3.2.40 \(\int e^{3 i \text {ArcTan}(a x)} x^m \, dx\) [140]

Optimal. Leaf size=159 \[ -\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};-a^2 x^2\right )}{1+m}+\frac {4 i a x^{2+m} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+m} \]

[Out]

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

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Rubi [A]
time = 0.66, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5168, 6874, 371, 864, 822} \begin {gather*} -\frac {3 x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m+2}+\frac {4 i a x^{m+2} \, _2F_1\left (\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};-a^2 x^2\right )}{m+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a*x])*x^m,x]

[Out]

(-3*x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(a^2*x^2)])/(1 + m) - (I*a*x^(2 + m)*Hypergeometri
c2F1[1/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m) + (4*x^(1 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/
2, -(a^2*x^2)])/(1 + m) + ((4*I)*a*x^(2 + m)*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -(a^2*x^2)])/(2 + m)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 864

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + c*(x/e))*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 5168

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.06, size = 113, normalized size = 0.71 \begin {gather*} -\frac {i x^{1+m} \sqrt {1-i a x} \sqrt {-i+a x} \left (F_1\left (1+m;-\frac {1}{2},\frac {1}{2};2+m;-i a x,i a x\right )-2 F_1\left (1+m;-\frac {1}{2},\frac {3}{2};2+m;-i a x,i a x\right )\right )}{(1+m) \sqrt {1+i a x} \sqrt {i+a x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((3*I)*ArcTan[a*x])*x^m,x]

[Out]

((-I)*x^(1 + m)*Sqrt[1 - I*a*x]*Sqrt[-I + a*x]*(AppellF1[1 + m, -1/2, 1/2, 2 + m, (-I)*a*x, I*a*x] - 2*AppellF
1[1 + m, -1/2, 3/2, 2 + m, (-I)*a*x, I*a*x]))/((1 + m)*Sqrt[1 + I*a*x]*Sqrt[I + a*x])

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Maple [A]
time = 0.08, size = 146, normalized size = 0.92

method	result	size
meijerg	\(\frac {x^{1+m} \hypergeom \left (\left [\frac {3}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {3 i a \,x^{2+m} \hypergeom \left (\left [\frac {3}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m}-\frac {3 a^{2} x^{m +3} \hypergeom \left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {m}{2}+\frac {5}{2}\right ], -a^{2} x^{2}\right )}{m +3}-\frac {i a^{3} x^{4+m} \hypergeom \left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [\frac {m}{2}+3\right ], -a^{2} x^{2}\right )}{4+m}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^m,x,method=_RETURNVERBOSE)

[Out]

x^(1+m)*hypergeom([3/2,1/2+1/2*m],[3/2+1/2*m],-a^2*x^2)/(1+m)+3*I*a/(2+m)*x^(2+m)*hypergeom([3/2,1+1/2*m],[2+1
/2*m],-a^2*x^2)-3*a^2/(m+3)*x^(m+3)*hypergeom([3/2,3/2+1/2*m],[1/2*m+5/2],-a^2*x^2)-I*a^3/(4+m)*x^(4+m)*hyperg
eom([3/2,2+1/2*m],[1/2*m+3],-a^2*x^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \left (\int \frac {i x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(I*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x*x**m/(a**2*x**
2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**3*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sq
rt(a**2*x**2 + 1)), x) + Integral(-3*I*a**2*x**2*x**m/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x
))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3}{{\left (a^2\,x^2+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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