3.42 \(\int \frac{e^{-i \tan ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=90 \[ \frac{2 a^2 \sqrt{a^2 x^2+1}}{3 x}+\frac{i a \sqrt{a^2 x^2+1}}{2 x^2}-\frac{\sqrt{a^2 x^2+1}}{3 x^3}-\frac{1}{2} i a^3 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

-Sqrt[1 + a^2*x^2]/(3*x^3) + ((I/2)*a*Sqrt[1 + a^2*x^2])/x^2 + (2*a^2*Sqrt[1 + a^2*x^2])/(3*x) - (I/2)*a^3*Arc
Tanh[Sqrt[1 + a^2*x^2]]

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Rubi [A]  time = 0.0695705, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5060, 835, 807, 266, 63, 208} \[ \frac{2 a^2 \sqrt{a^2 x^2+1}}{3 x}+\frac{i a \sqrt{a^2 x^2+1}}{2 x^2}-\frac{\sqrt{a^2 x^2+1}}{3 x^3}-\frac{1}{2} i a^3 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(I*ArcTan[a*x])*x^4),x]

[Out]

-Sqrt[1 + a^2*x^2]/(3*x^3) + ((I/2)*a*Sqrt[1 + a^2*x^2])/x^2 + (2*a^2*Sqrt[1 + a^2*x^2])/(3*x) - (I/2)*a^3*Arc
Tanh[Sqrt[1 + a^2*x^2]]

Rule 5060

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{e^{-i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac{1-i a x}{x^4 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}-\frac{1}{3} \int \frac{3 i a+2 a^2 x}{x^3 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{1}{6} \int \frac{-4 a^2+3 i a^3 x}{x^2 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{2 a^2 \sqrt{1+a^2 x^2}}{3 x}+\frac{1}{2} \left (i a^3\right ) \int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{2 a^2 \sqrt{1+a^2 x^2}}{3 x}+\frac{1}{4} \left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{2 a^2 \sqrt{1+a^2 x^2}}{3 x}+\frac{1}{2} (i a) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{2 a^2 \sqrt{1+a^2 x^2}}{3 x}-\frac{1}{2} i a^3 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0484899, size = 70, normalized size = 0.78 \[ \frac{1}{6} \left (\frac{\sqrt{a^2 x^2+1} \left (4 a^2 x^2+3 i a x-2\right )}{x^3}-3 i a^3 \log \left (\sqrt{a^2 x^2+1}+1\right )+3 i a^3 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(I*ArcTan[a*x])*x^4),x]

[Out]

((Sqrt[1 + a^2*x^2]*(-2 + (3*I)*a*x + 4*a^2*x^2))/x^3 + (3*I)*a^3*Log[x] - (3*I)*a^3*Log[1 + Sqrt[1 + a^2*x^2]
])/6

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Maple [B]  time = 0.081, size = 237, normalized size = 2.6 \begin{align*}{\frac{{\frac{i}{2}}a}{{x}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{i}{2}}{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +{\frac{i}{2}}{a}^{3}\sqrt{{a}^{2}{x}^{2}+1}+{\frac{{a}^{2}}{x} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{a}^{4}x\sqrt{{a}^{2}{x}^{2}+1}-{{a}^{4}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-i{a}^{3}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }+{{a}^{4}\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{1}{3\,{x}^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \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)/x^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{4}}\,{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)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72583, size = 221, normalized size = 2.46 \begin{align*} \frac{-3 i \, a^{3} x^{3} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) + 3 i \, a^{3} x^{3} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) + 4 \, a^{3} x^{3} +{\left (4 \, a^{2} x^{2} + 3 i \, a x - 2\right )} \sqrt{a^{2} x^{2} + 1}}{6 \, x^{3}} \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)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{x^{4} \left (i a x + 1\right )}\, dx \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)/x**4,x)

[Out]

Integral(sqrt(a**2*x**2 + 1)/(x**4*(I*a*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \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)/x^4,x, algorithm="giac")

[Out]

undef