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]

((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

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Rubi [A]  time = 0.0441652, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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]

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 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 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 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]

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*}

Mathematica [A]  time = 0.0302173, size = 42, normalized size = 0.7 \[ -\frac{3 \sinh ^{-1}(a x)}{a}+\frac{\sqrt{a^2 x^2+1} \left (\frac{4}{a x-i}+i\right )}{a} \]

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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Maple [B]  time = 0.055, size = 219, normalized size = 3.7 \begin{align*} -{\frac{1}{{a}^{4}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-3}}-{\frac{2\,i}{{a}^{3}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-2}}+{\frac{2\,i}{a} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{3}{2}}}}-3\,\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }x-3\,{\frac{1}{\sqrt{{a}^{2}}}\ln \left ({\frac{1}{\sqrt{{a}^{2}}} \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ) }+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ) } \end{align*}

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

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Maxima [A]  time = 1.506, size = 88, normalized size = 1.47 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.6467, size = 144, normalized size = 2.4 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}{\left (i a x + 1\right )^{3}}\, dx \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

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