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]

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

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 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]

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 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^{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*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 0.70 \[ \frac {(1+i a x)^{3/2} (a x+4 i)}{15 a \sqrt {1-i a x} (a x+i)^2} \]

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)

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fricas [A]  time = 0.44, size = 76, normalized size = 1.13 \[ -\frac {a^{3} x^{3} + 3 i \, a^{2} x^{2} - 3 \, a x + {\left (a^{2} x^{2} + 3 i \, a x + 4\right )} \sqrt {a^{2} x^{2} + 1} - i}{15 \, a^{4} x^{3} + 45 i \, a^{3} x^{2} - 45 \, a^{2} x - 15 i \, a} \]

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)

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giac [B]  time = 0.17, size = 111, normalized size = 1.66 \[ -\frac {2 \, {\left (5 \, a^{3} i {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )} - 15 \, a i {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{3} - 4 \, a^{4} + 25 \, a^{2} {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2} - 15 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{4}\right )}}{15 \, {\left (a i + \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{5}} \]

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

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maple [B]  time = 0.18, size = 269, normalized size = 4.01 \[ \frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}+a^{4} \left (-\frac {x^{3}}{2 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {-\frac {3 x}{8 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {3 \left (\frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}\right )}{8 a^{2}}}{a^{2}}\right )-4 i a^{3} \left (-\frac {x^{2}}{3 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}-\frac {2}{15 a^{4} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )-6 a^{2} \left (-\frac {x}{4 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {\frac {x}{5 \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {4 x}{15 \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {8 x}{15 \sqrt {a^{2} x^{2}+1}}}{4 a^{2}}\right )-\frac {4 i}{5 a \left (a^{2} x^{2}+1\right )^{\frac {5}{2}}} \]

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)

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maxima [B]  time = 0.33, size = 95, normalized size = 1.42 \[ -\frac {a^{2} x^{3}}{2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {x}{15 \, \sqrt {a^{2} x^{2} + 1}} + \frac {4 i \, a x^{2}}{3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {x}{30 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {11 \, x}{10 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} - \frac {4 i}{15 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a} \]

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)

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mupad [B]  time = 0.52, size = 41, normalized size = 0.61 \[ \frac {\sqrt {a^2\,x^2+1}\,\left (a^2\,x^2\,1{}\mathrm {i}-3\,a\,x+4{}\mathrm {i}\right )}{15\,a\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x - i\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{\frac {7}{2}}}\, dx \]

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

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