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

Optimal. Leaf size=60 \[ \frac {2 i (1-i a x)^2}{a \sqrt {a^2 x^2+1}}+\frac {3 i \sqrt {a^2 x^2+1}}{a}-\frac {3 \sinh ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5059, 853, 669, 641, 215} \[ \frac {2 i (1-i a x)^2}{a \sqrt {a^2 x^2+1}}+\frac {3 i \sqrt {a^2 x^2+1}}{a}-\frac {3 \sinh ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
 x] && GtQ[a, 0] && PosQ[b]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 669

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

Rule 853

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a^
m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[m, 0] && IntegerQ[n]

Rule 5059

Int[E^(ArcTan[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(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, x] && IntegerQ[(I*n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.70 \[ -\frac {3 \sinh ^{-1}(a x)}{a}+\frac {\sqrt {a^2 x^2+1} \left (\frac {4}{a x-i}+i\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 60, normalized size = 1.00 \[ \frac {4 \, a x + {\left (3 \, a x - 3 i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (i \, a x + 5\right )} - 4 i}{a^{2} x - i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

undef

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maple [B]  time = 0.11, size = 219, normalized size = 3.65 \[ -\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a^{4} \left (x -\frac {i}{a}\right )^{3}}-\frac {2 i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a^{3} \left (x -\frac {i}{a}\right )^{2}}+\frac {2 i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{a}-3 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\, x -\frac {3 \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 65, normalized size = 1.08 \[ \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{3} x^{2} - 2 i \, a^{2} x - a} - \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a} + \frac {6 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{2} x + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(a^2*x^2 + 1)^(3/2)/(a^3*x^2 - 2*I*a^2*x - a) - 3*arcsinh(a*x)/a + 6*I*sqrt(a^2*x^2 + 1)/(I*a^2*x + a)

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mupad [B]  time = 0.42, size = 73, normalized size = 1.22 \[ \frac {\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {4\,\sqrt {a^2\,x^2+1}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*x^2 + 1)^(1/2)*1i)/a - (3*asinh(x*(a^2)^(1/2)))/(a^2)^(1/2) - (4*(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 \[ i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x) + Integral(a**2*x**2*sqrt(a**2*x**
2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x))

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