3.2.44 \(\int e^{\frac {5}{2} i \text {ArcTan}(a x)} x^m \, dx\) [144]

Optimal. Leaf size=36 \[ \frac {x^{1+m} F_1\left (1+m;\frac {5}{4},-\frac {5}{4};2+m;i a x,-i a x\right )}{1+m} \]

[Out]

x^(1+m)*AppellF1(1+m,-5/4,5/4,2+m,-I*a*x,I*a*x)/(1+m)

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5170, 138} \begin {gather*} \frac {x^{m+1} F_1\left (m+1;\frac {5}{4},-\frac {5}{4};m+2;i a x,-i a x\right )}{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(((5*I)/2)*ArcTan[a*x])*x^m,x]

[Out]

(x^(1 + m)*AppellF1[1 + m, 5/4, -5/4, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

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*} \int e^{\frac {5}{2} i \tan ^{-1}(a x)} x^m \, dx &=\int \frac {x^m (1+i a x)^{5/4}}{(1-i a x)^{5/4}} \, dx\\ &=\frac {x^{1+m} F_1\left (1+m;\frac {5}{4},-\frac {5}{4};2+m;i a x,-i a x\right )}{1+m}\\ \end {align*}

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Mathematica [F]
time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int e^{\frac {5}{2} i \text {ArcTan}(a x)} x^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[E^(((5*I)/2)*ArcTan[a*x])*x^m,x]

[Out]

Integrate[E^(((5*I)/2)*ArcTan[a*x])*x^m, x]

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^m,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x^m,x)

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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)/(a^2*x^2+1)^(1/2))^(5/2)*x^m,x, algorithm="maxima")

[Out]

integrate(x^m*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), 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)/(a^2*x^2+1)^(1/2))^(5/2)*x^m,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)*x**m,x)

[Out]

Timed out

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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)/(a^2*x^2+1)^(1/2))^(5/2)*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 = -0.03 \begin {gather*} \int x^m\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2),x)

[Out]

int(x^m*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2), x)

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