3.234 \(\int \frac{e^{-\frac{3}{2} i \tan ^{-1}(a+b x)}}{x} \, dx\)

Optimal. Leaf size=427 \[ \frac{\log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt{2}}-\frac{\log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt{2}}-\frac{2 (a+i)^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{3/4}}-\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )+\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )-\frac{2 (a+i)^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{3/4}} \]

[Out]

(-2*(I + a)^(3/4)*ArcTan[((I + a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/(I
- a)^(3/4) - Sqrt[2]*ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)] + Sqrt[2]*ArcTan[1
+ (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)] - (2*(I + a)^(3/4)*ArcTanh[((I + a)^(1/4)*(1 + I*
a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/(I - a)^(3/4) + Log[1 + Sqrt[1 - I*a - I*b*x]/Sqrt
[1 + I*a + I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2] - Log[1 + Sqrt[1 - I*a
- I*b*x]/Sqrt[1 + I*a + I*b*x] + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2]

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Rubi [A]  time = 0.23995, antiderivative size = 427, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {5095, 105, 63, 331, 297, 1162, 617, 204, 1165, 628, 93, 212, 208, 205} \[ \frac{\log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt{2}}-\frac{\log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt{2}}-\frac{2 (a+i)^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{3/4}}-\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )+\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )-\frac{2 (a+i)^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x),x]

[Out]

(-2*(I + a)^(3/4)*ArcTan[((I + a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/(I
- a)^(3/4) - Sqrt[2]*ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)] + Sqrt[2]*ArcTan[1
+ (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)] - (2*(I + a)^(3/4)*ArcTanh[((I + a)^(1/4)*(1 + I*
a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/(I - a)^(3/4) + Log[1 + Sqrt[1 - I*a - I*b*x]/Sqrt
[1 + I*a + I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2] - Log[1 + Sqrt[1 - I*a
- I*b*x]/Sqrt[1 + I*a + I*b*x] + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2]

Rule 5095

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

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

Mathematica [C]  time = 0.0361035, size = 128, normalized size = 0.3 \[ \frac{2 (-i (a+b x+i))^{3/4} \left (\sqrt [4]{2} (i a+i b x+1)^{3/4} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};-\frac{1}{2} i (a+b x+i)\right )-2 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )\right )}{3 (i a+i b x+1)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x),x]

[Out]

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

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Maple [F]  time = 0.234, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ({(1+i \left ( bx+a \right ) ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)

[Out]

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\frac{i \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x, algorithm="maxima")

[Out]

integrate(1/(x*((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))^(3/2)), x)

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Fricas [B]  time = 2.85048, size = 1673, normalized size = 3.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/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)*lo
g(-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*sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + (-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*
log(((a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (a - I)*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^
3 - 3*I*a^2 - 3*a + I))^(1/4))/(a + I)) - (-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a
+ I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (a - I)*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I
*a^2 - 3*a + I))^(1/4))/(a + I)) + I*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a + I)
*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (I*a + 1)*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^
2 - 3*a + I))^(1/4))/(a + I)) - I*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a + I)*sq
rt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-I*a - 1)*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2
- 3*a + I))^(1/4))/(a + I))

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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/((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(3/2)/x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\frac{i \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x, algorithm="giac")

[Out]

integrate(1/(x*((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))^(3/2)), x)