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

Optimal. Leaf size=63 \[ \frac {i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a^2 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a^2 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + a^2*x^2]/(2*x^2) + (I*a*Sqrt[1 + a^2*x^2])/x + (a^2*ArcTanh[Sqrt[1 + a^2*x^2]])/2

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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.90 \[ \frac {1}{2} \left (\frac {(-1+2 i a x) \sqrt {a^2 x^2+1}}{x^2}+a^2 \log \left (\sqrt {a^2 x^2+1}+1\right )+a^2 (-\log (x))\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 83, normalized size = 1.32 \[ \frac {a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 2 i \, a^{2} x^{2} + \sqrt {a^{2} x^{2} + 1} {\left (2 i \, a x - 1\right )}}{2 \, x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

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^3,x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.17, size = 219, normalized size = 3.48 \[ \frac {a^{2} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}-\frac {a^{2} \sqrt {a^{2} x^{2}+1}}{2}+\frac {i a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}-i a^{3} x \sqrt {a^{2} x^{2}+1}-\frac {i a^{3} \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}+a^{2} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a^{3} \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 52, normalized size = 0.83 \[ \frac {a^2\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}+\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atanh((a^2*x^2 + 1)^(1/2)))/2 - (a^2*x^2 + 1)^(1/2)/(2*x^2) + (a*(a^2*x^2 + 1)^(1/2)*1i)/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{4} - i x^{3}}\, dx \]

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**3,x)

[Out]

-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**4 - I*x**3), x)

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