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

Optimal. Leaf size=159 \[ -\frac{3 x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};-a^2 x^2\right )}{m+1}+\frac{4 x^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};-a^2 x^2\right )}{m+1}+\frac{i a x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};-a^2 x^2\right )}{m+2}-\frac{4 i a x^{m+2} \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-a^2 x^2\right )}{m+2} \]

[Out]

(-3*x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) + (I*a*x^(2 + m)*Hypergeometri
c2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m) + (4*x^(1 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/
2, -(a^2*x^2)])/(1 + m) - ((4*I)*a*x^(2 + m)*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m)

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Rubi [A]  time = 0.704954, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5060, 6742, 364, 850, 808} \[ -\frac{3 x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},-a^2 x^2\right )}{m+1}+\frac{4 x^{m+1} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},-a^2 x^2\right )}{m+1}+\frac{i a x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m+2}-\frac{4 i a x^{m+2} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+2}{2},\frac{m+4}{2},-a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) + (I*a*x^(2 + m)*Hypergeometri
c2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m) + (4*x^(1 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/
2, -(a^2*x^2)])/(1 + m) - ((4*I)*a*x^(2 + m)*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m)

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rubi steps

\begin{align*} \int e^{-3 i \tan ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1-i a x)^2}{(1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=\int \left (-\frac{3 x^m}{\sqrt{1+a^2 x^2}}+\frac{i a x^{1+m}}{\sqrt{1+a^2 x^2}}+\frac{4 x^m}{(1+i a x) \sqrt{1+a^2 x^2}}\right ) \, dx\\ &=-\left (3 \int \frac{x^m}{\sqrt{1+a^2 x^2}} \, dx\right )+4 \int \frac{x^m}{(1+i a x) \sqrt{1+a^2 x^2}} \, dx+(i a) \int \frac{x^{1+m}}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};-a^2 x^2\right )}{1+m}+\frac{i a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};-a^2 x^2\right )}{2+m}+4 \int \frac{x^m (1-i a x)}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};-a^2 x^2\right )}{1+m}+\frac{i a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};-a^2 x^2\right )}{2+m}+4 \int \frac{x^m}{\left (1+a^2 x^2\right )^{3/2}} \, dx-(4 i a) \int \frac{x^{1+m}}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};-a^2 x^2\right )}{1+m}+\frac{i a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};-a^2 x^2\right )}{2+m}+\frac{4 x^{1+m} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};-a^2 x^2\right )}{1+m}-\frac{4 i a x^{2+m} \, _2F_1\left (\frac{3}{2},\frac{2+m}{2};\frac{4+m}{2};-a^2 x^2\right )}{2+m}\\ \end{align*}

Mathematica [C]  time = 0.0713403, size = 113, normalized size = 0.71 \[ \frac{i \sqrt{1+i a x} \sqrt{a x+i} x^{m+1} \left (F_1\left (m+1;\frac{1}{2},-\frac{1}{2};m+2;-i a x,i a x\right )-2 F_1\left (m+1;\frac{3}{2},-\frac{1}{2};m+2;-i a x,i a x\right )\right )}{(m+1) \sqrt{1-i a x} \sqrt{a x-i}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*x^(1 + m)*Sqrt[1 + I*a*x]*Sqrt[I + a*x]*(AppellF1[1 + m, 1/2, -1/2, 2 + m, (-I)*a*x, I*a*x] - 2*AppellF1[1
+ m, 3/2, -1/2, 2 + m, (-I)*a*x, I*a*x]))/((1 + m)*Sqrt[1 - I*a*x]*Sqrt[-I + a*x])

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Maple [F]  time = 0.41, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( 1+iax \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]

int(x^m/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x)

[Out]

int(x^m/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((a^2*x^2 + 1)^(3/2)*x^m/(I*a*x + 1)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} x^{2} + 1}{\left (i \, a x - 1\right )} x^{m}}{a^{2} x^{2} - 2 i \, a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*x^2 + 1)*(I*a*x - 1)*x^m/(a^2*x^2 - 2*I*a*x - 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*x^2 + 1)^(3/2)*x^m/(I*a*x + 1)^3, x)