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

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

[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 = 0.09, 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 = {5060, 835, 807, 266, 63, 208} \[ -\frac {2 i a^3 \sqrt {a^2 x^2+1}}{3 x}+\frac {3 a^2 \sqrt {a^2 x^2+1}}{8 x^2}+\frac {i a \sqrt {a^2 x^2+1}}{3 x^3}-\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(I*ArcTan[a*x])*x^5),x]

[Out]

-Sqrt[1 + a^2*x^2]/(4*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 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^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 ) \operatorname {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 ) \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}}{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 = 0.06, size = 76, normalized size = 0.67 \[ \frac {1}{24} \left (9 a^4 \log (x)-9 a^4 \log \left (\sqrt {a^2 x^2+1}+1\right )+\frac {\sqrt {a^2 x^2+1} \left (-16 i a^3 x^3+9 a^2 x^2+8 i a x-6\right )}{x^4}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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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fricas [A]  time = 0.46, size = 101, normalized size = 0.89 \[ -\frac {9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 16 i \, a^{4} x^{4} - {\left (-16 i \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 8 i \, a x - 6\right )} \sqrt {a^{2} x^{2} + 1}}{24 \, x^{4}} \]

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^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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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^5,x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.18, size = 259, normalized size = 2.29 \[ \frac {i a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 x^{3}}-\frac {3 a^{4} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{8}+\frac {3 a^{4} \sqrt {a^{2} x^{2}+1}}{8}-\frac {i a^{3} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+i a^{5} x \sqrt {a^{2} x^{2}+1}+\frac {i a^{5} \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-a^{4} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}-\frac {i a^{5} \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 x^{4}}+\frac {5 a^{2} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{8 x^{2}} \]

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^5,x)

[Out]

1/3*I*a/x^3*(a^2*x^2+1)^(3/2)-3/8*a^4*arctanh(1/(a^2*x^2+1)^(1/2))+3/8*a^4*(a^2*x^2+1)^(1/2)-I*a^3/x*(a^2*x^2+
1)^(3/2)+I*a^5*x*(a^2*x^2+1)^(1/2)+I*a^5*ln(x*a^2/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)-a^4*((x-I/a)^2*a^
2+2*I*a*(x-I/a))^(1/2)-I*a^5*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)
-1/4/x^4*(a^2*x^2+1)^(3/2)+5/8*a^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 \[ \int \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{5}}\,{d x} \]

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^5,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)^(1/2)/(x^5*(a*x*1i + 1)),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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{6} - i x^{5}}\, dx \]

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**5,x)

[Out]

-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**6 - I*x**5), x)

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