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

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 = 0.0448873, 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 = 0.0324315, 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 [A]  time = 0.054, size = 81, normalized size = 1.4 \begin{align*} 4\,{\frac{x}{\sqrt{{a}^{2}{x}^{2}+1}}}-{ia{x}^{2}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{5\,i}{a}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-3\,{\frac{1}{\sqrt{{a}^{2}}}\ln \left ({\frac{{a}^{2}x}{\sqrt{{a}^{2}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ) } \end{align*}

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)

[Out]

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

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Maxima [A]  time = 0.99551, size = 89, normalized size = 1.48 \begin{align*} -\frac{i \, a x^{2}}{\sqrt{a^{2} x^{2} + 1}} + \frac{4 \, x}{\sqrt{a^{2} x^{2} + 1}} - \frac{3 \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} - \frac{5 i}{\sqrt{a^{2} x^{2} + 1} a} \end{align*}

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, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.6227, size = 146, normalized size = 2.43 \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+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 (i a x + 1\right )^{3}}{\left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

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)

[Out]

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

[Out]

undef