3.138 \(\int e^{-4 i \tan ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=50 \[ -4 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,-i a x)+\frac{4 x^{m+1}}{1+i a x}+\frac{x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0390302, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5062, 89, 80, 64} \[ -4 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,-i a x)+\frac{4 x^{m+1}}{1+i a x}+\frac{x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^((4*I)*ArcTan[a*x]),x]

[Out]

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

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^{-4 i \tan ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1-i a x)^2}{(1+i a x)^2} \, dx\\ &=\frac{4 x^{1+m}}{1+i a x}+\frac{\int \frac{x^m \left (-a^2 (3+4 m)+i a^3 x\right )}{1+i a x} \, dx}{a^2}\\ &=\frac{x^{1+m}}{1+m}+\frac{4 x^{1+m}}{1+i a x}-(4 (1+m)) \int \frac{x^m}{1+i a x} \, dx\\ &=\frac{x^{1+m}}{1+m}+\frac{4 x^{1+m}}{1+i a x}-4 x^{1+m} \, _2F_1(1,1+m;2+m;-i a x)\\ \end{align*}

Mathematica [A]  time = 0.0207909, size = 58, normalized size = 1.16 \[ \frac{x^{m+1} (-4 (m+1) (a x-i) \text{Hypergeometric2F1}(1,m+1,m+2,-i a x)+a x-4 i m-5 i)}{(m+1) (a x-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/E^((4*I)*ArcTan[a*x]),x]

[Out]

(x^(1 + m)*(-5*I - (4*I)*m + a*x - 4*(1 + m)*(-I + a*x)*Hypergeometric2F1[1, 1 + m, 2 + m, (-I)*a*x]))/((1 + m
)*(-I + a*x))

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Maple [C]  time = 0.536, size = 428, normalized size = 8.6 \begin{align*}{\frac{-{\frac{i}{6}} \left ( ia \right ) ^{-m}}{a} \left ({\frac{{x}^{m} \left ( ia \right ) ^{m} \left ({a}^{2}{x}^{2}{m}^{4}+6\,{a}^{4}{x}^{4}m+11\,{a}^{2}{x}^{2}{m}^{3}-126\,iamx-2\,iax{m}^{4}+46\,{a}^{2}{x}^{2}{m}^{2}-72\,iax-{m}^{4}+24\,i{a}^{3}{x}^{3}+90\,{a}^{2}m{x}^{2}-21\,iax{m}^{3}-10\,{m}^{3}+72\,{a}^{2}{x}^{2}-79\,iax{m}^{2}-35\,{m}^{2}+6\,i{a}^{3}{x}^{3}m-50\,m-24 \right ) }{m \left ( 1+m \right ) \left ( 1+iax \right ) ^{3}}}+{x}^{m} \left ( ia \right ) ^{m} \left ({m}^{3}+9\,{m}^{2}+26\,m+24 \right ){\it LerchPhi} \left ( -iax,1,m \right ) \right ) }+{\frac{{\frac{i}{3}} \left ( ia \right ) ^{-m}}{a} \left ( -{\frac{{x}^{m} \left ( ia \right ) ^{m} \left ( -{a}^{2}{x}^{2}{m}^{2}-4\,{a}^{2}m{x}^{2}+2\,iax{m}^{2}-6\,{a}^{2}{x}^{2}+7\,iamx+{m}^{2}+6\,iax+3\,m+2 \right ) }{ \left ( 1+iax \right ) ^{3}}}+{x}^{m} \left ( ia \right ) ^{m}m \left ({m}^{2}+3\,m+2 \right ){\it LerchPhi} \left ( -iax,1,m \right ) \right ) }-{\frac{{\frac{i}{6}} \left ( ia \right ) ^{-m}}{a} \left ( -{\frac{{x}^{m} \left ( ia \right ) ^{m} \left ( -{a}^{2}{x}^{2}{m}^{2}+2\,{a}^{2}m{x}^{2}+2\,iax{m}^{2}-5\,iamx+{m}^{2}-3\,m+2 \right ) }{ \left ( 1+iax \right ) ^{3}}}+{x}^{m} \left ( ia \right ) ^{m} \left ({m}^{2}-3\,m+2 \right ) 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)^4*(a^2*x^2+1)^2,x)

[Out]

-1/6*I*(I*a)^(-m)/a*(x^m*(I*a)^m*(a^2*x^2*m^4+6*a^4*x^4*m+11*a^2*x^2*m^3-126*I*a*m*x-2*I*a*x*m^4+46*a^2*x^2*m^
2-72*I*a*x-m^4+24*I*a^3*x^3+90*a^2*m*x^2-21*I*a*x*m^3-10*m^3+72*a^2*x^2-79*I*a*x*m^2-35*m^2+6*I*a^3*x^3*m-50*m
-24)/(1+m)/m/(1+I*a*x)^3+x^m*(I*a)^m*(m^3+9*m^2+26*m+24)*LerchPhi(-I*a*x,1,m))+1/3*I*(I*a)^(-m)/a*(-x^m*(I*a)^
m*(-a^2*x^2*m^2-4*a^2*m*x^2+2*I*a*x*m^2-6*a^2*x^2+7*I*a*m*x+m^2+6*I*a*x+3*m+2)/(1+I*a*x)^3+x^m*(I*a)^m*m*(m^2+
3*m+2)*LerchPhi(-I*a*x,1,m))-1/6*I*(I*a)^(-m)/a*(-x^m*(I*a)^m*(-a^2*x^2*m^2+2*a^2*m*x^2+2*I*a*x*m^2-5*I*a*m*x+
m^2-3*m+2)/(1+I*a*x)^3+x^m*(I*a)^m*(m^2-3*m+2)*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 )}^{2} x^{m}}{{\left (i \, a x + 1\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^4*(a^2*x^2+1)^2,x, algorithm="maxima")

[Out]

integrate((a^2*x^2 + 1)^2*x^m/(I*a*x + 1)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^4*(a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(1+I*a*x)**4*(a**2*x**2+1)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} x^{2} + 1\right )}^{2} x^{m}}{{\left (i \, a x + 1\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^4*(a^2*x^2+1)^2,x, algorithm="giac")

[Out]

integrate((a^2*x^2 + 1)^2*x^m/(I*a*x + 1)^4, x)