Integrand size = 12, antiderivative size = 52 \[ \int e^{-i \arctan (a+b x)} \, dx=-\frac {i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}+\frac {\text {arcsinh}(a+b x)}{b} \]
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Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5201, 52, 55, 633, 221} \[ \int e^{-i \arctan (a+b x)} \, dx=\frac {\text {arcsinh}(a+b x)}{b}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b} \]
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Rule 52
Rule 55
Rule 221
Rule 633
Rule 5201
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx \\ & = -\frac {i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}+\int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx \\ & = -\frac {i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}+\int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx \\ & = -\frac {i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b^2} \\ & = -\frac {i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}+\frac {\text {arcsinh}(a+b x)}{b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.54 \[ \int e^{-i \arctan (a+b x)} \, dx=\frac {-i \sqrt {1+(a+b x)^2}+\text {arcsinh}(a+b x)}{b} \]
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Time = 0.47 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.33
method | result | size |
risch | \(-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b}+\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}\) | \(69\) |
default | \(-\frac {i \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b}\) | \(125\) |
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Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.15 \[ \int e^{-i \arctan (a+b x)} \, dx=\frac {-i \, a - 2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \, b} \]
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\[ \int e^{-i \arctan (a+b x)} \, dx=- i \int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a + b x - i}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int e^{-i \arctan (a+b x)} \, dx=\frac {\operatorname {arsinh}\left (b x + a\right )}{b} - \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b} \]
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Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int e^{-i \arctan (a+b x)} \, dx=-\frac {\log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} - \frac {i \, \sqrt {{\left (b x + a\right )}^{2} + 1}}{b} \]
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Timed out. \[ \int e^{-i \arctan (a+b x)} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2+1}}{1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}} \,d x \]
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