15*x/(a^2*x^2+1)^(1/2)+a^4*(-1/2*x^3/a^2/(a^2*x^2+1)^(5/2)+
3/2/a^2*(-1/4*x/a^2/(a^2*x^2+1)^(5/2)+1/4/a^2*(1/5*x/(a^2*x^2+1)^(5/2)+4/15*x/(a^2*x^2+1)^(3/2)+8/15*x/(a^2*x^
2+1)^(1/2))))-4*I*a^3*(-1/3*x^2/a^2/(a^2*x^2+1)^(5/2)-2/15/a^4/(a^2*x^2+1)^(5/2))-6*a^2*(-1/4*x/a^2/(a^2*x^2+1
)^(5/2)+1/4/a^2*(1/5*x/(a^2*x^2+1)^(5/2)+4/15*x/(a^2*x^2+1)^(3/2)+8/15*x/(a^2*x^2+1)^(1/2)))-4/5*I/a/(a^2*x^2+
1)^(5/2)

________________________________________________________________________________________

Maxima [B]  time = 0.99578, size = 128, normalized size = 1.91 \begin{align*} -\frac{a^{2} x^{3}}{2 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}}} - \frac{x}{15 \, \sqrt{a^{2} x^{2} + 1}} + \frac{4 i \, a x^{2}}{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}}} - \frac{x}{30 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}} + \frac{11 \, x}{10 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}}} - \frac{4 i}{15 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="maxima")

[Out]

-1/2*a^2*x^3/(a^2*x^2 + 1)^(5/2) - 1/15*x/sqrt(a^2*x^2 + 1) + 4/3*I*a*x^2/(a^2*x^2 + 1)^(5/2) - 1/30*x/(a^2*x^
2 + 1)^(3/2) + 11/10*x/(a^2*x^2 + 1)^(5/2) - 4/15*I/((a^2*x^2 + 1)^(5/2)*a)

________________________________________________________________________________________

Fricas [A]  time = 1.86898, size = 176, normalized size = 2.63 \begin{align*} -\frac{a^{3} x^{3} + 3 i \, a^{2} x^{2} - 3 \, a x +{\left (a^{2} x^{2} + 3 i \, a x + 4\right )} \sqrt{a^{2} x^{2} + 1} - i}{15 \, a^{4} x^{3} + 45 i \, a^{3} x^{2} - 45 \, a^{2} x - 15 i \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="fricas")

[Out]

-(a^3*x^3 + 3*I*a^2*x^2 - 3*a*x + (a^2*x^2 + 3*I*a*x + 4)*sqrt(a^2*x^2 + 1) - I)/(15*a^4*x^3 + 45*I*a^3*x^2 -
45*a^2*x - 15*I*a)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**(7/2),x)

[Out]

Integral((I*a*x + 1)**4/(a**2*x**2 + 1)**(7/2), x)

________________________________________________________________________________________

Giac [B]  time = 1.16208, size = 150, normalized size = 2.24 \begin{align*} -\frac{2 \,{\left (5 \, a^{3} i{\left (\sqrt{a^{2} + \frac{1}{x^{2}}} - \frac{1}{x}\right )} - 15 \, a i{\left (\sqrt{a^{2} + \frac{1}{x^{2}}} - \frac{1}{x}\right )}^{3} - 4 \, a^{4} + 25 \, a^{2}{\left (\sqrt{a^{2} + \frac{1}{x^{2}}} - \frac{1}{x}\right )}^{2} - 15 \,{\left (\sqrt{a^{2} + \frac{1}{x^{2}}} - \frac{1}{x}\right )}^{4}\right )}}{15 \,{\left (a i + \sqrt{a^{2} + \frac{1}{x^{2}}} - \frac{1}{x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^(7/2),x, algorithm="giac")

[Out]

-2/15*(5*a^3*i*(sqrt(a^2 + 1/x^2) - 1/x) - 15*a*i*(sqrt(a^2 + 1/x^2) - 1/x)^3 - 4*a^4 + 25*a^2*(sqrt(a^2 + 1/x
^2) - 1/x)^2 - 15*(sqrt(a^2 + 1/x^2) - 1/x)^4)/(a*i + sqrt(a^2 + 1/x^2) - 1/x)^5