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

Optimal. Leaf size=130 \[ \frac{2 i b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i} (a+i)^{3/2}}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{(1-i a) x} \]

[Out]

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

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Rubi [A]  time = 0.0640981, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5095, 94, 93, 208} \[ \frac{2 i b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i} (a+i)^{3/2}}-\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{(1-i a) x} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])/x^2,x]

[Out]

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

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 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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^2} \, dx &=\int \frac{\sqrt{1+i a+i b x}}{x^2 \sqrt{1-i a-i b x}} \, dx\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{(1-i a) x}-\frac{b \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{i+a}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{(1-i a) x}-\frac{(2 b) \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+a}\\ &=-\frac{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{(1-i a) x}+\frac{2 i b \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} (i+a)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0819284, size = 115, normalized size = 0.88 \[ -i \left (\frac{\sqrt{a^2+2 a b x+b^2 x^2+1}}{a x+i x}+\frac{2 b \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{(-1+i a)^{3/2} \sqrt{1+i a}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.109, size = 236, normalized size = 1.8 \begin{align*}{\frac{-ia}{ \left ({a}^{2}+1 \right ) x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{1}{ \left ({a}^{2}+1 \right ) x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{i{a}^{2}b\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 ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{ab\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 ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{ib\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 ){\frac{1}{\sqrt{{a}^{2}+1}}}} \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)/x^2,x)

[Out]

-I/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-1/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+I*a^2*b/(a^2+1)^(3/2)*l
n((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)+a*b/(a^2+1)^(3/2)*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)-I*b/(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: ValueError} \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)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.7045, size = 545, normalized size = 4.19 \begin{align*} -\frac{2 \,{\left (a + i\right )} \sqrt{\frac{b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}} x \log \left (-\frac{b^{2} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b +{\left (a^{3} + i \, a^{2} + a + i\right )} \sqrt{\frac{b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}}}{b}\right ) - 2 \,{\left (a + i\right )} \sqrt{\frac{b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}} x \log \left (-\frac{b^{2} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b -{\left (a^{3} + i \, a^{2} + a + i\right )} \sqrt{\frac{b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}}}{b}\right ) + 2 i \, b x + 2 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (2 \, a + 2 i\right )} 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)/x^2,x, algorithm="fricas")

[Out]

-(2*(a + I)*sqrt(b^2/(a^4 + 2*I*a^3 + 2*I*a - 1))*x*log(-(b^2*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b + (a^3 +
I*a^2 + a + I)*sqrt(b^2/(a^4 + 2*I*a^3 + 2*I*a - 1)))/b) - 2*(a + I)*sqrt(b^2/(a^4 + 2*I*a^3 + 2*I*a - 1))*x*
log(-(b^2*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b - (a^3 + I*a^2 + a + I)*sqrt(b^2/(a^4 + 2*I*a^3 + 2*I*a - 1)
))/b) + 2*I*b*x + 2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((2*a + 2*I)*x)

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

[Out]

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

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

[Out]

Exception raised: NotImplementedError