\(\int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx\) [107]

Optimal result

Integrand size = 16, antiderivative size = 371 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=-\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}-\frac {55 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \]

[Out]

-2*I*(1-I*a*x)^(9/4)/a^3/(1+I*a*x)^(1/4)-55/8*I*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a^3-11/4*I*(1-I*a*x)^(5/4)*(1+
I*a*x)^(3/4)/a^3-1/3*I*(1-I*a*x)^(9/4)*(1+I*a*x)^(3/4)/a^3-55/16*I*arctan(1-(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^
(1/4))/a^3*2^(1/2)+55/16*I*arctan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a^3*2^(1/2)-55/32*I*ln(1-(1-I*a*x
)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)^(1/2))/a^3*2^(1/2)+55/32*I*ln(1+(1-I*a*x)^(1/4)*2^(1
/2)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/2)/(1+I*a*x)^(1/2))/a^3*2^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5170, 91, 81, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=-\frac {55 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {i (1+i a x)^{3/4} (1-i a x)^{9/4}}{3 a^3}-\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {11 i (1+i a x)^{3/4} (1-i a x)^{5/4}}{4 a^3}-\frac {55 i (1+i a x)^{3/4} \sqrt [4]{1-i a x}}{8 a^3}-\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}+\frac {55 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3} \]

[In]

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

[Out]

((-2*I)*(1 - I*a*x)^(9/4))/(a^3*(1 + I*a*x)^(1/4)) - (((55*I)/8)*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/a^3 - ((
(11*I)/4)*(1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/a^3 - ((I/3)*(1 - I*a*x)^(9/4)*(1 + I*a*x)^(3/4))/a^3 - (((55*I
)/8)*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) + (((55*I)/8)*ArcTan[1 + (Sqrt[2
]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) - (((55*I)/16)*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1 + I*a*x]
- (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3) + (((55*I)/16)*Log[1 + Sqrt[1 - I*a*x]/Sqrt[1
+ I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a^3)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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*} \text {integral}& = \int \frac {x^2 (1-i a x)^{5/4}}{(1+i a x)^{5/4}} \, dx \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}+\frac {(2 i) \int \frac {(1-i a x)^{5/4} \left (-\frac {5 i a}{2}-\frac {a^2 x}{2}\right )}{\sqrt [4]{1+i a x}} \, dx}{a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {11 \int \frac {(1-i a x)^{5/4}}{\sqrt [4]{1+i a x}} \, dx}{2 a^2} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {55 \int \frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{8 a^2} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {55 \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{16 a^2} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{4 a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 a^3}-\frac {(55 i) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}-\frac {(55 i) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}-\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3} \\ & = -\frac {2 i (1-i a x)^{9/4}}{a^3 \sqrt [4]{1+i a x}}-\frac {55 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{8 a^3}-\frac {11 i (1-i a x)^{5/4} (1+i a x)^{3/4}}{4 a^3}-\frac {i (1-i a x)^{9/4} (1+i a x)^{3/4}}{3 a^3}-\frac {55 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {55 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.25 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=-\frac {\sqrt [4]{1-i a x} (i+a x)^2 \left (-21 i+3 a x+11 i 2^{3/4} \sqrt [4]{1+i a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-i a x)\right )\right )}{9 a^3 \sqrt [4]{1+i a x}} \]

[In]

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

[Out]

-1/9*((1 - I*a*x)^(1/4)*(I + a*x)^2*(-21*I + 3*a*x + (11*I)*2^(3/4)*(1 + I*a*x)^(1/4)*Hypergeometric2F1[1/4, 9
/4, 13/4, (1 - I*a*x)/2]))/(a^3*(1 + I*a*x)^(1/4))

Maple [F]

\[\int \frac {x^{2}}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.80 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=\frac {12 \, {\left (a^{4} x - i \, a^{3}\right )} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, {\left (a^{4} x - i \, a^{3}\right )} \sqrt {\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} \, a^{3} \sqrt {\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, {\left (a^{4} x - i \, a^{3}\right )} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (\frac {8}{55} \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, {\left (a^{4} x - i \, a^{3}\right )} \sqrt {-\frac {3025 i}{64 \, a^{6}}} \log \left (-\frac {8}{55} \, a^{3} \sqrt {-\frac {3025 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (8 i \, a^{3} x^{3} - 26 \, a^{2} x^{2} - 61 i \, a x - 287\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{24 \, {\left (a^{4} x - i \, a^{3}\right )}} \]

[In]

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

[Out]

1/24*(12*(a^4*x - I*a^3)*sqrt(3025/64*I/a^6)*log(8/55*a^3*sqrt(3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x
+ I))) - 12*(a^4*x - I*a^3)*sqrt(3025/64*I/a^6)*log(-8/55*a^3*sqrt(3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 + 1)/(
a*x + I))) - 12*(a^4*x - I*a^3)*sqrt(-3025/64*I/a^6)*log(8/55*a^3*sqrt(-3025/64*I/a^6) + sqrt(I*sqrt(a^2*x^2 +
1)/(a*x + I))) + 12*(a^4*x - I*a^3)*sqrt(-3025/64*I/a^6)*log(-8/55*a^3*sqrt(-3025/64*I/a^6) + sqrt(I*sqrt(a^2
*x^2 + 1)/(a*x + I))) + (8*I*a^3*x^3 - 26*a^2*x^2 - 61*I*a*x - 287)*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)
/(a*x + I)))/(a^4*x - I*a^3)

Sympy [F]

\[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int \frac {x^{2}}{\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(x**2/(I*(a*x - I)/sqrt(a**2*x**2 + 1))**(5/2), x)

Maxima [F]

\[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int { \frac {x^{2}}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=\text {Exception raised: TypeError} \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x^2 \, dx=\int \frac {x^2}{{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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