3.1.25 \(\int \frac {e^{3 i \text {ArcTan}(a x)}}{x^3} \, dx\) [25]

Optimal. Leaf size=92 \[ -\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+\frac {9}{2} a^2 \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right ) \]

[Out]

9/2*a^2*arctanh((a^2*x^2+1)^(1/2))-1/2*(a^2*x^2+1)^(1/2)/x^2-3*I*a*(a^2*x^2+1)^(1/2)/x-4*I*a^2*(a^2*x^2+1)^(1/
2)/(I+a*x)

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Rubi [A]
time = 0.51, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5168, 6874, 272, 44, 65, 214, 270, 665} \begin {gather*} -\frac {4 i a^2 \sqrt {a^2 x^2+1}}{a x+i}-\frac {3 i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}+\frac {9}{2} a^2 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 272

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 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 5168

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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 79, normalized size = 0.86 \begin {gather*} \sqrt {1+a^2 x^2} \left (-\frac {1}{2 x^2}-\frac {3 i a}{x}-\frac {4 i a^2}{i+a x}\right )-\frac {9}{2} a^2 \log (x)+\frac {9}{2} a^2 \log \left (1+\sqrt {1+a^2 x^2}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[1 + a^2*x^2]*(-1/2*1/x^2 - ((3*I)*a)/x - ((4*I)*a^2)/(I + a*x)) - (9*a^2*Log[x])/2 + (9*a^2*Log[1 + Sqrt[
1 + a^2*x^2]])/2

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Maple [A]
time = 0.13, size = 105, normalized size = 1.14

method	result	size
default	\(-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}-\frac {1}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {9 a^{2} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}+3 i a \left (-\frac {1}{x \sqrt {a^{2} x^{2}+1}}-\frac {2 a^{2} x}{\sqrt {a^{2} x^{2}+1}}\right )\)	\(105\)
risch	\(-\frac {i \left (6 a^{3} x^{3}-i a^{2} x^{2}+6 a x -i\right )}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {a^{2} \left (-9 \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {8 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a \left (x +\frac {i}{a}\right )}\right )}{2}\)	\(108\)
meijerg	\(\frac {a^{2} \left (\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \left (24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{\sqrt {\pi }}-\frac {3 i a \left (2 a^{2} x^{2}+1\right )}{x \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{2} \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }}-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a^3*x/(a^2*x^2+1)^(1/2)-1/2/x^2/(a^2*x^2+1)^(1/2)-9/2*a^2*(1/(a^2*x^2+1)^(1/2)-arctanh(1/(a^2*x^2+1)^(1/2))
)+3*I*a*(-1/x/(a^2*x^2+1)^(1/2)-2*a^2*x/(a^2*x^2+1)^(1/2))

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Maxima [A]
time = 0.27, size = 81, normalized size = 0.88 \begin {gather*} -\frac {7 i \, a^{3} x}{\sqrt {a^{2} x^{2} + 1}} + \frac {9}{2} \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {9 \, a^{2}}{2 \, \sqrt {a^{2} x^{2} + 1}} - \frac {3 i \, a}{\sqrt {a^{2} x^{2} + 1} x} - \frac {1}{2 \, \sqrt {a^{2} x^{2} + 1} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*I*a^3*x/sqrt(a^2*x^2 + 1) + 9/2*a^2*arcsinh(1/(a*abs(x))) - 9/2*a^2/sqrt(a^2*x^2 + 1) - 3*I*a/(sqrt(a^2*x^2
+ 1)*x) - 1/2/(sqrt(a^2*x^2 + 1)*x^2)

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Fricas [A]
time = 3.10, size = 130, normalized size = 1.41 \begin {gather*} \frac {-14 i \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-14 i \, a^{2} x^{2} + 5 \, a x - i\right )}}{2 \, {\left (a x^{3} + i \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-14*I*a^3*x^3 + 14*a^2*x^2 + 9*(a^3*x^3 + I*a^2*x^2)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 9*(a^3*x^3 + I*a
^2*x^2)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + sqrt(a^2*x^2 + 1)*(-14*I*a^2*x^2 + 5*a*x - I))/(a*x^3 + I*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \left (\int \frac {i}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(I/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x/(a**2*x**5*sqr
t(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**3/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sq
rt(a**2*x**2 + 1)), x) + Integral(-3*I*a**2*x**2/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x
))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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Mupad [B]
time = 0.43, size = 99, normalized size = 1.08 \begin {gather*} -\frac {a^2\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}-\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{x}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (a^2*atan((a^2*x^2 + 1)^(1/2)*1i)*9i)/2 - (a^2*x^2 + 1)^(1/2)/(2*x^2) - (a*(a^2*x^2 + 1)^(1/2)*3i)/x - (a^3*
(a^2*x^2 + 1)^(1/2)*4i)/((((a^2)^(1/2)*1i)/a + x*(a^2)^(1/2))*(a^2)^(1/2))

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