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 = 0.38 (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 | \(\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {i a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+i b \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )\) | \(164\) |
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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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Time = 0.77 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int e^{i \arctan (a+b x)} \, dx=\begin {cases} \frac {i \sqrt {\left (a + b x\right )^{2} + 1} + \operatorname {asinh}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\\frac {x \left (i a + 1\right )}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19 \[ \int e^{i \arctan (a+b x)} \, dx=\frac {\operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b} \]
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Time = 0.30 (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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Time = 1.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.87 \[ \int e^{i \arctan (a+b x)} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}\,1{}\mathrm {i}}{b}+\frac {\mathrm {asinh}\left (a+b\,x\right )}{b}+\frac {a\,\mathrm {asinh}\left (a+b\,x\right )\,1{}\mathrm {i}}{b}-\frac {a\,b^2\,\ln \left (\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}+\frac {x\,b^2+a\,b}{\sqrt {b^2}}\right )\,1{}\mathrm {i}}{{\left (b^2\right )}^{3/2}} \]
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