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]

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

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Rubi [A]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 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^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*}

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Mathematica [A]  time = 0.05, size = 70, normalized size = 0.78 \[ \frac {1}{6} \left (3 i a^3 \log (x)+\frac {\sqrt {a^2 x^2+1} \left (4 a^2 x^2+3 i a x-2\right )}{x^3}-3 i a^3 \log \left (\sqrt {a^2 x^2+1}+1\right )\right ) \]

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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fricas [A]  time = 0.45, size = 92, normalized size = 1.02 \[ \frac {-3 i \, a^{3} x^{3} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + 3 i \, a^{3} x^{3} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 4 \, a^{3} x^{3} + {\left (4 \, a^{2} x^{2} + 3 i \, a x - 2\right )} \sqrt {a^{2} x^{2} + 1}}{6 \, x^{3}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.17, size = 237, normalized size = 2.63 \[ -\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 x^{3}}-\frac {i a^{3} \arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}+\frac {i a^{3} \sqrt {a^{2} x^{2}+1}}{2}+\frac {a^{2} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}-a^{4} x \sqrt {a^{2} x^{2}+1}-\frac {a^{4} \ln \left (\frac {x \,a^{2}}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-i a^{3} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {a^{4} \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}+\frac {i a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}} \]

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^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{4}}\,{d x} \]

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^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 74, normalized size = 0.82 \[ \frac {2\,a^2\,\sqrt {a^2\,x^2+1}}{3\,x}-\frac {\sqrt {a^2\,x^2+1}}{3\,x^3}-\frac {a^3\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )}{2}+\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)^(1/2)/(x^4*(a*x*1i + 1)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{5} - i x^{4}}\, dx \]

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**4,x)

[Out]

-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**5 - I*x**4), x)

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