3.1.65 \(\int \frac {e^{\frac {1}{2} i \text {ArcTan}(a x)}}{x^2} \, dx\) [65]

Optimal. Leaf size=92 \[ -\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}-i a \text {ArcTan}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-i a \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5170, 96, 95, 218, 212, 209} \begin {gather*} -i a \text {ArcTan}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}-i a \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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^2} \, dx &=\int \frac {\sqrt [4]{1+i a x}}{x^2 \sqrt [4]{1-i a x}} \, dx\\ &=-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}+\frac {1}{2} (i a) \int \frac {1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}+(2 i a) \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ &=-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}-(i a) \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 )-(i a) \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-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}-i a \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-i a \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 = 71, normalized size = 0.77 \begin {gather*} -\frac {i (1-i a x)^{3/4} \left (-3 i+3 a x+2 a x \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {i+a x}{i-a x}\right )\right )}{3 x (1+i a x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*(1 - I*a*x)^(3/4)*(-3*I + 3*a*x + 2*a*x*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)]))/(x*(1
+ I*a*x)^(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
time = 3.08, size = 151, normalized size = 1.64 \begin {gather*} \frac {-i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (-i \, a x + 1\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*(a*x - I)/sqrt(a**2*x**2 + 1))/x**2, 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+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^2,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 {\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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