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

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 = 0.0231997, 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[x^m/E^((2*I)*ArcTan[a*x]),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 = 0.0082643, 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[x^m/E^((2*I)*ArcTan[a*x]),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 = 0.382, size = 158, normalized size = 4.1 \begin{align*}{\frac{i \left ( ia \right ) ^{-m}}{a} \left ({\frac{{x}^{m} \left ( ia \right ) ^{m} \left ( -{a}^{2}m{x}^{2}-iamx-{m}^{2}-2\,iax-3\,m-2 \right ) }{m \left ( 1+m \right ) \left ( 1+iax \right ) }}+{x}^{m} \left ( ia \right ) ^{m} \left ( 2+m \right ){\it LerchPhi} \left ( -iax,1,m \right ) \right ) }-{\frac{i \left ( ia \right ) ^{-m}}{a} \left ({\frac{{x}^{m} \left ( ia \right ) ^{m} \left ( -1-m \right ) }{ \left ( 1+m \right ) \left ( 1+iax \right ) }}+{x}^{m} \left ( ia \right ) ^{m}m{\it LerchPhi} \left ( -iax,1,m \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integral(-(a*x + I)*x^m/(a*x - I), x)

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Sympy [B]  time = 31.5172, size = 578, normalized size = 14.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(-I*a*m**2*x**4*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + gamma
(m + 4)) - 5*I*a*m*x**4*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + ga
mma(m + 4)) - 6*I*a*x**4*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + g
amma(m + 4)) - m**2*x**3*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + g
amma(m + 4)) - 5*m*x**3*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + ga
mma(m + 4)) + m*x**3*x**m*gamma(m + 3)/(I*a*x*gamma(m + 4) + gamma(m + 4)) - 6*x**3*x**m*lerchphi(a*x*exp_pola
r(3*I*pi/2), 1, m + 3)*gamma(m + 3)/(I*a*x*gamma(m + 4) + gamma(m + 4)) + 3*x**3*x**m*gamma(m + 3)/(I*a*x*gamm
a(m + 4) + gamma(m + 4))) - I*a*m**2*x**2*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 1)*gamma(m + 1)/(I*a*x
*gamma(m + 2) + gamma(m + 2)) - I*a*m*x**2*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 1)*gamma(m + 1)/(I*a*
x*gamma(m + 2) + gamma(m + 2)) - m**2*x*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 1)*gamma(m + 1)/(I*a*x*g
amma(m + 2) + gamma(m + 2)) - m*x*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 1)*gamma(m + 1)/(I*a*x*gamma(m
+ 2) + gamma(m + 2)) + m*x*x**m*gamma(m + 1)/(I*a*x*gamma(m + 2) + gamma(m + 2)) + x*x**m*gamma(m + 1)/(I*a*x
*gamma(m + 2) + gamma(m + 2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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