Integrand size = 18, antiderivative size = 395 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 \sqrt [4]{i-a} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac {2 \sqrt [4]{i-a} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {\log \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}} \]
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Time = 0.18 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5202, 456, 492, 217, 1179, 642, 1176, 631, 210, 218, 214, 211} \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 \sqrt [4]{-a+i} \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac {2 \sqrt [4]{-a+i} \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{a+i}}-\frac {\log \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt {2}}+\frac {\log \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt {2}} \]
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Rule 210
Rule 211
Rule 214
Rule 217
Rule 218
Rule 456
Rule 492
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 5202
Rubi steps \begin{align*} \text {integral}& = 8 \text {Subst}\left (\int \frac {1}{\left (1+\frac {1}{x^4}\right ) \left (1-i a-\frac {1+i a}{x^4}\right ) x^4} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = 8 \text {Subst}\left (\int \frac {x^4}{\left (1+x^4\right ) \left (-1-i a+(1-i a) x^4\right )} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+(4 (1+i a)) \text {Subst}\left (\int \frac {1}{-1-i a+(1-i a) x^4} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = 2 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+2 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\left (2 \sqrt {i-a}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {i-a}-\sqrt {i+a} x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\left (2 \sqrt {i-a}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {i-a}+\sqrt {i+a} x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i-a} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {2 \sqrt [4]{i-a} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}+\text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i-a} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {2 \sqrt [4]{i-a} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {\log \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}}+\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right ) \\ & = -\frac {2 \sqrt [4]{i-a} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac {2 \sqrt [4]{i-a} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\frac {\log \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{\sqrt {2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.31 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\frac {2}{3} (-i (i+a+b x))^{3/4} \left (-\sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )+\frac {2 (-i+a) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )}{(i+a) (1+i a+i b x)^{3/4}}\right ) \]
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\[\int \frac {\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}{x}d x\]
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none
Time = 0.28 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\frac {1}{2} \, \sqrt {4 i} \log \left (\frac {1}{2} \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac {1}{2} \, \sqrt {4 i} \log \left (-\frac {1}{2} \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \frac {1}{2} \, \sqrt {-4 i} \log \left (\frac {1}{2} \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac {1}{2} \, \sqrt {-4 i} \log \left (-\frac {1}{2} \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) - i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) + i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) + \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) \]
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\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {\sqrt {\frac {i \left (a + b x - i\right )}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}}}{x}\, dx \]
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\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int { \frac {\sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}}{x} \,d x } \]
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Exception generated. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}}{x} \,d x \]
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