3.84 \(\int \frac{e^{\frac{5}{2} i \tan ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=121 \[ -\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-5 i a \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0381298, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5062, 94, 93, 212, 206, 203} \[ -\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-5 i a \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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{5}{2} i \tan ^{-1}(a x)}}{x^2} \, dx &=\int \frac{(1+i a x)^{5/4}}{x^2 (1-i a x)^{5/4}} \, dx\\ &=-\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac{1}{2} (5 i a) \int \frac{\sqrt [4]{1+i a x}}{x (1-i a x)^{5/4}} \, dx\\ &=\frac{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+\frac{1}{2} (5 i a) \int \frac{1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=\frac{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}+(10 i a) \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{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-(5 i a) \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 )-(5 i a) \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{10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac{(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-5 i a \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0205397, size = 87, normalized size = 0.72 \[ \frac{-10 a x (a x+i) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{a x+i}{-a x+i}\right )-3 \left (9 a^2 x^2-8 i a x+1\right )}{3 x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(1 - (8*I)*a*x + 9*a^2*x^2) - 10*a*x*(I + a*x)*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)])/(3*x*(
1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))

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Maple [F]  time = 0.15, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.63947, size = 377, normalized size = 3.12 \begin{align*} \frac{-5 i \, a x \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 5 \, a x \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 5 \, a x \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 5 i \, a x \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \,{\left (-9 i \, a x + 1\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-5*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 5*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I
) - 5*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 5*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1)
- 2*(-9*I*a*x + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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