3.24 \(\int \frac {e^{3 i \tan ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=63 \[ -\frac {4 a \sqrt {a^2 x^2+1}}{a x+i}-\frac {\sqrt {a^2 x^2+1}}{x}-3 i a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5060, 6742, 264, 266, 63, 208, 651} \[ -\frac {4 a \sqrt {a^2 x^2+1}}{a x+i}-\frac {\sqrt {a^2 x^2+1}}{x}-3 i a \tanh ^{-1}\left (\sqrt {a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a*x])/x^2,x]

[Out]

-(Sqrt[1 + a^2*x^2]/x) - (4*a*Sqrt[1 + a^2*x^2])/(I + a*x) - (3*I)*a*ArcTanh[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 264

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 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 651

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 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]

Rule 6742

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

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Mathematica [A]  time = 0.05, size = 61, normalized size = 0.97 \[ \sqrt {a^2 x^2+1} \left (-\frac {1}{x}-\frac {4 a}{a x+i}\right )-3 i a \log \left (\sqrt {a^2 x^2+1}+1\right )+3 i a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((3*I)*ArcTan[a*x])/x^2,x]

[Out]

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

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fricas [B]  time = 0.44, size = 109, normalized size = 1.73 \[ -\frac {5 \, a^{2} x^{2} + 5 i \, a x + 3 \, {\left (i \, a^{2} x^{2} - a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + 3 \, {\left (-i \, a^{2} x^{2} + a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (5 \, a x + i\right )}}{a x^{2} + i \, x} \]

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^2,x, algorithm="fricas")

[Out]

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

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

[Out]

undef

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maple [A]  time = 0.15, size = 80, normalized size = 1.27 \[ \frac {i a}{\sqrt {a^{2} x^{2}+1}}-\frac {5 a^{2} x}{\sqrt {a^{2} x^{2}+1}}+3 i a \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )-\frac {1}{x \sqrt {a^{2} x^{2}+1}} \]

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^2,x)

[Out]

I*a/(a^2*x^2+1)^(1/2)-5*a^2*x/(a^2*x^2+1)^(1/2)+3*I*a*(1/(a^2*x^2+1)^(1/2)-arctanh(1/(a^2*x^2+1)^(1/2)))-1/x/(
a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.31, size = 60, normalized size = 0.95 \[ -\frac {5 \, a^{2} x}{\sqrt {a^{2} x^{2} + 1}} - 3 i \, a \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {4 i \, a}{\sqrt {a^{2} x^{2} + 1}} - \frac {1}{\sqrt {a^{2} x^{2} + 1} x} \]

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^2,x, algorithm="maxima")

[Out]

-5*a^2*x/sqrt(a^2*x^2 + 1) - 3*I*a*arcsinh(1/(a*abs(x))) + 4*I*a/sqrt(a^2*x^2 + 1) - 1/(sqrt(a^2*x^2 + 1)*x)

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mupad [B]  time = 0.06, size = 75, normalized size = 1.19 \[ -a\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )\,3{}\mathrm {i}-\frac {\sqrt {a^2\,x^2+1}}{x}-\frac {4\,a^2\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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**2,x)

[Out]

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

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