3.1.10 \(\int \frac {e^{i \text {ArcTan}(a x)}}{x^5} \, dx\) [10]

Optimal. Leaf size=113 \[ -\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right ) \]

[Out]

-3/8*a^4*arctanh((a^2*x^2+1)^(1/2))-1/4*(a^2*x^2+1)^(1/2)/x^4-1/3*I*a*(a^2*x^2+1)^(1/2)/x^3+3/8*a^2*(a^2*x^2+1
)^(1/2)/x^2+2/3*I*a^3*(a^2*x^2+1)^(1/2)/x

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Rubi [A]
time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5168, 849, 821, 272, 65, 214} \begin {gather*} \frac {3 a^2 \sqrt {a^2 x^2+1}}{8 x^2}-\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {i a \sqrt {a^2 x^2+1}}{3 x^3}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right )+\frac {2 i a^3 \sqrt {a^2 x^2+1}}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Sqrt[1 + a^2*x^2]/x^4 - ((I/3)*a*Sqrt[1 + a^2*x^2])/x^3 + (3*a^2*Sqrt[1 + a^2*x^2])/(8*x^2) + (((2*I)/3)*
a^3*Sqrt[1 + a^2*x^2])/x - (3*a^4*ArcTanh[Sqrt[1 + a^2*x^2]])/8

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 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 821

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 849

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 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]

Rubi steps

\begin {align*} \int \frac {e^{i \tan ^{-1}(a x)}}{x^5} \, dx &=\int \frac {1+i a x}{x^5 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {1}{4} \int \frac {-4 i a+3 a^2 x}{x^4 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {1}{12} \int \frac {-9 a^2-8 i a^3 x}{x^3 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}-\frac {1}{24} \int \frac {16 i a^3-9 a^4 x}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}+\frac {1}{8} \left (3 a^4\right ) \int \frac {1}{x \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}+\frac {1}{16} \left (3 a^4\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}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}+\frac {1}{8} \left (3 a^2\right ) \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}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 76, normalized size = 0.67 \begin {gather*} \frac {1}{24} \left (\frac {\sqrt {1+a^2 x^2} \left (-6-8 i a x+9 a^2 x^2+16 i a^3 x^3\right )}{x^4}+9 a^4 \log (x)-9 a^4 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Sqrt[1 + a^2*x^2]*(-6 - (8*I)*a*x + 9*a^2*x^2 + (16*I)*a^3*x^3))/x^4 + 9*a^4*Log[x] - 9*a^4*Log[1 + Sqrt[1 +
a^2*x^2]])/24

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Maple [A]
time = 0.08, size = 97, normalized size = 0.86

method	result	size
risch	\(\frac {i \left (16 a^{5} x^{5}-9 i a^{4} x^{4}+8 a^{3} x^{3}-3 i a^{2} x^{2}-8 a x +6 i\right )}{24 x^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{4} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{8}\)	\(77\)
default	\(i a \left (-\frac {\sqrt {a^{2} x^{2}+1}}{3 x^{3}}+\frac {2 a^{2} \sqrt {a^{2} x^{2}+1}}{3 x}\right )-\frac {\sqrt {a^{2} x^{2}+1}}{4 x^{4}}-\frac {3 a^{2} \left (-\frac {\sqrt {a^{2} x^{2}+1}}{2 x^{2}}+\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}\right )}{4}\)	\(97\)
meijerg	\(\frac {a^{4} \left (\frac {\sqrt {\pi }\, \left (-7 a^{4} x^{4}-8 a^{2} x^{2}+8\right )}{16 a^{4} x^{4}}-\frac {\sqrt {\pi }\, \left (-12 a^{2} x^{2}+8\right ) \sqrt {a^{2} x^{2}+1}}{16 a^{4} x^{4}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{4}+\frac {3 \left (\frac {7}{6}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{8}-\frac {\sqrt {\pi }}{2 x^{4} a^{4}}+\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{2 \sqrt {\pi }}-\frac {i a \left (-2 a^{2} x^{2}+1\right ) \sqrt {a^{2} x^{2}+1}}{3 x^{3}}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 86, normalized size = 0.76 \begin {gather*} -\frac {3}{8} \, a^{4} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {2 i \, \sqrt {a^{2} x^{2} + 1} a^{3}}{3 \, x} + \frac {3 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{8 \, x^{2}} - \frac {i \, \sqrt {a^{2} x^{2} + 1} a}{3 \, x^{3}} - \frac {\sqrt {a^{2} x^{2} + 1}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*a^4*arcsinh(1/(a*abs(x))) + 2/3*I*sqrt(a^2*x^2 + 1)*a^3/x + 3/8*sqrt(a^2*x^2 + 1)*a^2/x^2 - 1/3*I*sqrt(a^
2*x^2 + 1)*a/x^3 - 1/4*sqrt(a^2*x^2 + 1)/x^4

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Fricas [A]
time = 4.92, size = 101, normalized size = 0.89 \begin {gather*} -\frac {9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - 16 i \, a^{4} x^{4} - {\left (16 i \, a^{3} x^{3} + 9 \, a^{2} x^{2} - 8 i \, a x - 6\right )} \sqrt {a^{2} x^{2} + 1}}{24 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(9*a^4*x^4*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 9*a^4*x^4*log(-a*x + sqrt(a^2*x^2 + 1) - 1) - 16*I*a^4*x^
4 - (16*I*a^3*x^3 + 9*a^2*x^2 - 8*I*a*x - 6)*sqrt(a^2*x^2 + 1))/x^4

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Sympy [A]
time = 3.50, size = 122, normalized size = 1.08 \begin {gather*} \frac {2 i a^{4} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3} - \frac {3 a^{4} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {i a^{2} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3 x^{2}} + \frac {a}{8 x^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*a**4*sqrt(1 + 1/(a**2*x**2))/3 - 3*a**4*asinh(1/(a*x))/8 + 3*a**3/(8*x*sqrt(1 + 1/(a**2*x**2))) - I*a**2*s
qrt(1 + 1/(a**2*x**2))/(3*x**2) + a/(8*x**3*sqrt(1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sqrt(1 + 1/(a**2*x**2)))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (89) = 178\).
time = 0.43, size = 237, normalized size = 2.10 \begin {gather*} -\frac {3}{8} \, a^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + \frac {3}{8} \, a^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) - \frac {9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{7} a^{4} - 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{5} a^{4} - 48 i \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{4} a^{3} {\left | a \right |} - 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{3} a^{4} + 64 i \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} a^{3} {\left | a \right |} + 9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )} a^{4} - 16 i \, a^{3} {\left | a \right |}}{12 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*a^4*log(abs(-x*abs(a) + sqrt(a^2*x^2 + 1) + 1)) + 3/8*a^4*log(abs(-x*abs(a) + sqrt(a^2*x^2 + 1) - 1)) - 1
/12*(9*(x*abs(a) - sqrt(a^2*x^2 + 1))^7*a^4 - 33*(x*abs(a) - sqrt(a^2*x^2 + 1))^5*a^4 - 48*I*(x*abs(a) - sqrt(
a^2*x^2 + 1))^4*a^3*abs(a) - 33*(x*abs(a) - sqrt(a^2*x^2 + 1))^3*a^4 + 64*I*(x*abs(a) - sqrt(a^2*x^2 + 1))^2*a
^3*abs(a) + 9*(x*abs(a) - sqrt(a^2*x^2 + 1))*a^4 - 16*I*a^3*abs(a))/((x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)^4

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Mupad [B]
time = 0.03, size = 95, normalized size = 0.84 \begin {gather*} \frac {a^4\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8}-\frac {\sqrt {a^2\,x^2+1}}{4\,x^4}-\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{3\,x^3}+\frac {3\,a^2\,\sqrt {a^2\,x^2+1}}{8\,x^2}+\frac {a^3\,\sqrt {a^2\,x^2+1}\,2{}\mathrm {i}}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*atan((a^2*x^2 + 1)^(1/2)*1i)*3i)/8 - (a^2*x^2 + 1)^(1/2)/(4*x^4) - (a*(a^2*x^2 + 1)^(1/2)*1i)/(3*x^3) + (
3*a^2*(a^2*x^2 + 1)^(1/2))/(8*x^2) + (a^3*(a^2*x^2 + 1)^(1/2)*2i)/(3*x)

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