Integrand size = 16, antiderivative size = 130 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=-\frac {\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{(1+i a) x}-\frac {2 i b \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{3/2} \sqrt {i+a}} \]
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Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5203, 96, 95, 214} \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=-\frac {2 i b \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{3/2} \sqrt {a+i}}-\frac {\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{(1+i a) x} \]
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Rule 95
Rule 96
Rule 214
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \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) \text {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 \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 )}{(i-a)^{3/2} \sqrt {i+a}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=i \left (\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{(-i+a) x}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{(-1-i a)^{3/2} \sqrt {-1+i a}}\right ) \]
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Time = 0.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a -i\right ) x}+\frac {b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a -i\right ) \sqrt {a^{2}+1}}\) | \(93\) |
default | \(\frac {i \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {a b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{a^{2}+1}+\frac {2 b^{2} \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{a^{2}+1}\right )}{i-a}+\frac {i b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{\left (i-a \right )^{2}}-\frac {i b \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{\left (i-a \right )^{2}}\) | \(546\) |
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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 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=-\frac {{\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b + {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - {\left (a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}} x \log \left (-\frac {b^{2} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b - {\left (a^{3} - i \, a^{2} + a - i\right )} \sqrt {\frac {b^{2}}{a^{4} - 2 i \, a^{3} - 2 i \, a - 1}}}{b}\right ) - i \, b x - i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a - i\right )} x} \]
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\[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=- i \int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a x^{2} + b x^{3} - i x^{2}}\, dx \]
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\[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {\sqrt {{\left (b x + a\right )}^{2} + 1}}{{\left (i \, b x + i \, a + 1\right )} x^{2}} \,d x } \]
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none
Time = 0.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=\frac {b \log \left (\frac {{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | 2 \, x {\left | b \right |} - 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1} {\left (a - i\right )}} - \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a b + a^{2} {\left | b \right |} + {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} - a^{2} - 1\right )} {\left (-i \, a - 1\right )}} \]
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Timed out. \[ \int \frac {e^{-i \arctan (a+b x)}}{x^2} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2+1}}{x^2\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )} \,d x \]
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