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

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[E^(I*ArcTan[a*x])*x^m,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\\ &=(i a) \int \frac {x^{1+m}}{\sqrt {1+a^2 x^2}} \, dx+\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[E^(I*ArcTan[a*x])*x^m,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)*Sqrt[
1 + I*a*x]*Sqrt[I + a*x])

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fricas [F]  time = 0.46, 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((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^m,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 \[ \int \frac {{\left (i \, a x + 1\right )} x^{m}}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.31, size = 71, normalized size = 0.90 \[ \frac {x^{1+m} \hypergeom \left (\left [\frac {1}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {i a \,x^{2+m} \hypergeom \left (\left [\frac {1}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.13, size = 95, normalized size = 1.20 \[ \frac {i a x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {a^{2} x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {a^{2} x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*x**2*x**m*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), a**2*x**2*exp_polar(I*pi))/(2*gamma(m/2 + 2)) +
x*x**m*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,), a**2*x**2*exp_polar(I*pi))/(2*gamma(m/2 + 3/2))

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