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

Optimal. Leaf size=79 \[ \frac {x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5060, 808, 364} \[ \frac {x^{m+1} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) - (I*a*x^(2 + m)*Hypergeometric2F
1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m)

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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 85, normalized size = 1.08 \[ -\frac {i \sqrt {1+i a x} \sqrt {a x+i} x^{m+1} F_1\left (m+1;\frac {1}{2},-\frac {1}{2};m+2;-i a x,i a x\right )}{(m+1) \sqrt {1-i a x} \sqrt {a x-i}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m/E^(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])/((1 + m)*Sqr
t[1 - I*a*x]*Sqrt[-I + a*x])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{m}}{a x - i}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1} x^{m}}{i \, a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \sqrt {a^{2} x^{2}+1}}{i a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1} x^{m}}{i \, a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m\,\sqrt {a^2\,x^2+1}}{1+a\,x\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {x^{m} \sqrt {a^{2} x^{2} + 1}}{a x - i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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