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

Optimal. Leaf size=90 \[ \frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}+\frac{x^2 \sqrt{a^2 x^2+1}}{3 a^2}-\frac{(16+9 i a x) \sqrt{a^2 x^2+1}}{24 a^4}+\frac{3 i \sinh ^{-1}(a x)}{8 a^4} \]

[Out]

(x^2*Sqrt[1 + a^2*x^2])/(3*a^2) + ((I/4)*x^3*Sqrt[1 + a^2*x^2])/a - ((16 + (9*I)*a*x)*Sqrt[1 + a^2*x^2])/(24*a
^4) + (((3*I)/8)*ArcSinh[a*x])/a^4

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Rubi [A]  time = 0.0642709, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5060, 833, 780, 215} \[ \frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}+\frac{x^2 \sqrt{a^2 x^2+1}}{3 a^2}-\frac{(16+9 i a x) \sqrt{a^2 x^2+1}}{24 a^4}+\frac{3 i \sinh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[1 + a^2*x^2])/(3*a^2) + ((I/4)*x^3*Sqrt[1 + a^2*x^2])/a - ((16 + (9*I)*a*x)*Sqrt[1 + a^2*x^2])/(24*a
^4) + (((3*I)/8)*ArcSinh[a*x])/a^4

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 833

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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^{i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac{x^3 (1+i a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}+\frac{\int \frac{x^2 \left (-3 i a+4 a^2 x\right )}{\sqrt{1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac{x^2 \sqrt{1+a^2 x^2}}{3 a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}+\frac{\int \frac{x \left (-8 a^2-9 i a^3 x\right )}{\sqrt{1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{x^2 \sqrt{1+a^2 x^2}}{3 a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{(16+9 i a x) \sqrt{1+a^2 x^2}}{24 a^4}+\frac{(3 i) \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{8 a^3}\\ &=\frac{x^2 \sqrt{1+a^2 x^2}}{3 a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{(16+9 i a x) \sqrt{1+a^2 x^2}}{24 a^4}+\frac{3 i \sinh ^{-1}(a x)}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0385748, size = 56, normalized size = 0.62 \[ \frac{\sqrt{a^2 x^2+1} \left (6 i a^3 x^3+8 a^2 x^2-9 i a x-16\right )+9 i \sinh ^{-1}(a x)}{24 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*(-16 - (9*I)*a*x + 8*a^2*x^2 + (6*I)*a^3*x^3) + (9*I)*ArcSinh[a*x])/(24*a^4)

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Maple [A]  time = 0.072, size = 109, normalized size = 1.2 \begin{align*}{\frac{{\frac{i}{4}}{x}^{3}}{a}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{{\frac{3\,i}{8}}x}{{a}^{3}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{{\frac{3\,i}{8}}}{{a}^{3}}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{{x}^{2}}{3\,{a}^{2}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{2}{3\,{a}^{4}}\sqrt{{a}^{2}{x}^{2}+1}} \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^3,x)

[Out]

1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-3/8*I/a^3*x*(a^2*x^2+1)^(1/2)+3/8*I/a^3*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/
(a^2)^(1/2)+1/3*x^2*(a^2*x^2+1)^(1/2)/a^2-2/3*(a^2*x^2+1)^(1/2)/a^4

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Maxima [A]  time = 0.990504, size = 126, normalized size = 1.4 \begin{align*} \frac{i \, \sqrt{a^{2} x^{2} + 1} x^{3}}{4 \, a} + \frac{\sqrt{a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac{3 i \, \sqrt{a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac{3 i \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{8 \, \sqrt{a^{2}} a^{3}} - \frac{2 \, \sqrt{a^{2} x^{2} + 1}}{3 \, a^{4}} \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^3,x, algorithm="maxima")

[Out]

1/4*I*sqrt(a^2*x^2 + 1)*x^3/a + 1/3*sqrt(a^2*x^2 + 1)*x^2/a^2 - 3/8*I*sqrt(a^2*x^2 + 1)*x/a^3 + 3/8*I*arcsinh(
a^2*x/sqrt(a^2))/(sqrt(a^2)*a^3) - 2/3*sqrt(a^2*x^2 + 1)/a^4

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Fricas [A]  time = 1.71779, size = 144, normalized size = 1.6 \begin{align*} \frac{{\left (6 i \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 9 i \, a x - 16\right )} \sqrt{a^{2} x^{2} + 1} - 9 i \, \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right )}{24 \, a^{4}} \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^3,x, algorithm="fricas")

[Out]

1/24*((6*I*a^3*x^3 + 8*a^2*x^2 - 9*I*a*x - 16)*sqrt(a^2*x^2 + 1) - 9*I*log(-a*x + sqrt(a^2*x^2 + 1)))/a^4

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Sympy [A]  time = 5.4138, size = 119, normalized size = 1.32 \begin{align*} \frac{i a x^{5}}{4 \sqrt{a^{2} x^{2} + 1}} + \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} - \frac{i x^{3}}{8 a \sqrt{a^{2} x^{2} + 1}} - \frac{3 i x}{8 a^{3} \sqrt{a^{2} x^{2} + 1}} + \frac{3 i \operatorname{asinh}{\left (a x \right )}}{8 a^{4}} \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**3,x)

[Out]

I*a*x**5/(4*sqrt(a**2*x**2 + 1)) + 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)) - I*x**3/(8*a*sqrt(a**2*x**2 + 1)) - 3*I*x/(8*a**3*sqrt(a**2*x**2 + 1)) + 3*I*a
sinh(a*x)/(8*a**4)

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Giac [A]  time = 1.11541, size = 99, normalized size = 1.1 \begin{align*} \frac{1}{24} \, \sqrt{a^{2} x^{2} + 1}{\left ({\left (2 \,{\left (\frac{3 \, i x}{a} + \frac{4}{a^{2}}\right )} x - \frac{9 \, i}{a^{3}}\right )} x - \frac{16}{a^{4}}\right )} - \frac{3 \, i \log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right )}{8 \, a^{3}{\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^3,x, algorithm="giac")

[Out]

1/24*sqrt(a^2*x^2 + 1)*((2*(3*i*x/a + 4/a^2)*x - 9*i/a^3)*x - 16/a^4) - 3/8*i*log(-x*abs(a) + sqrt(a^2*x^2 + 1
))/(a^3*abs(a))