Integrand size = 14, antiderivative size = 110 \[ \int e^{-i \arctan (a+b x)} x \, dx=\frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \text {arcsinh}(a+b x)}{2 b^2} \]
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Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5203, 81, 52, 55, 633, 221} \[ \int e^{-i \arctan (a+b x)} x \, dx=\frac {(-2 a+i) \text {arcsinh}(a+b x)}{2 b^2}+\frac {\sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{2 b^2}+\frac {(1+2 i a) \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{2 b^2} \]
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Rule 52
Rule 55
Rule 81
Rule 221
Rule 633
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {x \sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx \\ & = \frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx}{2 b} \\ & = \frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b} \\ & = \frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b} \\ & = \frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{4 b^3} \\ & = \frac {(1+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}+\frac {(i-2 a) \text {arcsinh}(a+b x)}{2 b^2} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int e^{-i \arctan (a+b x)} x \, dx=\frac {\sqrt {1+i a+i b x} \left (2-i a+a^2-3 i b x-b^2 x^2\right )}{2 b^2 \sqrt {-i (i+a+b x)}}+\frac {(-1)^{3/4} (1+2 i a) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{5/2}} \]
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Time = 0.52 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {i \left (-b x +a -2 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {\left (-i+2 a \right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b \sqrt {b^{2}}}\) | \(87\) |
default | \(-\frac {i \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}+\frac {\left (i a +1\right ) \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b^{2}}\) | \(237\) |
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Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int e^{-i \arctan (a+b x)} x \, dx=\frac {3 i \, a^{2} + 4 \, {\left (2 \, a - i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, b x - i \, a - 2\right )} + 4 \, a}{8 \, b^{2}} \]
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\[ \int e^{-i \arctan (a+b x)} x \, dx=- i \int \frac {x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a + b x - i}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int e^{-i \arctan (a+b x)} x \, dx=-\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b} - \frac {a \operatorname {arsinh}\left (b x + a\right )}{b^{2}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{2}} + \frac {i \, \operatorname {arsinh}\left (b x + a\right )}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int e^{-i \arctan (a+b x)} x \, dx=-\frac {1}{2} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (\frac {i \, x}{b} + \frac {-i \, a b - 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a - i\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b {\left | b \right |}} \]
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Timed out. \[ \int e^{-i \arctan (a+b x)} x \, dx=\int \frac {x\,\sqrt {{\left (a+b\,x\right )}^2+1}}{1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}} \,d x \]
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