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

Optimal. Leaf size=39 \[ -\frac {x^{m+1}}{m+1}+\frac {2 x^{m+1} \, _2F_1(1,m+1;m+2;-i a x)}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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])))

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.74 \[ \frac {x^{m+1} (-1+2 \, _2F_1(1,m+1;m+2;-i a x))}{m+1} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a x + i\right )} x^{m}}{a x - i}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*x + I)*x^m/(a*x - I), x)

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

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)

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maple [C]  time = 0.42, size = 158, normalized size = 4.05 \[ \frac {i \left (i a \right )^{-m} \left (\frac {x^{m} \left (i a \right )^{m} \left (-a^{2} m \,x^{2}-i a m x -2 i a x -m^{2}-3 m -2\right )}{\left (1+m \right ) m \left (i a x +1\right )}+x^{m} \left (i a \right )^{m} \left (2+m \right ) \Phi \left (-i a x , 1, m\right )\right )}{a}-\frac {i \left (i a \right )^{-m} \left (\frac {x^{m} \left (i a \right )^{m} \left (-1-m \right )}{\left (1+m \right ) \left (i a x +1\right )}+x^{m} \left (i a \right )^{m} m \Phi \left (-i a x , 1, m\right )\right )}{a} \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^m\,\left (a^2\,x^2+1\right )}{{\left (1+a\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 4.06, size = 136, normalized size = 3.49 \[ - \frac {i a m x^{2} x^{m} \Phi \left (a x e^{\frac {3 i \pi }{2}}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac {2 i a x^{2} x^{m} \Phi \left (a x e^{\frac {3 i \pi }{2}}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac {m x x^{m} \Phi \left (a x e^{\frac {3 i \pi }{2}}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac {x x^{m} \Phi \left (a x e^{\frac {3 i \pi }{2}}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} \]

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]

-I*a*m*x**2*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 2)*gamma(m + 2)/gamma(m + 3) - 2*I*a*x**2*x**m*lerch
phi(a*x*exp_polar(3*I*pi/2), 1, m + 2)*gamma(m + 2)/gamma(m + 3) + m*x*x**m*lerchphi(a*x*exp_polar(3*I*pi/2),
1, m + 1)*gamma(m + 1)/gamma(m + 2) + x*x**m*lerchphi(a*x*exp_polar(3*I*pi/2), 1, m + 1)*gamma(m + 1)/gamma(m
+ 2)

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