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

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 = 0.0420944, 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[E^((4*I)*ArcTan[a*x])*x^m,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 = 0.0213528, 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[E^((4*I)*ArcTan[a*x])*x^m,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 = 0.516, size = 417, normalized size = 8.3 \begin{align*}{\frac{1}{2} \left ({a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ({a}^{2} \right ) ^{1/2+m/2}}{2\,{a}^{2}{x}^{2}+2}}+2\,{\frac{{x}^{1+m} \left ({a}^{2} \right ) ^{1/2+m/2} \left ( -1/4\,{m}^{2}+1/4 \right ){\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,1/2+m/2 \right ) }{1+m}} \right ) }+{\frac{2\,i}{a} \left ({a}^{2} \right ) ^{-{\frac{m}{2}}} \left ({\frac{{x}^{m} \left ( -m-2 \right ) }{ \left ( 2+m \right ) \left ({a}^{2}{x}^{2}+1 \right ) } \left ({a}^{2} \right ) ^{{\frac{m}{2}}}}+{\frac{{x}^{m}m}{2} \left ({a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,{\frac{m}{2}} \right ) } \right ) }-3\, \left ({a}^{2} \right ) ^{-1/2-m/2} \left ({\frac{{x}^{1+m} \left ({a}^{2} \right ) ^{3/2+m/2} \left ( -3-m \right ) }{{a}^{2} \left ( 3+m \right ) \left ({a}^{2}{x}^{2}+1 \right ) }}+1/2\,{\frac{{x}^{1+m} \left ({a}^{2} \right ) ^{3/2+m/2} \left ( 1+m \right ){\it LerchPhi} \left ( -{a}^{2}{x}^{2},1,1/2+m/2 \right ) }{{a}^{2}}} \right ) -{\frac{2\,i}{a} \left ({a}^{2} \right ) ^{-{\frac{m}{2}}} \left ({\frac{{x}^{m} \left ( 2\,{a}^{2}{x}^{2}+m+2 \right ) }{ \left ({a}^{2}{x}^{2}+1 \right ) m} \left ({a}^{2} \right ) ^{{\frac{m}{2}}}}-{\frac{{x}^{m} \left ( 2+m \right ) }{2} \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 ({\frac{{x}^{1+m} \left ( 2\,{a}^{2}{x}^{2}+m+3 \right ) }{ \left ({a}^{2}{x}^{2}+1 \right ){a}^{4} \left ( 1+m \right ) } \left ({a}^{2} \right ) ^{{\frac{5}{2}}+{\frac{m}{2}}}}-{\frac{{x}^{1+m} \left ( 3+m \right ) }{2\,{a}^{4}} \left ({a}^{2} \right ) ^{{\frac{5}{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)^4/(a^2*x^2+1)^2*x^m,x)

[Out]

1/2*(a^2)^(-1/2-1/2*m)*(2*x^(1+m)*(a^2)^(1/2+1/2*m)/(2*a^2*x^2+2)+2/(1+m)*x^(1+m)*(a^2)^(1/2+1/2*m)*(-1/4*m^2+
1/4)*LerchPhi(-a^2*x^2,1,1/2+1/2*m))+2*I/a*(a^2)^(-1/2*m)*(1/(2+m)*x^m*(a^2)^(1/2*m)*(-m-2)/(a^2*x^2+1)+1/2*x^
m*(a^2)^(1/2*m)*m*LerchPhi(-a^2*x^2,1,1/2*m))-3*(a^2)^(-1/2-1/2*m)*(1/(3+m)*x^(1+m)*(a^2)^(3/2+1/2*m)*(-3-m)/a
^2/(a^2*x^2+1)+1/2*x^(1+m)*(a^2)^(3/2+1/2*m)*(1+m)/a^2*LerchPhi(-a^2*x^2,1,1/2+1/2*m))-2*I/a*(a^2)^(-1/2*m)*(x
^m*(a^2)^(1/2*m)*(2*a^2*x^2+m+2)/(a^2*x^2+1)/m-1/2*x^m*(a^2)^(1/2*m)*(2+m)*LerchPhi(-a^2*x^2,1,1/2*m))+1/2*(a^
2)^(-1/2-1/2*m)*(x^(1+m)*(a^2)^(5/2+1/2*m)*(2*a^2*x^2+m+3)/(a^2*x^2+1)/a^4/(1+m)-1/2*x^(1+m)*(a^2)^(5/2+1/2*m)
*(3+m)/a^4*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 )}^{4} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*a*x + 1)^4*x^m/(a^2*x^2 + 1)^2, 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((1+I*a*x)^4/(a^2*x^2+1)^2*x^m,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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m} \left (i a x + 1\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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