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

Optimal. Leaf size=267 \[ -\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt{2}}-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

-2*ArcTan[(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]*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)] - 2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*
a*x)^(1/4)] - 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]
+ 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]

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Rubi [A]  time = 0.173959, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {5062, 105, 63, 331, 297, 1162, 617, 204, 1165, 628, 93, 212, 206, 203} \[ -\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{\sqrt{2}}-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTan[(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]*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)] - 2*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*
a*x)^(1/4)] - 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]
+ 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]

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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 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 297

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

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])

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

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

Rule 206

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

Rule 203

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

Rubi steps

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

Mathematica [C]  time = 0.0338281, size = 97, normalized size = 0.36 \[ -\frac{2 (1-i a x)^{3/4} \left (\sqrt [4]{2} (1+i a x)^{3/4} \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{3}{4},\frac{7}{4},\frac{1}{2} (1-i a x)\right )+2 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{a x+i}{-a x+i}\right )\right )}{3 (1+i a x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(1 - I*a*x)^(3/4)*(2^(1/4)*(1 + I*a*x)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, (1 - I*a*x)/2] + 2*Hypergeom
etric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)]))/(3*(1 + I*a*x)^(3/4))

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Maple [F]  time = 0.12, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\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(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70595, size = 660, normalized size = 2.47 \begin{align*} \frac{1}{2} \, \sqrt{4 i} \log \left (\frac{1}{2} \, \sqrt{4 i} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac{1}{2} \, \sqrt{4 i} \log \left (-\frac{1}{2} \, \sqrt{4 i} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + \frac{1}{2} \, \sqrt{-4 i} \log \left (\frac{1}{2} \, \sqrt{-4 i} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac{1}{2} \, \sqrt{-4 i} \log \left (-\frac{1}{2} \, \sqrt{-4 i} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - i \, \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + i\right ) + i \, \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - i\right ) + \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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