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

Optimal. Leaf size=75 \[ -\frac{i \left (a^2 x^2+1\right )^{3/2}}{3 a^3}+\frac{x \sqrt{a^2 x^2+1}}{2 a^2}+\frac{i \sqrt{a^2 x^2+1}}{a^3}-\frac{\sinh ^{-1}(a x)}{2 a^3} \]

[Out]

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

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Rubi [A]  time = 0.0473174, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5060, 797, 641, 195, 215} \[ -\frac{i \left (a^2 x^2+1\right )^{3/2}}{3 a^3}+\frac{x \sqrt{a^2 x^2+1}}{2 a^2}+\frac{i \sqrt{a^2 x^2+1}}{a^3}-\frac{\sinh ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.0311688, size = 46, normalized size = 0.61 \[ \frac{-3 \sinh ^{-1}(a x)+\left (-2 i a^2 x^2+3 a x+4 i\right ) \sqrt{a^2 x^2+1}}{6 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*I + 3*a*x - (2*I)*a^2*x^2)*Sqrt[1 + a^2*x^2] - 3*ArcSinh[a*x])/(6*a^3)

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Maple [B]  time = 0.077, size = 168, normalized size = 2.2 \begin{align*}{\frac{-{\frac{i}{3}}}{{a}^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{x}{2\,{a}^{2}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{1}{2\,{a}^{2}}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{i}{{a}^{3}}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}-{\frac{1}{{a}^{2}}\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*(a^2*x^2+1)^(3/2)/a^3+1/2*x*(a^2*x^2+1)^(1/2)/a^2+1/2/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)
^(1/2)+I/a^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)-1/a^2*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 = 1.54917, size = 80, normalized size = 1.07 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1} x}{2 \, a^{2}} - \frac{i \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{3 \, a^{3}} - \frac{\operatorname{arsinh}\left (a x\right )}{2 \, a^{3}} + \frac{i \, \sqrt{a^{2} x^{2} + 1}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67148, size = 124, normalized size = 1.65 \begin{align*} \frac{\sqrt{a^{2} x^{2} + 1}{\left (-2 i \, a^{2} x^{2} + 3 \, a x + 4 i\right )} + 3 \, \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right )}{6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(sqrt(a^2*x^2 + 1)*(-2*I*a^2*x^2 + 3*a*x + 4*I) + 3*log(-a*x + sqrt(a^2*x^2 + 1)))/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10836, size = 85, normalized size = 1.13 \begin{align*} -\frac{1}{6} \, \sqrt{a^{2} x^{2} + 1}{\left ({\left (\frac{2 \, i x}{a} - \frac{3}{a^{2}}\right )} x - \frac{4 \, i}{a^{3}}\right )} + \frac{\log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{2 \, a^{2}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(a^2*x^2 + 1)*((2*i*x/a - 3/a^2)*x - 4*i/a^3) + 1/2*log(-x*abs(a) + sqrt(a^2*x^2 + 1))/(a^2*abs(a))