3.130 \(\int e^{\frac {1}{4} i \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=674 \[ \frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {i \sqrt {2-\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2-\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}+\frac {i \sqrt {2+\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}-\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a} \]

[Out]

I*(1-I*a*x)^(7/8)*(1+I*a*x)^(1/8)/a-1/4*I*arctan((-2*(1-I*a*x)^(1/8)/(1+I*a*x)^(1/8)+(2+2^(1/2))^(1/2))/(2-2^(
1/2))^(1/2))*(2-2^(1/2))^(1/2)/a+1/4*I*arctan((2*(1-I*a*x)^(1/8)/(1+I*a*x)^(1/8)+(2+2^(1/2))^(1/2))/(2-2^(1/2)
)^(1/2))*(2-2^(1/2))^(1/2)/a+1/8*I*ln(1+(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)-(1-I*a*x)^(1/8)*(2-2^(1/2))^(1/2)/(1+I
*a*x)^(1/8))*(2-2^(1/2))^(1/2)/a-1/8*I*ln(1+(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/8)*(2-2^(1/2))^(1/2)/
(1+I*a*x)^(1/8))*(2-2^(1/2))^(1/2)/a-1/4*I*arctan((-2*(1-I*a*x)^(1/8)/(1+I*a*x)^(1/8)+(2-2^(1/2))^(1/2))/(2+2^
(1/2))^(1/2))*(2+2^(1/2))^(1/2)/a+1/4*I*arctan((2*(1-I*a*x)^(1/8)/(1+I*a*x)^(1/8)+(2-2^(1/2))^(1/2))/(2+2^(1/2
))^(1/2))*(2+2^(1/2))^(1/2)/a+1/8*I*ln(1+(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)-(1-I*a*x)^(1/8)*(2+2^(1/2))^(1/2)/(1+
I*a*x)^(1/8))*(2+2^(1/2))^(1/2)/a-1/8*I*ln(1+(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)+(1-I*a*x)^(1/8)*(2+2^(1/2))^(1/2)
/(1+I*a*x)^(1/8))*(2+2^(1/2))^(1/2)/a

________________________________________________________________________________________

Rubi [A]  time = 0.43, antiderivative size = 674, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {5061, 50, 63, 331, 299, 1122, 1169, 634, 618, 204, 628} \[ \frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {i \sqrt {2-\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2-\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}+\frac {i \sqrt {2+\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \log \left (\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}+1\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}-\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[E^((I/4)*ArcTan[a*x]),x]

[Out]

(I*(1 - I*a*x)^(7/8)*(1 + I*a*x)^(1/8))/a - ((I/4)*Sqrt[2 + Sqrt[2]]*ArcTan[(Sqrt[2 - Sqrt[2]] - (2*(1 - I*a*x
)^(1/8))/(1 + I*a*x)^(1/8))/Sqrt[2 + Sqrt[2]]])/a - ((I/4)*Sqrt[2 - Sqrt[2]]*ArcTan[(Sqrt[2 + Sqrt[2]] - (2*(1
- I*a*x)^(1/8))/(1 + I*a*x)^(1/8))/Sqrt[2 - Sqrt[2]]])/a + ((I/4)*Sqrt[2 + Sqrt[2]]*ArcTan[(Sqrt[2 - Sqrt[2]]
+ (2*(1 - I*a*x)^(1/8))/(1 + I*a*x)^(1/8))/Sqrt[2 + Sqrt[2]]])/a + ((I/4)*Sqrt[2 - Sqrt[2]]*ArcTan[(Sqrt[2 +
Sqrt[2]] + (2*(1 - I*a*x)^(1/8))/(1 + I*a*x)^(1/8))/Sqrt[2 - Sqrt[2]]])/a + ((I/8)*Sqrt[2 - Sqrt[2]]*Log[1 + (
1 - I*a*x)^(1/4)/(1 + I*a*x)^(1/4) - (Sqrt[2 - Sqrt[2]]*(1 - I*a*x)^(1/8))/(1 + I*a*x)^(1/8)])/a - ((I/8)*Sqrt
[2 - Sqrt[2]]*Log[1 + (1 - I*a*x)^(1/4)/(1 + I*a*x)^(1/4) + (Sqrt[2 - Sqrt[2]]*(1 - I*a*x)^(1/8))/(1 + I*a*x)^
(1/8)])/a + ((I/8)*Sqrt[2 + Sqrt[2]]*Log[1 + (1 - I*a*x)^(1/4)/(1 + I*a*x)^(1/4) - (Sqrt[2 + Sqrt[2]]*(1 - I*a
*x)^(1/8))/(1 + I*a*x)^(1/8)])/a - ((I/8)*Sqrt[2 + Sqrt[2]]*Log[1 + (1 - I*a*x)^(1/4)/(1 + I*a*x)^(1/4) + (Sqr
t[2 + Sqrt[2]]*(1 - I*a*x)^(1/8))/(1 + I*a*x)^(1/8)])/a

Rule 50

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 63

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 204

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

Rule 299

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b,
4]]}, Dist[s^3/(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] - Dist[s^3/
(2*Sqrt[2]*b*r), Int[x^(m - n/4)/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] && IGt
Q[n/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && GtQ[a/b, 0]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1122

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d^3*(d*x)^(m - 3)*(a + b*
x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 1)), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 5061

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*} \int e^{\frac {1}{4} i \tan ^{-1}(a x)} \, dx &=\int \frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}} \, dx\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {1}{4} \int \frac {1}{\sqrt [8]{1-i a x} (1+i a x)^{7/8}} \, dx\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {x^6}{\left (2-x^8\right )^{7/8}} \, dx,x,\sqrt [8]{1-i a x}\right )}{a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {x^6}{1+x^8} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {i \operatorname {Subst}\left (\int \frac {x^4}{1-\sqrt {2} x^2+x^4} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{\sqrt {2} a}-\frac {i \operatorname {Subst}\left (\int \frac {x^4}{1+\sqrt {2} x^2+x^4} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{\sqrt {2} a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}-\frac {i \operatorname {Subst}\left (\int \frac {1-\sqrt {2} x^2}{1-\sqrt {2} x^2+x^4} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{\sqrt {2} a}+\frac {i \operatorname {Subst}\left (\int \frac {1+\sqrt {2} x^2}{1+\sqrt {2} x^2+x^4} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{\sqrt {2} a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2-\sqrt {2}}-\left (1-\sqrt {2}\right ) x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 \sqrt {2 \left (2-\sqrt {2}\right )} a}+\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2-\sqrt {2}}+\left (1-\sqrt {2}\right ) x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 \sqrt {2 \left (2-\sqrt {2}\right )} a}-\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2+\sqrt {2}}-\left (1+\sqrt {2}\right ) x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 \sqrt {2 \left (2+\sqrt {2}\right )} a}-\frac {i \operatorname {Subst}\left (\int \frac {\sqrt {2+\sqrt {2}}+\left (1+\sqrt {2}\right ) x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 \sqrt {2 \left (2+\sqrt {2}\right )} a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {\left (i \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{4 a}+\frac {\left (i \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{4 a}+\frac {\left (i \sqrt {2-\sqrt {2}}\right ) \operatorname {Subst}\left (\int \frac {-\sqrt {2-\sqrt {2}}+2 x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {\left (i \sqrt {2-\sqrt {2}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2-\sqrt {2}}+2 x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}+\frac {\left (i \sqrt {2+\sqrt {2}}\right ) \operatorname {Subst}\left (\int \frac {-\sqrt {2+\sqrt {2}}+2 x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {\left (i \sqrt {2+\sqrt {2}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2+\sqrt {2}}+2 x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}+\frac {\left (i \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{4 a}+\frac {\left (i \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx,x,\frac {\sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{4 a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}+\frac {i \sqrt {2-\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {i \sqrt {2-\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}+\frac {i \sqrt {2+\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {\left (i \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,-\sqrt {2+\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 a}-\frac {\left (i \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,\sqrt {2+\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 a}-\frac {\left (i \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,-\sqrt {2-\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 a}-\frac {\left (i \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )}\right ) \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,\sqrt {2-\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{2 a}\\ &=\frac {i (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{a}-\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}-\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2+\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+\frac {2 \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}}{\sqrt {2-\sqrt {2}}}\right )}{4 a}+\frac {i \sqrt {2-\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {i \sqrt {2-\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}+\frac {i \sqrt {2+\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}-\frac {i \sqrt {2+\sqrt {2}} \log \left (1+\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{1-i a x}}{\sqrt [8]{1+i a x}}\right )}{8 a}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 41, normalized size = 0.06 \[ -\frac {16 i e^{\frac {9}{4} i \tan ^{-1}(a x)} \, _2F_1\left (\frac {9}{8},2;\frac {17}{8};-e^{2 i \tan ^{-1}(a x)}\right )}{9 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((I/4)*ArcTan[a*x]),x]

[Out]

(((-16*I)/9)*E^(((9*I)/4)*ArcTan[a*x])*Hypergeometric2F1[9/8, 2, 17/8, -E^((2*I)*ArcTan[a*x])])/a

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fricas [A]  time = 0.68, size = 383, normalized size = 0.57 \[ \frac {-i \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (4 \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) + a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (4 i \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) - a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (-4 i \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) + i \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (-4 \, a \left (\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) - i \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (4 \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) + a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (4 i \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) - a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (-4 i \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) + i \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} \log \left (-4 \, a \left (-\frac {i}{256 \, a^{4}}\right )^{\frac {1}{4}} + \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}\right ) + {\left (a x + i\right )} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4),x, algorithm="fricas")

[Out]

(-I*a*(1/256*I/a^4)^(1/4)*log(4*a*(1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)) + a*(1/256*I/a^
4)^(1/4)*log(4*I*a*(1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)) - a*(1/256*I/a^4)^(1/4)*log(-4
*I*a*(1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)) + I*a*(1/256*I/a^4)^(1/4)*log(-4*a*(1/256*I/
a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)) - I*a*(-1/256*I/a^4)^(1/4)*log(4*a*(-1/256*I/a^4)^(1/4) +
(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)) + a*(-1/256*I/a^4)^(1/4)*log(4*I*a*(-1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x
^2 + 1)/(a*x + I))^(1/4)) - a*(-1/256*I/a^4)^(1/4)*log(-4*I*a*(-1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x
+ I))^(1/4)) + I*a*(-1/256*I/a^4)^(1/4)*log(-4*a*(-1/256*I/a^4)^(1/4) + (I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4)
) + (a*x + I)*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4))/a

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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(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for
the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc in
dex_m operator + Error: Bad Argument ValueDone

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {1}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4),x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4),x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4),x)

[Out]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt [4]{\frac {i a x + 1}{\sqrt {a^{2} x^{2} + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/4),x)

[Out]

Integral(((I*a*x + 1)/sqrt(a**2*x**2 + 1))**(1/4), x)

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