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

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Rubi [A]  time = 0.24, antiderivative size = 427, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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 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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.04, size = 128, normalized size = 0.30 \[ \frac {2 (-i (a+b x+i))^{3/4} \left (\sqrt [4]{2} (i a+i b x+1)^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {1}{2} i (a+b x+i)\right )-2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )\right )}{3 (i a+i b x+1)^{3/4}} \]

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, (-1/2*I)*(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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fricas [B]  time = 0.54, size = 629, normalized size = 1.47 \[ -\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^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (a - i\right )} \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}}}{a + i}\right ) - \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (a - i\right )} \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}}}{a + i}\right ) + i \, \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (i \, a + 1\right )} \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}}}{a + i}\right ) - i \, \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (-i \, a - 1\right )} \left (-\frac {a^{3} + 3 i \, a^{2} - 3 \, a - i}{a^{3} - 3 i \, a^{2} - 3 \, a + i}\right )^{\frac {1}{4}}}{a + i}\right ) \]

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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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/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,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=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch f
or the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.c
c index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial wi
th parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Ba
d Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be w
rong.The choice was done assuming 0=[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 assum
ing 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the roo
t of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m
operator + Error: Bad Argument ValueEvaluation time: 1.81Done

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )^{\frac {3}{2}} x}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x\,{\left (\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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