\(\int e^{-\frac {5}{2} i \arctan (a x)} \, dx\) [109]

Optimal result

Integrand size = 12, antiderivative size = 299 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=\frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}+\frac {5 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {5 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}+\frac {5 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{2 \sqrt {2} a}-\frac {5 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{2 \sqrt {2} a} \]

[Out]

4*I*(1-I*a*x)^(5/4)/a/(1+I*a*x)^(1/4)+5*I*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/a+5/2*I*arctan(1-(1-I*a*x)^(1/4)*2^(
1/2)/(1+I*a*x)^(1/4))/a*2^(1/2)-5/2*I*arctan(1+(1-I*a*x)^(1/4)*2^(1/2)/(1+I*a*x)^(1/4))/a*2^(1/2)+5/4*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*2^(1/2)-5/4*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*2^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {5169, 49, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=\frac {5 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {5 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}+\frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i (1+i a x)^{3/4} \sqrt [4]{1-i a x}}{a}+\frac {5 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2} a}-\frac {5 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt {2} a} \]

[In]

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

[Out]

((4*I)*(1 - I*a*x)^(5/4))/(a*(1 + I*a*x)^(1/4)) + ((5*I)*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/a + ((5*I)*ArcTa
n[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) - ((5*I)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/
4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) + (((5*I)/2)*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) - (((5*I)/2)*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)

Rule 49

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

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 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 5169

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a x)^{5/4}}{(1+i a x)^{5/4}} \, dx \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}-5 \int \frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}-\frac {5}{2} \int \frac {1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}-\frac {(10 i) \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{a} \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}-\frac {(10 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 )}{a} \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}-\frac {(5 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 )}{a}-\frac {(5 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 )}{a} \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}-\frac {(5 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 )}{2 a}-\frac {(5 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 )}{2 a}+\frac {(5 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 )}{2 \sqrt {2} a}+\frac {(5 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 )}{2 \sqrt {2} a} \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}+\frac {5 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 )}{2 \sqrt {2} a}-\frac {5 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 )}{2 \sqrt {2} a}-\frac {(5 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 )}{\sqrt {2} a}+\frac {(5 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 )}{\sqrt {2} a} \\ & = \frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}+\frac {5 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {5 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}+\frac {5 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 )}{2 \sqrt {2} a}-\frac {5 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 )}{2 \sqrt {2} a} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.13 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=\frac {8 i e^{-\frac {1}{2} i \arctan (a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},-e^{2 i \arctan (a x)}\right )}{a} \]

[In]

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

[Out]

((8*I)*Hypergeometric2F1[-1/4, 2, 3/4, -E^((2*I)*ArcTan[a*x])])/(a*E^((I/2)*ArcTan[a*x]))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.87 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=-\frac {{\left (a^{2} x - i \, a\right )} \sqrt {\frac {25 i}{a^{2}}} \log \left (\frac {1}{5} \, a \sqrt {\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (a^{2} x - i \, a\right )} \sqrt {\frac {25 i}{a^{2}}} \log \left (-\frac {1}{5} \, a \sqrt {\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (a^{2} x - i \, a\right )} \sqrt {-\frac {25 i}{a^{2}}} \log \left (\frac {1}{5} \, a \sqrt {-\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (a^{2} x - i \, a\right )} \sqrt {-\frac {25 i}{a^{2}}} \log \left (-\frac {1}{5} \, a \sqrt {-\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-i \, a x - 9\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, {\left (a^{2} x - i \, a\right )}} \]

[In]

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

[Out]

-1/2*((a^2*x - I*a)*sqrt(25*I/a^2)*log(1/5*a*sqrt(25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (a^2*x -
I*a)*sqrt(25*I/a^2)*log(-1/5*a*sqrt(25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (a^2*x - I*a)*sqrt(-25*
I/a^2)*log(1/5*a*sqrt(-25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + (a^2*x - I*a)*sqrt(-25*I/a^2)*log(-1
/5*a*sqrt(-25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 2*sqrt(a^2*x^2 + 1)*(-I*a*x - 9)*sqrt(I*sqrt(a^2
*x^2 + 1)/(a*x + I)))/(a^2*x - I*a)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate(1/((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:The choice was done assuming 0=[0,0]Warning, replacing 0 by 81, a substitution variable should perhaps be p
urged.Warni

Mupad [F(-1)]

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

[In]

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

[Out]

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