3.143 \(\int e^{-3 i \tan ^{-1}(a x)} x^m \, dx\)

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

[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 = 0.70, 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 = {5060, 6742, 364, 850, 808} \[ -\frac {3 x^{m+1} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \text {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}+\frac {i a x^{m+2} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}-\frac {4 i a x^{m+2} \text {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^((3*I)*ArcTan[a*x]),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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

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 850

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 5060

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 6742

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]  time = 0.08, size = 113, normalized size = 0.71 \[ \frac {i \sqrt {1+i a x} \sqrt {a x+i} x^{m+1} \left (F_1\left (m+1;\frac {1}{2},-\frac {1}{2};m+2;-i a x,i a x\right )-2 F_1\left (m+1;\frac {3}{2},-\frac {1}{2};m+2;-i a x,i a x\right )\right )}{(m+1) \sqrt {1-i a x} \sqrt {a x-i}} \]

Warning: Unable to verify antiderivative.

[In]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} x^{2} + 1} {\left (i \, a x - 1\right )} x^{m}}{a^{2} x^{2} - 2 i \, a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(1+I*a*x)^3*(a^2*x^2+1)^(3/2),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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (i a x +1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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