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

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Rubi [A]  time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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^{-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*}

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Mathematica [A]  time = 0.08, size = 142, normalized size = 1.60 \[ \frac {2 \sqrt [4]{-1} (-i b)^{3/2} \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (a+b x+i)}}{\sqrt {-i b}}\right )}{b^{3/2}}-\frac {2 \sqrt {-1+i a} \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{\sqrt {-1-i a}} \]

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[-1 + I*a]*ArcTanh[(Sqrt[-1 - I*a]*Sqrt[(-I)*(I + a + b*x)])/(Sqrt[-1 + I*a]*Sqrt[1 + I*a + I*b*x])])/Sq
rt[-1 - I*a]

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fricas [B]  time = 0.47, size = 155, normalized size = 1.74 \[ -\frac {1}{2} \, \sqrt {-\frac {4 \, a + 4 i}{a - i}} \log \left (-b x + \frac {1}{2} \, {\left (i \, a + 1\right )} \sqrt {-\frac {4 \, a + 4 i}{a - i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 \, a + 4 i}{a - i}} \log \left (-b x + \frac {1}{2} \, {\left (-i \, a - 1\right )} \sqrt {-\frac {4 \, a + 4 i}{a - i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + i \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \]

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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giac [A]  time = 0.15, size = 77, normalized size = 0.87 \[ \frac {b i \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + \frac {2 \, {\left (a + i\right )} \arctan \left (-\frac {{\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} i}{\sqrt {a^{2} + 1}}\right )}{\sqrt {a^{2} + 1}} \]

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)

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maple [B]  time = 0.18, size = 283, normalized size = 3.18 \[ \frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{i-a}+\frac {i a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right ) \sqrt {b^{2}}}-\frac {i \sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{i-a}-\frac {i \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{i-a}+\frac {b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\left (i-a \right ) \sqrt {b^{2}}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2+1}}{x\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a x + b x^{2} - i x}\, dx \]

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]

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

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