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

Optimal. Leaf size=395 \[ -\frac{\log \left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}-\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}+\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt{2}}-\frac{2 \sqrt [4]{-a+i} \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}}-\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac{2 \sqrt [4]{-a+i} \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}} \]

[Out]

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

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Rubi [A]  time = 0.245591, antiderivative size = 395, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5094, 445, 481, 211, 1165, 628, 1162, 617, 204, 212, 208, 205} \[ -\frac{\log \left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}-\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1+i (a+b x)}}{\sqrt{1-i (a+b x)}}+\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt{2}}-\frac{2 \sqrt [4]{-a+i} \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}}-\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac{2 \sqrt [4]{-a+i} \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}} \]

Antiderivative was successfully verified.

[In]

Int[E^((I/2)*ArcTan[a + b*x])/x,x]

[Out]

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

Rule 5094

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Dist[4/(I^m*n*b^(m + 1)*c^(m + 1)), Sub
st[Int[(x^(2/(I*n))*(1 - I*a*c - (1 + I*a*c)*x^(2/(I*n)))^m)/(1 + x^(2/(I*n)))^(m + 2), x], x, (1 - I*c*(a + b
*x))^((I*n)/2)/(1 + I*c*(a + b*x))^((I*n)/2)], x] /; FreeQ[{a, b, c}, x] && ILtQ[m, 0] && LtQ[-1, I*n, 1]

Rule 445

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[x^(m + n*(p + q
))*(b + a/x^n)^p*(d + c/x^n)^q, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[p, q] &&
NegQ[n]

Rule 481

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

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])

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 208

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

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.0874389, size = 124, normalized size = 0.31 \[ \frac{2}{3} (-i (a+b x+i))^{3/4} \left (\frac{2 (a-i) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )}{(a+i) (i a+i b x+1)^{3/4}}-\sqrt [4]{2} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};-\frac{1}{2} i (a+b x+i)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.215, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sqrt{{(1+i \left ( bx+a \right ) ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(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 \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.44152, size = 1141, normalized size = 2.89 \begin{align*} \frac{1}{2} \, \sqrt{4 i} \log \left (\frac{1}{2} \, \sqrt{4 i} + \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac{1}{2} \, \sqrt{4 i} \log \left (-\frac{1}{2} \, \sqrt{4 i} + \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \frac{1}{2} \, \sqrt{-4 i} \log \left (\frac{1}{2} \, \sqrt{-4 i} + \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac{1}{2} \, \sqrt{-4 i} \log \left (-\frac{1}{2} \, \sqrt{-4 i} + \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}} \log \left (\sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}}\right ) - i \, \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}} \log \left (\sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + i \, \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}}\right ) + i \, \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}} \log \left (\sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - i \, \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}}\right ) + \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}} \log \left (\sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - \left (-\frac{a - i}{a + i}\right )^{\frac{1}{4}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(4*I)*log(1/2*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - 1/2*sqrt(4*I)*log
(-1/2*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 1/2*sqrt(-4*I)*log(1/2*sqrt(-4*I)
+ sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - 1/2*sqrt(-4*I)*log(-1/2*sqrt(-4*I) + sqrt(I*sqrt
(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - (-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a
^2 + 1)/(b*x + a + I)) + (-(a - I)/(a + I))^(1/4)) - I*(-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x^2 + 2*a*
b*x + a^2 + 1)/(b*x + a + I)) + I*(-(a - I)/(a + I))^(1/4)) + I*(-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x
^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - I*(-(a - I)/(a + I))^(1/4)) + (-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (-(a - I)/(a + I))^(1/4))

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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*(b*x+a))/(1+(b*x+a)**2)**(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 \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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