3.1.94 \(\int \frac {e^{-\frac {1}{2} i \text {ArcTan}(a x)}}{x^3} \, dx\) [94]

Optimal. Leaf size=132 \[ \frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \text {ArcTan}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

1/4*I*a*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x-1/2*(1-I*a*x)^(5/4)*(1+I*a*x)^(3/4)/x^2-1/4*a^2*arctan((1+I*a*x)^(1/
4)/(1-I*a*x)^(1/4))+1/4*a^2*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))

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Rubi [A]
time = 0.03, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5170, 98, 96, 95, 304, 209, 212} \begin {gather*} -\frac {1}{4} a^2 \text {ArcTan}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}+\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*a*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/x - ((1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(2*x^2) - (a^2*ArcTan[
(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4 + (a^2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

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

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

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 \frac {e^{-\frac {1}{2} i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac {\sqrt [4]{1-i a x}}{x^3 \sqrt [4]{1+i a x}} \, dx\\ &=-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} (i a) \int \frac {\sqrt [4]{1-i a x}}{x^2 \sqrt [4]{1+i a x}} \, dx\\ &=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{8} a^2 \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{2} a^2 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ &=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}+\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ &=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 81, normalized size = 0.61 \begin {gather*} \frac {\sqrt [4]{1-i a x} \left (-2+i a x-3 a^2 x^2+2 a^2 x^2 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {i+a x}{i-a x}\right )\right )}{4 x^2 \sqrt [4]{1+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I*a*x)^(1/4)*(-2 + I*a*x - 3*a^2*x^2 + 2*a^2*x^2*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x)]))/(
4*x^2*(1 + I*a*x)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.47, size = 178, normalized size = 1.35 \begin {gather*} \frac {a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-3 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*sqrt(I*(a*x - I)/sqrt(a**2*x**2 + 1))), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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