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

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 = 0.0471978, 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[E^(I*ArcTan[a*x])*x^2,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 = 0.0312297, 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[E^(I*ArcTan[a*x])*x^2,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 [A]  time = 0.074, size = 89, normalized size = 1.2 \begin{align*}{\frac{{\frac{i}{3}}{x}^{2}}{a}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{{\frac{2\,i}{3}}}{{a}^{3}}\sqrt{{a}^{2}{x}^{2}+1}}+{\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01625, size = 100, normalized size = 1.33 \begin{align*} \frac{i \, \sqrt{a^{2} x^{2} + 1} x^{2}}{3 \, a} + \frac{\sqrt{a^{2} x^{2} + 1} x}{2 \, a^{2}} - \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{2 \, \sqrt{a^{2}} a^{2}} - \frac{2 i \, \sqrt{a^{2} x^{2} + 1}}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68213, size = 123, normalized size = 1.64 \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((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^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 [A]  time = 3.82505, size = 75, normalized size = 1. \begin{align*} i a \left (\begin{cases} \frac{x^{2} \sqrt{a^{2} x^{2} + 1}}{3 a^{2}} - \frac{2 \sqrt{a^{2} x^{2} + 1}}{3 a^{4}} & \text{for}\: a \neq 0 \\\frac{x^{4}}{4} & \text{otherwise} \end{cases}\right ) + \frac{x \sqrt{a^{2} x^{2} + 1}}{2 a^{2}} - \frac{\operatorname{asinh}{\left (a x \right )}}{2 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*Piecewise((x**2*sqrt(a**2*x**2 + 1)/(3*a**2) - 2*sqrt(a**2*x**2 + 1)/(3*a**4), Ne(a, 0)), (x**4/4, True))
+ x*sqrt(a**2*x**2 + 1)/(2*a**2) - asinh(a*x)/(2*a**3)

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Giac [A]  time = 1.13049, 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((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^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))