3.4.19 \(\int \frac {e^{5 i \text {ArcTan}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [319]

Optimal. Leaf size=35 \[ -\frac {2}{3 a (i+a x)^3}-\frac {i}{2 a (i+a x)^2} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 35, 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 = {5181, 45} \begin {gather*} -\frac {i}{2 a (a x+i)^2}-\frac {2}{3 a (a x+i)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((5*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

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

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 5181

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

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Mathematica [A]
time = 0.01, size = 24, normalized size = 0.69 \begin {gather*} -\frac {1+3 i a x}{6 a (i+a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((5*I)*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

-1/6*(1 + (3*I)*a*x)/(a*(I + a*x)^3)

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Maple [A]
time = 0.09, size = 20, normalized size = 0.57

method	result	size
default	\(\frac {-\frac {i x}{2}-\frac {1}{6 a}}{\left (a x +i\right )^{3}}\)	\(20\)
risch	\(\frac {-\frac {i x}{2}-\frac {1}{6 a}}{\left (a x +i\right )^{3}}\)	\(20\)
norman	\(\frac {x +\frac {5}{2} i a \,x^{2}+\frac {1}{6} i a^{5} x^{6}-\frac {5}{3} a^{2} x^{3}}{\left (a^{2} x^{2}+1\right )^{3}}\)	\(39\)
meijerg	\(\frac {\frac {x \sqrt {a^{2}}\, \left (15 a^{4} x^{4}+40 a^{2} x^{2}+33\right )}{4 \left (a^{2} x^{2}+1\right )^{3}}+\frac {15 \sqrt {a^{2}}\, \arctan \left (a x \right )}{4 a}}{12 \sqrt {a^{2}}}+\frac {5 i a \,x^{2} \left (a^{4} x^{4}+3 a^{2} x^{2}+3\right )}{6 \left (a^{2} x^{2}+1\right )^{3}}-\frac {5 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}} \left (-3 a^{4} x^{4}-8 a^{2} x^{2}+3\right )}{4 a^{2} \left (a^{2} x^{2}+1\right )^{3}}+\frac {3 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{4 a^{3}}\right )}{6 \sqrt {a^{2}}}-\frac {5 i a^{3} x^{4} \left (a^{2} x^{2}+3\right )}{6 \left (a^{2} x^{2}+1\right )^{3}}+\frac {-\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (-15 a^{4} x^{4}+40 a^{2} x^{2}+15\right )}{48 a^{4} \left (a^{2} x^{2}+1\right )^{3}}+\frac {5 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{16 a^{5}}}{\sqrt {a^{2}}}+\frac {i a^{5} x^{6}}{6 \left (a^{2} x^{2}+1\right )^{3}}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^5/(a^2*x^2+1)^4,x,method=_RETURNVERBOSE)

[Out]

(-1/2*I*x-1/6/a)/(I+a*x)^3

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
time = 0.48, size = 59, normalized size = 1.69 \begin {gather*} -\frac {3 i \, a^{4} x^{4} + 10 \, a^{3} x^{3} - 12 i \, a^{2} x^{2} - 6 \, a x + i}{6 \, {\left (a^{7} x^{6} + 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="maxima")

[Out]

-1/6*(3*I*a^4*x^4 + 10*a^3*x^3 - 12*I*a^2*x^2 - 6*a*x + I)/(a^7*x^6 + 3*a^5*x^4 + 3*a^3*x^2 + a)

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Fricas [A]
time = 5.41, size = 35, normalized size = 1.00 \begin {gather*} \frac {-3 i \, a x - 1}{6 \, {\left (a^{4} x^{3} + 3 i \, a^{3} x^{2} - 3 \, a^{2} x - i \, a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.14, size = 39, normalized size = 1.11 \begin {gather*} \frac {- 3 i a x - 1}{6 a^{4} x^{3} + 18 i a^{3} x^{2} - 18 a^{2} x - 6 i a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**5/(a**2*x**2+1)**4,x)

[Out]

(-3*I*a*x - 1)/(6*a**4*x**3 + 18*I*a**3*x**2 - 18*a**2*x - 6*I*a)

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Giac [A]
time = 0.40, size = 18, normalized size = 0.51 \begin {gather*} -\frac {3 i \, a x + 1}{6 \, {\left (a x + i\right )}^{3} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^4,x, algorithm="giac")

[Out]

-1/6*(3*I*a*x + 1)/((a*x + I)^3*a)

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Mupad [B]
time = 0.10, size = 21, normalized size = 0.60 \begin {gather*} -\frac {3\,a\,x-\mathrm {i}}{6\,a\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^5/(a^2*x^2 + 1)^4,x)

[Out]

-(3*a*x - 1i)/(6*a*(a*x*1i - 1)^3)

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