3.139 \(\int e^{-6 i \tan ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=115 \[ \frac {2 \left (2 m^2+4 m+3\right ) x^{m+1} \, _2F_1(1,m+1;m+2;-i a x)}{m+1}+\frac {4 i x^{m+1} \left (-a \left (m^2+3 m+3\right ) x+i (m+1)^2\right )}{(m+1) (1+i a x)^2}-\frac {(1-i a x)^2 x^{m+1}}{(m+1) (1+i a x)^2} \]

[Out]

-x^(1+m)*(1-I*a*x)^2/(1+m)/(1+I*a*x)^2+4*I*x^(1+m)*(I*(1+m)^2-a*(m^2+3*m+3)*x)/(1+m)/(1+I*a*x)^2+2*(2*m^2+4*m+
3)*x^(1+m)*hypergeom([1, 1+m],[2+m],-I*a*x)/(1+m)

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, 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, 100, 145, 64} \[ \frac {2 \left (2 m^2+4 m+3\right ) x^{m+1} \text {Hypergeometric2F1}(1,m+1,m+2,-i a x)}{m+1}+\frac {4 i x^{m+1} \left (-a \left (m^2+3 m+3\right ) x+i (m+1)^2\right )}{(m+1) (1+i a x)^2}-\frac {(1-i a x)^2 x^{m+1}}{(m+1) (1+i a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^((6*I)*ArcTan[a*x]),x]

[Out]

-((x^(1 + m)*(1 - I*a*x)^2)/((1 + m)*(1 + I*a*x)^2)) + ((4*I)*x^(1 + m)*(I*(1 + m)^2 - a*(3 + 3*m + m^2)*x))/(
(1 + m)*(1 + I*a*x)^2) + (2*(3 + 4*m + 2*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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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^{-6 i \tan ^{-1}(a x)} x^m \, dx &=\int \frac {x^m (1-i a x)^3}{(1+i a x)^3} \, dx\\ &=-\frac {x^{1+m} (1-i a x)^2}{(1+m) (1+i a x)^2}-\frac {i \int \frac {x^m (1-i a x) \left (2 i a (1+m)+2 a^2 (3+m) x\right )}{(1+i a x)^3} \, dx}{a (1+m)}\\ &=-\frac {x^{1+m} (1-i a x)^2}{(1+m) (1+i a x)^2}+\frac {4 i x^{1+m} \left (i (1+m)^2-a \left (3+3 m+m^2\right ) x\right )}{(1+m) (1+i a x)^2}+\left (2 \left (3+4 m+2 m^2\right )\right ) \int \frac {x^m}{1+i a x} \, dx\\ &=-\frac {x^{1+m} (1-i a x)^2}{(1+m) (1+i a x)^2}+\frac {4 i x^{1+m} \left (i (1+m)^2-a \left (3+3 m+m^2\right ) x\right )}{(1+m) (1+i a x)^2}+\frac {2 \left (3+4 m+2 m^2\right ) x^{1+m} \, _2F_1(1,1+m;2+m;-i a x)}{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 94, normalized size = 0.82 \[ \frac {x^{m+1} \left (-a^2 x^2+2 \left (2 m^2+4 m+3\right ) (a x-i)^2 \, _2F_1(1,m+1;m+2;-i a x)+m^2 (4+4 i a x)+4 m (2+3 i a x)+10 i a x+5\right )}{(m+1) (a x-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/E^((6*I)*ArcTan[a*x]),x]

[Out]

(x^(1 + m)*(5 + (10*I)*a*x - a^2*x^2 + 4*m*(2 + (3*I)*a*x) + m^2*(4 + (4*I)*a*x) + 2*(3 + 4*m + 2*m^2)*(-I + a
*x)^2*Hypergeometric2F1[1, 1 + m, 2 + m, (-I)*a*x]))/((1 + m)*(-I + a*x)^2)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{3} x^{3} + 3 i \, a^{2} x^{2} - 3 \, a x - i\right )} x^{m}}{a^{3} x^{3} - 3 i \, a^{2} x^{2} - 3 \, a x + i}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^6*(a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

integral(-(a^3*x^3 + 3*I*a^2*x^2 - 3*a*x - I)*x^m/(a^3*x^3 - 3*I*a^2*x^2 - 3*a*x + I), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} x^{2} + 1\right )}^{3} x^{m}}{{\left (i \, a x + 1\right )}^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^6*(a^2*x^2+1)^3,x, algorithm="giac")

[Out]

integrate((a^2*x^2 + 1)^3*x^m/(I*a*x + 1)^6, x)

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maple [C]  time = 0.74, size = 1196, normalized size = 10.40 \[ \frac {i \left (i a \right )^{-m} \left (\frac {x^{m} \left (i a \right )^{m} \left (-a^{4} x^{4} m^{6}-120 a^{6} x^{6} m -22 a^{4} x^{4} m^{5}+4 i a^{3} x^{3} m^{6}-120 i a^{5} x^{5} m -197 a^{4} x^{4} m^{4}+87 i a^{3} x^{3} m^{5}-720 i a^{5} x^{5}-932 a^{4} x^{4} m^{3}+764 i a^{3} x^{3} m^{4}-2556 a^{4} x^{4} m^{2}+3483 i a^{3} x^{3} m^{3}+6 a^{2} x^{2} m^{6}-4200 a^{4} x^{4} m +8802 i a^{3} x^{3} m^{2}+129 a^{2} x^{2} m^{5}-4 i a x \,m^{6}-3600 a^{4} x^{4}+12000 i a^{3} x^{3} m +1112 a^{2} x^{2} m^{4}-85 i a x \,m^{5}+7200 i a^{3} x^{3}+4911 a^{2} x^{2} m^{3}-720 i a x \,m^{4}+11722 a^{2} m^{2} x^{2}-3095 i a x \,m^{3}-m^{6}+14400 a^{2} m \,x^{2}-7076 i a x \,m^{2}-21 m^{5}+7200 a^{2} x^{2}-8100 i a m x -175 m^{4}-3600 i a x -735 m^{3}-1624 m^{2}-1764 m -720\right )}{\left (1+m \right ) m \left (i a x +1\right )^{5}}+x^{m} \left (i a \right )^{m} \left (m^{5}+20 m^{4}+155 m^{3}+580 m^{2}+1044 m +720\right ) \Phi \left (-i a x , 1, m\right )\right )}{120 a}-\frac {i \left (i a \right )^{-m} \left (-\frac {x^{m} \left (i a \right )^{m} \left (a^{4} x^{4} m^{4}+11 a^{4} x^{4} m^{3}-4 i a^{3} x^{3} m^{4}+46 a^{4} x^{4} m^{2}-43 i a^{3} x^{3} m^{3}+96 a^{4} x^{4} m -171 i a^{3} x^{3} m^{2}+120 a^{4} x^{4}-312 i a^{3} x^{3} m -6 a^{2} x^{2} m^{4}-240 i a^{3} x^{3}-63 a^{2} x^{2} m^{3}+4 i a x \,m^{4}-239 a^{2} m^{2} x^{2}+41 i a x \,m^{3}-392 a^{2} m \,x^{2}+149 i a x \,m^{2}-240 a^{2} x^{2}+226 i a m x +m^{4}+120 i a x +10 m^{3}+35 m^{2}+50 m +24\right )}{\left (i a x +1\right )^{5}}+x^{m} \left (i a \right )^{m} m \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right ) \Phi \left (-i a x , 1, m\right )\right )}{40 a}+\frac {i \left (i a \right )^{-m} \left (-\frac {x^{m} \left (i a \right )^{m} \left (a^{4} x^{4} m^{4}+a^{4} x^{4} m^{3}-4 i a^{3} x^{3} m^{4}-4 a^{4} x^{4} m^{2}-3 i a^{3} x^{3} m^{3}-4 a^{4} x^{4} m +19 i a^{3} x^{3} m^{2}+18 i a^{3} x^{3} m -6 a^{2} x^{2} m^{4}-3 a^{2} x^{2} m^{3}+4 i a x \,m^{4}+31 a^{2} m^{2} x^{2}+i a x \,m^{3}+18 a^{2} m \,x^{2}-21 i a x \,m^{2}-40 a^{2} x^{2}-4 i a m x +m^{4}+20 i a x -5 m^{2}+4\right )}{\left (i a x +1\right )^{5}}+x^{m} \left (i a \right )^{m} \left (m^{2}-3 m +2\right ) m \left (m^{2}+3 m +2\right ) \Phi \left (-i a x , 1, m\right )\right )}{40 a}-\frac {i \left (i a \right )^{-m} \left (-\frac {x^{m} \left (i a \right )^{m} \left (a^{4} x^{4} m^{4}-9 a^{4} x^{4} m^{3}-4 i a^{3} x^{3} m^{4}+26 a^{4} x^{4} m^{2}+37 i a^{3} x^{3} m^{3}-24 a^{4} x^{4} m -111 i a^{3} x^{3} m^{2}+108 i a^{3} x^{3} m -6 a^{2} x^{2} m^{4}+57 a^{2} x^{2} m^{3}+4 i a x \,m^{4}-179 a^{2} m^{2} x^{2}-39 i a x \,m^{3}+188 a^{2} m \,x^{2}+129 i a x \,m^{2}-154 i a m x +m^{4}-10 m^{3}+35 m^{2}-50 m +24\right )}{\left (i a x +1\right )^{5}}+x^{m} \left (i a \right )^{m} \left (m^{4}-10 m^{3}+35 m^{2}-50 m +24\right ) m \Phi \left (-i a x , 1, m\right )\right )}{120 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(1+I*a*x)^6*(a^2*x^2+1)^3,x)

[Out]

1/120*I*(I*a)^(-m)/a*(x^m*(I*a)^m*(-720+764*I*a^3*x^3*m^4+12000*I*a^3*x^3*m-3095*I*a*x*m^3+11722*a^2*m^2*x^2+6
*a^2*x^2*m^6-120*a^6*x^6*m+129*a^2*x^2*m^5-720*I*a^5*x^5+7200*I*a^3*x^3-3600*I*a*x-197*a^4*x^4*m^4-m^6-175*m^4
-4*I*a*x*m^6-120*I*a^5*x^5*m-932*a^4*x^4*m^3-22*a^4*x^4*m^5+4*I*a^3*x^3*m^6+87*I*a^3*x^3*m^5+8802*I*a^3*x^3*m^
2-720*I*a*x*m^4-8100*I*a*m*x-1764*m+7200*a^2*x^2+14400*a^2*m*x^2-85*I*a*x*m^5-21*m^5-7076*I*a*x*m^2-2556*a^4*x
^4*m^2-4200*a^4*x^4*m-735*m^3+1112*a^2*x^2*m^4+4911*a^2*x^2*m^3+3483*I*a^3*x^3*m^3-a^4*x^4*m^6-1624*m^2-3600*a
^4*x^4)/(1+m)/m/(1+I*a*x)^5+x^m*(I*a)^m*(m^5+20*m^4+155*m^3+580*m^2+1044*m+720)*LerchPhi(-I*a*x,1,m))-1/40*I*(
I*a)^(-m)/a*(-x^m*(I*a)^m*(24-312*I*a^3*x^3*m+41*I*a*x*m^3+149*I*a*x*m^2+226*I*a*m*x-239*a^2*m^2*x^2+a^4*x^4*m
^4-240*I*a^3*x^3+120*I*a*x+m^4+11*a^4*x^4*m^3-4*I*a^3*x^3*m^4-171*I*a^3*x^3*m^2+4*I*a*x*m^4-43*I*a^3*x^3*m^3+5
0*m-240*a^2*x^2-392*a^2*m*x^2+46*a^4*x^4*m^2+96*a^4*x^4*m+10*m^3-6*a^2*x^2*m^4-63*a^2*x^2*m^3+35*m^2+120*a^4*x
^4)/(1+I*a*x)^5+x^m*(I*a)^m*m*(m^4+10*m^3+35*m^2+50*m+24)*LerchPhi(-I*a*x,1,m))+1/40*I*(I*a)^(-m)/a*(-x^m*(I*a
)^m*(a^4*x^4*m^4+a^4*x^4*m^3+20*I*a*x-4*a^4*x^4*m^2+18*I*a^3*x^3*m-6*a^2*x^2*m^4-4*a^4*x^4*m+4*I*a*x*m^4-3*a^2
*x^2*m^3+19*I*a^3*x^3*m^2-4*I*a*m*x+31*a^2*m^2*x^2-3*I*a^3*x^3*m^3+m^4+18*a^2*m*x^2-21*I*a*x*m^2-40*a^2*x^2+I*
a*x*m^3-5*m^2-4*I*a^3*x^3*m^4+4)/(1+I*a*x)^5+x^m*(I*a)^m*(m^2-3*m+2)*m*(m^2+3*m+2)*LerchPhi(-I*a*x,1,m))-1/120
*I*(I*a)^(-m)/a*(-x^m*(I*a)^m*(a^4*x^4*m^4-9*a^4*x^4*m^3+108*I*a^3*x^3*m+26*a^4*x^4*m^2+4*I*a*x*m^4-6*a^2*x^2*
m^4-24*a^4*x^4*m-111*I*a^3*x^3*m^2+57*a^2*x^2*m^3-154*I*a*m*x+37*I*a^3*x^3*m^3-179*a^2*m^2*x^2+129*I*a*x*m^2+m
^4+188*a^2*m*x^2-39*I*a*x*m^3-10*m^3-4*I*a^3*x^3*m^4+35*m^2-50*m+24)/(1+I*a*x)^5+x^m*(I*a)^m*(m^4-10*m^3+35*m^
2-50*m+24)*m*LerchPhi(-I*a*x,1,m))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} x^{2} + 1\right )}^{3} x^{m}}{{\left (i \, a x + 1\right )}^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^6*(a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

integrate((a^2*x^2 + 1)^3*x^m/(I*a*x + 1)^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m\,{\left (a^2\,x^2+1\right )}^3}{{\left (1+a\,x\,1{}\mathrm {i}\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(a^2*x^2 + 1)^3)/(a*x*1i + 1)^6,x)

[Out]

int((x^m*(a^2*x^2 + 1)^3)/(a*x*1i + 1)^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{m}}{a^{6} x^{6} - 6 i a^{5} x^{5} - 15 a^{4} x^{4} + 20 i a^{3} x^{3} + 15 a^{2} x^{2} - 6 i a x - 1}\, dx - \int \frac {3 a^{2} x^{2} x^{m}}{a^{6} x^{6} - 6 i a^{5} x^{5} - 15 a^{4} x^{4} + 20 i a^{3} x^{3} + 15 a^{2} x^{2} - 6 i a x - 1}\, dx - \int \frac {3 a^{4} x^{4} x^{m}}{a^{6} x^{6} - 6 i a^{5} x^{5} - 15 a^{4} x^{4} + 20 i a^{3} x^{3} + 15 a^{2} x^{2} - 6 i a x - 1}\, dx - \int \frac {a^{6} x^{6} x^{m}}{a^{6} x^{6} - 6 i a^{5} x^{5} - 15 a^{4} x^{4} + 20 i a^{3} x^{3} + 15 a^{2} x^{2} - 6 i a x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(1+I*a*x)**6*(a**2*x**2+1)**3,x)

[Out]

-Integral(x**m/(a**6*x**6 - 6*I*a**5*x**5 - 15*a**4*x**4 + 20*I*a**3*x**3 + 15*a**2*x**2 - 6*I*a*x - 1), x) -
Integral(3*a**2*x**2*x**m/(a**6*x**6 - 6*I*a**5*x**5 - 15*a**4*x**4 + 20*I*a**3*x**3 + 15*a**2*x**2 - 6*I*a*x
- 1), x) - Integral(3*a**4*x**4*x**m/(a**6*x**6 - 6*I*a**5*x**5 - 15*a**4*x**4 + 20*I*a**3*x**3 + 15*a**2*x**2
- 6*I*a*x - 1), x) - Integral(a**6*x**6*x**m/(a**6*x**6 - 6*I*a**5*x**5 - 15*a**4*x**4 + 20*I*a**3*x**3 + 15*
a**2*x**2 - 6*I*a*x - 1), x)

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