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

Optimal. Leaf size=39 \[ -\frac{x^{m+1}}{m+1}+\frac{2 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,i a x)}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0226873, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5062, 80, 64} \[ -\frac{x^{m+1}}{m+1}+\frac{2 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,i a x)}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^((2*I)*ArcTan[a*x])*x^m,x]

[Out]

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

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

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

Mathematica [A]  time = 0.0066885, size = 29, normalized size = 0.74 \[ \frac{x^{m+1} (-1+2 \text{Hypergeometric2F1}(1,m+1,m+2,i a x))}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((2*I)*ArcTan[a*x])*x^m,x]

[Out]

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

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Maple [C]  time = 0.463, size = 175, normalized size = 4.5 \begin{align*}{\frac{{x}^{1+m}}{1+m} \left ({\frac{1}{2}}+{\frac{m}{2}} \right ){\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,{\frac{1}{2}}+{\frac{m}{2}} \right ) }+{\frac{i}{a} \left ({a}^{2} \right ) ^{-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{m} \left ({a}^{2} \right ) ^{m/2}}{m}}+{\frac{{x}^{m} \left ( -m-2 \right ) }{2+m} \left ({a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,{\frac{m}{2}} \right ) } \right ) }-{\frac{1}{2} \left ({a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ({a}^{2} \right ) ^{3/2+m/2}}{{a}^{2} \left ( 1+m \right ) }}+{\frac{{x}^{1+m} \left ( -3-m \right ) }{{a}^{2} \left ( 3+m \right ) } \left ({a}^{2} \right ) ^{{\frac{3}{2}}+{\frac{m}{2}}}{\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,{\frac{1}{2}}+{\frac{m}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*x - I)*x^m/(a*x + I), x)

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Sympy [B]  time = 3.93472, size = 269, normalized size = 6.9 \begin{align*} - \frac{a^{2} m x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} - \frac{3 a^{2} x^{3} x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + \frac{3}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}{4 \Gamma \left (\frac{m}{2} + \frac{5}{2}\right )} + \frac{i a m x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + 1\right ) \Gamma \left (\frac{m}{2} + 1\right )}{2 \Gamma \left (\frac{m}{2} + 2\right )} + \frac{i a x^{2} x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + 1\right ) \Gamma \left (\frac{m}{2} + 1\right )}{\Gamma \left (\frac{m}{2} + 2\right )} + \frac{m x x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{x x^{m} \Phi \left (a^{2} x^{2} e^{i \pi }, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*m*x**3*x**m*lerchphi(a**2*x**2*exp_polar(I*pi), 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*gamma(m/2 + 5/2)) - 3*
a**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(I*pi), 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*gamma(m/2 + 5/2)) + I*a*m
*x**2*x**m*lerchphi(a**2*x**2*exp_polar(I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(2*gamma(m/2 + 2)) + I*a*x**2*x**m*l
erchphi(a**2*x**2*exp_polar(I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/gamma(m/2 + 2) + m*x*x**m*lerchphi(a**2*x**2*exp
_polar(I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*gamma(m/2 + 3/2)) + x*x**m*lerchphi(a**2*x**2*exp_polar(I*pi),
1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*gamma(m/2 + 3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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