Integrand size = 16, antiderivative size = 89 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=i \text {arcsinh}(a+b x)-\frac {2 \sqrt {i-a} \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{\sqrt {i+a}} \]
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Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5203, 132, 55, 633, 221, 12, 95, 214} \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=i \text {arcsinh}(a+b x)-\frac {2 \sqrt {-a+i} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{\sqrt {a+i}} \]
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Rule 12
Rule 55
Rule 95
Rule 132
Rule 214
Rule 221
Rule 633
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1+i a+i b x}}{x \sqrt {1-i a-i b x}} \, dx \\ & = (i b) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx+\int \frac {1+i a}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx \\ & = (1+i a) \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx+(i b) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx \\ & = (2 (1+i a)) \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 )+\frac {i \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} \\ & = i \text {arcsinh}(a+b x)-\frac {2 \sqrt {i-a} \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}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\frac {2 (-1)^{3/4} \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{\sqrt {b}}-\frac {2 \sqrt {-1-i a} \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1+i a}} \]
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Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20
method | result | size |
default | \(\frac {i b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {\left (i a +1\right ) \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\sqrt {a^{2}+1}}\) | \(107\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (59) = 118\).
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.62 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\sqrt {-\frac {a - i}{a + i}} \log \left (-b x + {\left (i \, a - 1\right )} \sqrt {-\frac {a - i}{a + i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt {-\frac {a - i}{a + i}} \log \left (-b x + {\left (-i \, a + 1\right )} \sqrt {-\frac {a - i}{a + i}} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - i \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) \]
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\[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=i \left (\int \frac {b}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx + \int \left (- \frac {i}{x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\right )\, dx + \int \frac {a}{x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx\right ) \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (59) = 118\).
Time = 0.18 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.62 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=-\frac {i \, a \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {\operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} + i \, \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right ) \]
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Time = 0.34 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=-\frac {{\left (-i \, a - 1\right )} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {{\left (b x + a\right )}^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {i \, b \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} \]
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Time = 1.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.33 \[ \int \frac {e^{i \arctan (a+b x)}}{x} \, dx=\mathrm {asinh}\left (a+b\,x\right )\,1{}\mathrm {i}-\frac {\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )}{\sqrt {a^2+1}}-\frac {a\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )\,1{}\mathrm {i}}{\sqrt {a^2+1}} \]
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