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

Optimal. Leaf size=89 \[ -i \sinh ^{-1}(a+b x)-\frac{2 \sqrt{a+i} \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i}} \]

[Out]

(-I)*ArcSinh[a + b*x] - (2*Sqrt[I + a]*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a -
I*b*x])])/Sqrt[I - a]

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Rubi [A]  time = 0.0651262, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5095, 105, 53, 619, 215, 93, 208} \[ -i \sinh ^{-1}(a+b x)-\frac{2 \sqrt{a+i} \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(I*ArcTan[a + b*x])*x),x]

[Out]

(-I)*ArcSinh[a + b*x] - (2*Sqrt[I + a]*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a -
I*b*x])])/Sqrt[I - a]

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 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0785007, size = 132, normalized size = 1.48 \[ \frac{2 \sqrt [4]{-1} (-i b)^{3/2} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{3/2}}-2 \sqrt{\frac{a+i}{a-i}} \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.116, size = 283, normalized size = 3.2 \begin{align*}{\frac{-i}{i-a}\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}}+{\frac{b}{i-a}\ln \left ({ \left ( ib+ \left ( x-{\frac{i-a}{b}} \right ){b}^{2} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{i}{i-a}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{iab}{i-a}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{i}{i-a}\sqrt{{a}^{2}+1}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) } \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)/x,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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)/x,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.32046, size = 404, normalized size = 4.54 \begin{align*} -\frac{1}{2} \, \sqrt{-\frac{4 \, a + 4 i}{a - i}} \log \left (-b x + \frac{1}{2} \,{\left (i \, a + 1\right )} \sqrt{-\frac{4 \, a + 4 i}{a - i}} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \frac{1}{2} \, \sqrt{-\frac{4 \, a + 4 i}{a - i}} \log \left (-b x + \frac{1}{2} \,{\left (-i \, a - 1\right )} \sqrt{-\frac{4 \, a + 4 i}{a - i}} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + i \, \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \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)/x,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.16925, size = 104, normalized size = 1.17 \begin{align*} \frac{b i \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{{\left | b \right |}} + \frac{2 \,{\left (a + i\right )} \arctan \left (-\frac{{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} i}{\sqrt{a^{2} + 1}}\right )}{\sqrt{a^{2} + 1}} \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)/x,x, algorithm="giac")

[Out]

b*i*log(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/abs(b) + 2*(a + i)*arctan(-(x*abs(b) - sqrt((b*x + a
)^2 + 1))*i/sqrt(a^2 + 1))/sqrt(a^2 + 1)