Integrand size = 14, antiderivative size = 324 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {25 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}-\frac {5 (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}+\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2}+\frac {25 \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^2} \]
-2*(1-I*a*x)^(9/4)/a^2/(1+I*a*x)^(1/4)-25/4*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4 )/a^2-5/2*(1-I*a*x)^(5/4)*(1+I*a*x)^(3/4)/a^2-25/8*arctan(1-(1-I*a*x)^(1/4 )*2^(1/2)/(1+I*a*x)^(1/4))/a^2*2^(1/2)+25/8*arctan(1+(1-I*a*x)^(1/4)*2^(1/ 2)/(1+I*a*x)^(1/4))/a^2*2^(1/2)-25/16*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^2*2^(1/2)+25/16*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^2*2^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.19 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 (1-i a x)^{9/4} \left (-\frac {9}{\sqrt [4]{1+i a x}}+5\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {9}{4},\frac {13}{4},\frac {1}{2} (1-i a x)\right )\right )}{9 a^2} \]
(2*(1 - I*a*x)^(9/4)*(-9/(1 + I*a*x)^(1/4) + 5*2^(3/4)*Hypergeometric2F1[1 /4, 9/4, 13/4, (1 - I*a*x)/2]))/(9*a^2)
Time = 0.43 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5585, 87, 60, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x e^{-\frac {5}{2} i \arctan (a x)} \, dx\) |
| \(\Big \downarrow \) 5585 |
| \(\displaystyle \int \frac {x (1-i a x)^{5/4}}{(1+i a x)^{5/4}}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {5 i \int \frac {(1-i a x)^{5/4}}{\sqrt [4]{i a x+1}}dx}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \int \frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}dx-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {1}{2} \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \int \frac {1}{\sqrt [4]{i a x+1}}d\sqrt [4]{1-i a x}}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \int \frac {1}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \int \frac {\sqrt {1-i a x}+1}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {1-i a x}-1}d\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \int \frac {1-\sqrt {1-i a x}}{2-i a x}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1\right )}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}{\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}+1}d\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{i a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}-\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {2 (1-i a x)^{9/4}}{a^2 \sqrt [4]{1+i a x}}-\frac {5 i \left (\frac {5}{4} \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-i a x}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-i a x}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}-\frac {i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}\right )-\frac {i (1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a}\right )}{a}\) |
(-2*(1 - I*a*x)^(9/4))/(a^2*(1 + I*a*x)^(1/4)) - ((5*I)*(((-1/2*I)*(1 - I* a*x)^(5/4)*(1 + I*a*x)^(3/4))/a + (5*(((-I)*(1 - I*a*x)^(1/4)*(1 + I*a*x)^ (3/4))/a + ((2*I)*((-(ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^( 1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)] /Sqrt[2])/2 + (-1/2*Log[1 + Sqrt[1 - I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/ (1 + I*a*x)^(1/4)]/Sqrt[2] + Log[1 + Sqrt[1 - I*a*x] + (Sqrt[2]*(1 - I*a*x )^(1/4))/(1 + I*a*x)^(1/4)]/(2*Sqrt[2]))/2))/a))/4))/a
3.2.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[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]]))
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[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]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
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] && !Intege rQ[(I*n - 1)/2]
\[\int \frac {x}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}d x\]
Time = 0.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.89 \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} i \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} i \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} i \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, {\left (a^{3} x - i \, a^{2}\right )} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} i \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \sqrt {a^{2} x^{2} + 1} {\left (2 i \, a^{2} x^{2} - 9 \, a x + 43 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{4 \, {\left (a^{3} x - i \, a^{2}\right )}} \]
-1/4*(2*(a^3*x - I*a^2)*sqrt(625/16*I/a^4)*log(4/25*I*a^2*sqrt(625/16*I/a^ 4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*(a^3*x - I*a^2)*sqrt(625/16* I/a^4)*log(-4/25*I*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*(a^3*x - I*a^2)*sqrt(-625/16*I/a^4)*log(4/25*I*a^2*sqrt(-625/16 *I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 2*(a^3*x - I*a^2)*sqrt(-6 25/16*I/a^4)*log(-4/25*I*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1 )/(a*x + I))) - sqrt(a^2*x^2 + 1)*(2*I*a^2*x^2 - 9*a*x + 43*I)*sqrt(I*sqrt (a^2*x^2 + 1)/(a*x + I)))/(a^3*x - I*a^2)
\[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\int \frac {x}{\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \]
\[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\int { \frac {x}{\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ]Warning, replacing 0 by 81, a substitution variable should perhaps be pur ged.Warni
Timed out. \[ \int e^{-\frac {5}{2} i \arctan (a x)} x \, dx=\int \frac {x}{{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \]