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]

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

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Rubi [A]  time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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 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 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^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*}

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Mathematica [A]  time = 0.08, size = 119, normalized size = 0.92 \[ i \left (\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{(a-i) x}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{(-1-i a)^{3/2} \sqrt {-1+i a}}\right ) \]

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

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fricas [B]  time = 0.49, size = 227, normalized size = 1.75 \[ -\frac {2 \, {\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b + {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - 2 \, {\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b - {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - 2 i \, b x - 2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (2 \, a - 2 i\right )} x} \]

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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giac [A]  time = 0.23, size = 147, normalized size = 1.13 \[ \frac {b \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a - i\right )}} + \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b + a^{2} {\left | b \right |} + {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )} {\left (a i + 1\right )}} \]

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^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.18, size = 602, normalized size = 4.63 \[ \frac {i b \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i-a \right )^{2}}+\frac {i b^{2} a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right )^{2} \sqrt {b^{2}}}-\frac {i b \sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (i-a \right )^{2}}-\frac {i \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (i-a \right ) \left (a^{2}+1\right ) x}+\frac {2 i a b \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i-a \right ) \left (a^{2}+1\right )}+\frac {i a^{2} b^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right ) \left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {i a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (i-a \right ) \sqrt {a^{2}+1}}+\frac {i b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{\left (i-a \right ) \left (a^{2}+1\right )}+\frac {i b^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (i-a \right ) \left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {i b \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{\left (i-a \right )^{2}}+\frac {b^{2} \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\left (i-a \right )^{2} \sqrt {b^{2}}} \]

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^2,x)

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a x^{2} + b x^{3} - i x^{2}}\, dx \]

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**2,x)

[Out]

-I*Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a*x**2 + b*x**3 - I*x**2), x)

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