\(\int e^{4 i \arctan (a x)} x^m \, dx\) [135]

Optimal result

Integrand size = 14, antiderivative size = 50 \[ \int e^{4 i \arctan (a x)} x^m \, dx=\frac {x^{1+m}}{1+m}+\frac {4 x^{1+m}}{1-i a x}-4 x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5170, 91, 81, 66} \[ \int e^{4 i \arctan (a x)} x^m \, dx=-4 x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,i a x)+\frac {4 x^{m+1}}{1-i a x}+\frac {x^{m+1}}{m+1} \]

[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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 81

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 91

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 5170

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*} \text {integral}& = \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} \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.16 \[ \int e^{4 i \arctan (a x)} x^m \, dx=\frac {x^{1+m} (5 i+4 i m+a x-4 (1+m) (i+a x) \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x))}{(1+m) (i+a x)} \]

[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))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5.

Time = 0.45 (sec) , antiderivative size = 417, normalized size of antiderivative = 8.34

method	result	size
meijerg	\(\frac {\left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (a^{2}\right )^{\frac {1}{2}+\frac {m}{2}}}{2 a^{2} x^{2}+2}+\frac {2 x^{1+m} \left (a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-\frac {m^{2}}{4}+\frac {1}{4}\right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}\right )}{2}+\frac {2 i \left (a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (-2-m \right )}{\left (2+m \right ) \left (a^{2} x^{2}+1\right )}+\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} m \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{2}\right )}{a}-3 \left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {x^{1+m} \left (a^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (-3-m \right )}{\left (3+m \right ) a^{2} \left (a^{2} x^{2}+1\right )}+\frac {x^{1+m} \left (a^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (1+m \right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{2 a^{2}}\right )-\frac {2 i \left (a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (2 a^{2} x^{2}+m +2\right )}{\left (a^{2} x^{2}+1\right ) m}-\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (2+m \right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{2}\right )}{a}+\frac {\left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {x^{1+m} \left (a^{2}\right )^{\frac {m}{2}+\frac {5}{2}} \left (2 a^{2} x^{2}+m +3\right )}{\left (a^{2} x^{2}+1\right ) a^{4} \left (1+m \right )}-\frac {x^{1+m} \left (a^{2}\right )^{\frac {m}{2}+\frac {5}{2}} \left (3+m \right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{2 a^{4}}\right )}{2}\)	\(417\)

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2*x^m,x,method=_RETURNVERBOSE)

[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)*(-2-m)/(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)^(1/2*m+5/2)*(2*a^2*x^2+m+3)/(a^2*x^2+1)/a^4/(1+m)-1/2*x^(1+m)*(a^2)^(1/2*m+5/2)
*(3+m)/a^4*LerchPhi(-a^2*x^2,1,1/2+1/2*m))

Fricas [F]

\[ \int e^{4 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{4} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{2}} \,d x } \]

[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)

Sympy [F]

\[ \int e^{4 i \arctan (a x)} x^m \, dx=\int \frac {x^{m} \left (a x - i\right )^{4}}{\left (a^{2} x^{2} + 1\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int e^{4 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{4} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{2}} \,d x } \]

[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)

Giac [F]

\[ \int e^{4 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{4} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int e^{4 i \arctan (a x)} x^m \, dx=\int \frac {x^m\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^4}{{\left (a^2\,x^2+1\right )}^2} \,d x \]

[In]

int((x^m*(a*x*1i + 1)^4)/(a^2*x^2 + 1)^2,x)

[Out]

int((x^m*(a*x*1i + 1)^4)/(a^2*x^2 + 1)^2, x)