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

Optimal. Leaf size=114 \[ \frac {2 \left (2 m^2+4 m+3\right ) x^{m+1} \, _2F_1(1,m+1;m+2;i a x)}{m+1}+\frac {4 i x^{m+1} \left (a \left (m^2+3 m+3\right ) x+i (m+1)^2\right )}{(m+1) (1-i a x)^2}-\frac {(1+i a x)^2 x^{m+1}}{(m+1) (1-i a x)^2} \]

[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 = 0.10, antiderivative size = 114, 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[E^((6*I)*ArcTan[a*x])*x^m,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 = 0.05, 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[E^((6*I)*ArcTan[a*x])*x^m,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 = 0.59, 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((1+I*a*x)^6/(a^2*x^2+1)^3*x^m,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 \[ \int \frac {{\left (i \, a x + 1\right )}^{6} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.75, size = 748, normalized size = 6.56 \[ \frac {\left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {x^{1+m} \left (a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-a^{2} m^{2} x^{2}+2 a^{2} m \,x^{2}+3 a^{2} x^{2}-m^{2}+4 m +5\right )}{2 \left (1+m \right ) \left (a^{2} x^{2}+1\right )^{2}}+\frac {4 x^{1+m} \left (a^{2}\right )^{\frac {1}{2}+\frac {m}{2}} \left (\frac {1}{16} m^{3}-\frac {3}{16} m^{2}-\frac {1}{16} m +\frac {3}{16}\right ) \Phi \left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}\right )}{4}+\frac {3 i \left (a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (a^{2} m \,x^{2}+m -2\right )}{2 \left (a^{2} x^{2}+1\right )^{2}}-\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (-2+m \right ) m \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{4}\right )}{2 a}-\frac {15 \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 (a^{2} m \,x^{2}+a^{2} x^{2}+m -1\right )}{2 \left (a^{2} x^{2}+1\right )^{2} a^{2}}-\frac {x^{1+m} \left (a^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (1+m \right ) \left (-1+m \right ) \Phi \left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{4 a^{2}}\right )}{4}-\frac {5 i \left (a^{2}\right )^{-\frac {m}{2}} \left (-\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (a^{2} m \,x^{2}+4 a^{2} x^{2}+m +2\right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (2+m \right ) m \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{4}\right )}{a}+\frac {15 \left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (-\frac {x^{1+m} \left (a^{2}\right )^{\frac {5}{2}+\frac {m}{2}} \left (a^{2} m \,x^{2}+5 a^{2} x^{2}+m +3\right )}{2 a^{4} \left (a^{2} x^{2}+1\right )^{2}}+\frac {x^{1+m} \left (a^{2}\right )^{\frac {5}{2}+\frac {m}{2}} \left (m^{2}+4 m +3\right ) \Phi \left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{4 a^{4}}\right )}{4}+\frac {3 i \left (a^{2}\right )^{-\frac {m}{2}} \left (\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (8 a^{4} x^{4}+a^{2} m^{2} x^{2}+8 a^{2} m \,x^{2}+16 a^{2} x^{2}+m^{2}+6 m +8\right )}{2 \left (a^{2} x^{2}+1\right )^{2} m}-\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (m^{2}+6 m +8\right ) \Phi \left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{4}\right )}{2 a}-\frac {\left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {x^{1+m} \left (a^{2}\right )^{\frac {7}{2}+\frac {m}{2}} \left (8 a^{4} x^{4}+a^{2} m^{2} x^{2}+10 a^{2} m \,x^{2}+25 a^{2} x^{2}+m^{2}+8 m +15\right )}{2 \left (a^{2} x^{2}+1\right )^{2} \left (1+m \right ) a^{6}}-\frac {x^{1+m} \left (a^{2}\right )^{\frac {7}{2}+\frac {m}{2}} \left (m^{2}+8 m +15\right ) \Phi \left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{4 a^{6}}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2)^(-1/2-1/2*m)*(1/2/(1+m)*x^(1+m)*(a^2)^(1/2+1/2*m)*(-a^2*m^2*x^2+2*a^2*m*x^2+3*a^2*x^2-m^2+4*m+5)/(a^
2*x^2+1)^2+4/(1+m)*x^(1+m)*(a^2)^(1/2+1/2*m)*(1/16*m^3-3/16*m^2-1/16*m+3/16)*LerchPhi(-a^2*x^2,1,1/2+1/2*m))+3
/2*I/a*(a^2)^(-1/2*m)*(1/2*x^m*(a^2)^(1/2*m)*(a^2*m*x^2+m-2)/(a^2*x^2+1)^2-1/4*x^m*(a^2)^(1/2*m)*(-2+m)*m*Lerc
hPhi(-a^2*x^2,1,1/2*m))-15/4*(a^2)^(-1/2-1/2*m)*(1/2*x^(1+m)*(a^2)^(3/2+1/2*m)*(a^2*m*x^2+a^2*x^2+m-1)/(a^2*x^
2+1)^2/a^2-1/4*x^(1+m)*(a^2)^(3/2+1/2*m)*(1+m)*(-1+m)/a^2*LerchPhi(-a^2*x^2,1,1/2+1/2*m))-5*I/a*(a^2)^(-1/2*m)
*(-1/2*x^m*(a^2)^(1/2*m)*(a^2*m*x^2+4*a^2*x^2+m+2)/(a^2*x^2+1)^2+1/4*x^m*(a^2)^(1/2*m)*(2+m)*m*LerchPhi(-a^2*x
^2,1,1/2*m))+15/4*(a^2)^(-1/2-1/2*m)*(-1/2*x^(1+m)*(a^2)^(5/2+1/2*m)*(a^2*m*x^2+5*a^2*x^2+m+3)/a^4/(a^2*x^2+1)
^2+1/4*x^(1+m)*(a^2)^(5/2+1/2*m)*(m^2+4*m+3)/a^4*LerchPhi(-a^2*x^2,1,1/2+1/2*m))+3/2*I/a*(a^2)^(-1/2*m)*(1/2*x
^m*(a^2)^(1/2*m)*(8*a^4*x^4+a^2*m^2*x^2+8*a^2*m*x^2+16*a^2*x^2+m^2+6*m+8)/(a^2*x^2+1)^2/m-1/4*x^m*(a^2)^(1/2*m
)*(m^2+6*m+8)*LerchPhi(-a^2*x^2,1,1/2*m))-1/4*(a^2)^(-1/2-1/2*m)*(1/2*x^(1+m)*(a^2)^(7/2+1/2*m)*(8*a^4*x^4+a^2
*m^2*x^2+10*a^2*m*x^2+25*a^2*x^2+m^2+8*m+15)/(a^2*x^2+1)^2/(1+m)/a^6-1/4*x^(1+m)*(a^2)^(7/2+1/2*m)*(m^2+8*m+15
)/a^6*LerchPhi(-a^2*x^2,1,1/2+1/2*m))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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