3.103 \(\int \frac{e^{-\frac{3}{2} i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=132 \[ \frac{9}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{9}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}+\frac{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x} \]

[Out]

(((3*I)/4)*a*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x - ((1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(2*x^2) + (9*a^2*A
rcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4 + (9*a^2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4

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Rubi [A]  time = 0.0425153, antiderivative size = 132, normalized size of antiderivative = 1., 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 = {5062, 96, 94, 93, 212, 206, 203} \[ \frac{9}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{9}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}+\frac{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/4)*a*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x - ((1 - I*a*x)^(7/4)*(1 + I*a*x)^(1/4))/(2*x^2) + (9*a^2*A
rcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4 + (9*a^2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4

Rule 5062

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]

Rule 96

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 94

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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

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 212

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] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{e^{-\frac{3}{2} i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac{(1-i a x)^{3/4}}{x^3 (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}-\frac{1}{4} (3 i a) \int \frac{(1-i a x)^{3/4}}{x^2 (1+i a x)^{3/4}} \, dx\\ &=\frac{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x}-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}-\frac{1}{8} \left (9 a^2\right ) \int \frac{1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=\frac{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x}-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}-\frac{1}{2} \left (9 a^2\right ) \operatorname{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{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x}-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}+\frac{1}{4} \left (9 a^2\right ) \operatorname{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} \left (9 a^2\right ) \operatorname{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{3 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x}-\frac{(1-i a x)^{7/4} \sqrt [4]{1+i a x}}{2 x^2}+\frac{9}{4} a^2 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{9}{4} a^2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0189489, size = 81, normalized size = 0.61 \[ \frac{(1-i a x)^{3/4} \left (6 a^2 x^2 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{a x+i}{-a x+i}\right )-5 a^2 x^2+3 i a x-2\right )}{4 x^2 (1+i a x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*a*x)^(3/4)*(-2 + (3*I)*a*x - 5*a^2*x^2 + 6*a^2*x^2*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)]
))/(4*x^2*(1 + I*a*x)^(3/4))

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Maple [F]  time = 0.158, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1067, size = 414, normalized size = 3.14 \begin{align*} \frac{9 \, a^{2} x^{2} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 9 i \, a^{2} x^{2} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 9 i \, a^{2} x^{2} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 9 \, a^{2} x^{2} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - 1\right ) +{\left (10 \, a^{2} x^{2} + 14 i \, a x - 4\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(9*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 9*I*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x +
I)) + I) - 9*I*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) - 9*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(
a*x + I)) - 1) + (10*a^2*x^2 + 14*I*a*x - 4)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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