3.90 \(\int e^{-\frac{1}{2} i \tan ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=295 \[ \frac{(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 a^2}+\frac{(1+i a x)^{3/4} \sqrt [4]{1-i a x}}{4 a^2}+\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}-\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2} \]

[Out]

((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(4*a^2) + ((1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(2*a^2) + ArcTan[1 - (Sq
rt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/(4*Sqrt[2]*a^2) - ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*
x)^(1/4)]/(4*Sqrt[2]*a^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)]/(8*Sqrt[2]*a^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)]/(8*Sqrt[2]*a^2)

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Rubi [A]  time = 0.172965, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5062, 80, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac{(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 a^2}+\frac{(1+i a x)^{3/4} \sqrt [4]{1-i a x}}{4 a^2}+\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}-\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(4*a^2) + ((1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/(2*a^2) + ArcTan[1 - (Sq
rt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/(4*Sqrt[2]*a^2) - ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*
x)^(1/4)]/(4*Sqrt[2]*a^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)]/(8*Sqrt[2]*a^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)]/(8*Sqrt[2]*a^2)

Rule 5062

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]

Rule 80

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 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 240

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 211

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 1165

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 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 1162

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 617

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 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])

Rubi steps

\begin{align*} \int e^{-\frac{1}{2} i \tan ^{-1}(a x)} x \, dx &=\int \frac{x \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx\\ &=\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac{i \int \frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \, dx}{4 a}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac{i \int \frac{1}{(1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{8 a}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{2 a^2}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{2 a^2}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac{\operatorname{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 )}{4 a^2}-\frac{\operatorname{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 )}{4 a^2}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}-\frac{\operatorname{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 )}{8 a^2}-\frac{\operatorname{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 )}{8 a^2}+\frac{\operatorname{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 )}{8 \sqrt{2} a^2}+\frac{\operatorname{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 )}{8 \sqrt{2} a^2}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac{\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{\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{\operatorname{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 )}{4 \sqrt{2} a^2}+\frac{\operatorname{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 )}{4 \sqrt{2} a^2}\\ &=\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 a^2}+\frac{(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 a^2}+\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}+\frac{\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{\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}\\ \end{align*}

Mathematica [C]  time = 0.0210959, size = 63, normalized size = 0.21 \[ \frac{(1-i a x)^{5/4} \left (5 (1+i a x)^{3/4}-2^{3/4} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{5}{4},\frac{9}{4},\frac{1}{2} (1-i a x)\right )\right )}{10 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*a*x)^(5/4)*(5*(1 + I*a*x)^(3/4) - 2^(3/4)*Hypergeometric2F1[1/4, 5/4, 9/4, (1 - I*a*x)/2]))/(10*a^2)

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Maple [F]  time = 0.141, size = 0, normalized size = 0. \begin{align*} \int{x{\frac{1}{\sqrt{{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15505, size = 618, normalized size = 2.09 \begin{align*} \frac{2 \, a^{2} \sqrt{\frac{i}{16 \, a^{4}}} \log \left (4 i \, a^{2} \sqrt{\frac{i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt{\frac{i}{16 \, a^{4}}} \log \left (-4 i \, a^{2} \sqrt{\frac{i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt{-\frac{i}{16 \, a^{4}}} \log \left (4 i \, a^{2} \sqrt{-\frac{i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, a^{2} \sqrt{-\frac{i}{16 \, a^{4}}} \log \left (-4 i \, a^{2} \sqrt{-\frac{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 x + 3\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^2*sqrt(1/16*I/a^4)*log(4*I*a^2*sqrt(1/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*a^2*sqrt(1
/16*I/a^4)*log(-4*I*a^2*sqrt(1/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*a^2*sqrt(-1/16*I/a^4)*log(
4*I*a^2*sqrt(-1/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 2*a^2*sqrt(-1/16*I/a^4)*log(-4*I*a^2*sqrt(-
1/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + sqrt(a^2*x^2 + 1)*(-2*I*a*x + 3)*sqrt(I*sqrt(a^2*x^2 + 1)
/(a*x + I)))/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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