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

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

[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])])/((I - a)^(3/2)*Sqrt[I + a])

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Rubi [A]  time = 0.0635947, 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} \[ -\frac{\sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{(1+i a) x}-\frac{2 i b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{3/2} \sqrt{a+i}} \]

Antiderivative was successfully verified.

[In]

Int[1/(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])])/((I - a)^(3/2)*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 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 )}{(i-a)^{3/2} \sqrt{i+a}}\\ \end{align*}

Mathematica [A]  time = 0.0529437, size = 114, normalized size = 0.88 \[ \frac{\frac{i \sqrt{a^2+2 a b x+b^2 x^2+1}}{x}-\frac{2 b \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{\sqrt{-1+i a} \sqrt{1+i a}}}{a-i} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.145, size = 602, normalized size = 4.6 \begin{align*}{\frac{-ib}{ \left ( i-a \right ) ^{2}}\sqrt{ \left ( x-{\frac{i-a}{b}} \right ) ^{2}{b}^{2}+2\,i \left ( x-{\frac{i-a}{b}} \right ) b}}+{\frac{{b}^{2}}{ \left ( i-a \right ) ^{2}}\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}{ \left ( i-a \right ) \left ({a}^{2}+1 \right ) x} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{2\,iab}{ \left ( i-a \right ) \left ({a}^{2}+1 \right ) }\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{i{a}^{2}{b}^{2}}{ \left ( i-a \right ) \left ({a}^{2}+1 \right ) }\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{iab}{i-a}\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}}}}+{\frac{i{b}^{2}x}{ \left ( i-a \right ) \left ({a}^{2}+1 \right ) }\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{i{b}^{2}}{ \left ( i-a \right ) \left ({a}^{2}+1 \right ) }\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{ib}{ \left ( i-a \right ) ^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{ia{b}^{2}}{ \left ( i-a \right ) ^{2}}\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{ib}{ \left ( i-a \right ) ^{2}}\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^2,x)

[Out]

-I*b/(I-a)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)+b^2/(I-a)^2*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)/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+2*I/(I-a)
*a*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+I/(I-a)*a^2*b^2/(a^2+1)*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*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)+I/(I-a)*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I/(I-a)*b^2/(a^2+1)*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*b/(I-a)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+I*b^2/(I-a)^2*a*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*b/(I-a)^2*(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{{\left (b x + a\right )}^{2} + 1}}{{\left (i \, b x + i \, a + 1\right )} x^{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)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.24796, 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/(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{\sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x^{2} \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**2,x)

[Out]

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

[Out]

Exception raised: NotImplementedError