Integrand size = 16, antiderivative size = 130 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=-\frac {\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{(1-i a) x}+\frac {2 i b \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{\sqrt {i-a} (i+a)^{3/2}} \]
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
Time = 0.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=-i \left (\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{i x+a x}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a} (-1+i a)^{3/2}}\right ) \]
(-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)))
Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5618, 105, 104, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 5618 |
\(\displaystyle \int \frac {\sqrt {i a+i b x+1}}{x^2 \sqrt {-i a-i b x+1}}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -\frac {b \int \frac {1}{x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{a+i}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1-i a) x}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {2 b \int \frac {1}{-i a+\frac {(1-i a) (i a+i b x+1)}{-i a-i b x+1}-1}d\frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}{a+i}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1-i a) x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 i b \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {-a+i} (a+i)^{3/2}}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1-i a) x}\) |
-((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))
3.2.68.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[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[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
Time = 0.68 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (i+a \right ) x}+\frac {b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (i+a \right ) \sqrt {a^{2}+1}}\) | \(93\) |
default | \(-\frac {i b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\sqrt {a^{2}+1}}+\left (i a +1\right ) \left (-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a^{2}+1\right ) x}+\frac {a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}+1\right )^{\frac {3}{2}}}\right )\) | \(152\) |
-I*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/(I+a)/x+1/(I+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)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (86) = 172\).
Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.72 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=-\frac {{\left (a + i\right )} \sqrt {\frac {b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b + {\left (a^{3} + i \, a^{2} + a + i\right )} \sqrt {\frac {b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}}}{b}\right ) - {\left (a + i\right )} \sqrt {\frac {b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b - {\left (a^{3} + i \, a^{2} + a + i\right )} \sqrt {\frac {b^{2}}{a^{4} + 2 i \, a^{3} + 2 i \, a - 1}}}{b}\right ) + i \, b x + i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a + i\right )} x} \]
-((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) - (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) + I*b*x + I*sqrt(b^2*x^2 + 2*a*b*x + a ^2 + 1))/((a + I)*x)
\[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=i \left (\int \left (- \frac {i}{x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\right )\, dx + \int \frac {a}{x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx + \int \frac {b}{x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx\right ) \]
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))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (86) = 172\).
Time = 0.18 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.84 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=\frac {a {\left (i \, a + 1\right )} b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {i \, b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )}}{{\left (a^{2} + 1\right )} x} \]
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)
Time = 0.35 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=\frac {b \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a + i\right )}} - \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b + a^{2} {\left | b \right |} + {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )} {\left (i \, a - 1\right )}} \]
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)*(I*a - 1))
Time = 1.85 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.68 \[ \int \frac {e^{i \arctan (a+b x)}}{x^2} \, dx=\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )}{{\left (a^2+1\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x\,\left (a^2+1\right )}-\frac {b\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )\,1{}\mathrm {i}}{\sqrt {a^2+1}}+\frac {a^2\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )\,1{}\mathrm {i}}{{\left (a^2+1\right )}^{3/2}}-\frac {a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}\,1{}\mathrm {i}}{x\,\left (a^2+1\right )} \]
(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)