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]

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} \[ \frac {2 i b \tanh ^{-1}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i} (a+i)^{3/2}}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1-i a) x} \]

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 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 )}{\sqrt {i-a} (i+a)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 120, normalized size = 0.92 \[ -i \left (\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{a x+i x}+\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (a+b x+i)}}{\sqrt {-1+i a} \sqrt {i a+i b x+1}}\right )}{\sqrt {-1-i a} (-1+i a)^{3/2}}\right ) \]

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

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fricas [B]  time = 0.50, 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+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 = 145, normalized size = 1.12 \[ \frac {b \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a + i\right )}} - \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b + a^{2} {\left | b \right |} + {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )} {\left (a i - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+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 = 0.17, size = 236, normalized size = 1.82 \[ -\frac {i b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\sqrt {a^{2}+1}}-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a}{\left (a^{2}+1\right ) x}-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a^{2}+1\right ) x}+\frac {i a^{2} b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}+1\right )^{\frac {3}{2}}}+\frac {a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}+1\right )^{\frac {3}{2}}} \]

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*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/(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)*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)+a*b/(a^2+1)^(3/2)*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)

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maxima [B]  time = 0.32, size = 239, normalized size = 1.84 \[ \frac {a {\left (i \, a + 1\right )} b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {i \, b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )}}{{\left (a^{2} + 1\right )} x} \]

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]

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

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mupad [B]  time = 1.68, size = 218, normalized size = 1.68 \[ \frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )}{{\left (a^2+1\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x\,\left (a^2+1\right )}-\frac {b\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )\,1{}\mathrm {i}}{\sqrt {a^2+1}}+\frac {a^2\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )\,1{}\mathrm {i}}{{\left (a^2+1\right )}^{3/2}}-\frac {a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}\,1{}\mathrm {i}}{x\,\left (a^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*b*atanh((a^2 + a*b*x + 1)/((a^2 + 1)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2)))*1i)/(a^2 + 1)^(3/2) - (a
^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2)/(x*(a^2 + 1)) - (b*log(a*b + (a^2 + 1)/x + ((a^2 + 1)^(1/2)*(a^2 + b^2*x^2 +
2*a*b*x + 1)^(1/2))/x)*1i)/(a^2 + 1)^(1/2) - (a*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2)*1i)/(x*(a^2 + 1)) + (a*b*
atanh((a^2 + a*b*x + 1)/((a^2 + 1)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2))))/(a^2 + 1)^(3/2)

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

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]

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

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