\(\int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx\) [233]

Optimal result

Integrand size = 18, antiderivative size = 319 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 \sqrt [4]{i-a} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )-\frac {2 \sqrt [4]{i-a} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{\sqrt [4]{i+a}}+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)} \left (1+\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}\right ) \] Output:

-2*(I-a)^(1/4)*arctan((I+a)^(1/4)*(1+I*(b*x+a))^(1/4)/(I-a)^(1/4)/(1-I*(b* 
x+a))^(1/4))/(I+a)^(1/4)-2^(1/2)*arctan(1-2^(1/2)*(1+I*(b*x+a))^(1/4)/(1-I 
*(b*x+a))^(1/4))+2^(1/2)*arctan(1+2^(1/2)*(1+I*(b*x+a))^(1/4)/(1-I*(b*x+a) 
)^(1/4))-2*(I-a)^(1/4)*arctanh((I+a)^(1/4)*(1+I*(b*x+a))^(1/4)/(I-a)^(1/4) 
/(1-I*(b*x+a))^(1/4))/(I+a)^(1/4)+2^(1/2)*arctanh(2^(1/2)*(1+I*(b*x+a))^(1 
/4)/(1-I*(b*x+a))^(1/4)/(1+(1+I*(b*x+a))^(1/2)/(1-I*(b*x+a))^(1/2)))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.39 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\frac {2}{3} (-i (i+a+b x))^{3/4} \left (-\sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )+\frac {2 (-i+a) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )}{(i+a) (1+i a+i b x)^{3/4}}\right ) \] Input:

Integrate[E^((I/2)*ArcTan[a + b*x])/x,x]
 

Output:

(2*((-I)*(I + a + b*x))^(3/4)*(-(2^(1/4)*Hypergeometric2F1[3/4, 3/4, 7/4, 
(-1/2*I)*(I + a + b*x)]) + (2*(-I + a)*Hypergeometric2F1[3/4, 1, 7/4, (1 + 
 a^2 - I*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)])/((I + a)*(1 + I*a + I*b* 
x)^(3/4))))/3
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.37, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {5617, 947, 981, 755, 756, 218, 221, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx\)

\(\Big \downarrow \) 5617

\(\displaystyle 8 \int \frac {1-i (a+b x)}{(i (a+b x)+1) \left (\frac {1-i (a+b x)}{i (a+b x)+1}+1\right ) \left (-i a-\frac {(i a+1) (1-i (a+b x))}{i (a+b x)+1}+1\right )}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\)

\(\Big \downarrow \) 947

\(\displaystyle 8 \int \frac {i (a+b x)+1}{(1-i (a+b x)) \left (\frac {i (a+b x)+1}{1-i (a+b x)}+1\right ) \left (-i a+\frac {(1-i a) (i (a+b x)+1)}{1-i (a+b x)}-1\right )}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\)

\(\Big \downarrow \) 981

\(\displaystyle 8 \left (\frac {1}{2} \int \frac {1}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} (1+i a) \int \frac {1}{-i a+\frac {(1-i a) (i (a+b x)+1)}{1-i (a+b x)}-1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )\)

\(\Big \downarrow \) 755

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \int \frac {\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+1}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} (1+i a) \int \frac {1}{-i a+\frac {(1-i a) (i (a+b x)+1)}{1-i (a+b x)}-1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )\)

\(\Big \downarrow \) 756

\(\displaystyle 8 \left (\frac {1}{2} (1+i a) \left (-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}-\frac {i \int \frac {1}{\sqrt {i-a}+\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \int \frac {\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+1}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )\right )\)

\(\Big \downarrow \) 218

\(\displaystyle 8 \left (\frac {1}{2} (1+i a) \left (-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \int \frac {\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+1}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )\right )\)

\(\Big \downarrow \) 221

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \int \frac {\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+1}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-1}d\left (\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}{\frac {i (a+b x)+1}{1-i (a+b x)}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 8 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}{\frac {\sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}+1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 8 \left (\frac {1}{2} (1+i a) \left (-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}\right )+\frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}+\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}-\frac {\sqrt {2} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}+1\right )}{2 \sqrt {2}}\right )\right )\right )\)

Input:

Int[E^((I/2)*ArcTan[a + b*x])/x,x]
 

Output:

8*(((1 + I*a)*(((-1/2*I)*ArcTan[((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))/(( 
I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(1/4)) - (( 
I/2)*ArcTanh[((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))/((I - a)^(1/4)*(1 - I 
*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(1/4))))/2 + ((-(ArcTan[1 - (S 
qrt[2]*(1 + I*(a + b*x))^(1/4))/(1 - I*(a + b*x))^(1/4)]/Sqrt[2]) + ArcTan 
[1 + (Sqrt[2]*(1 + I*(a + b*x))^(1/4))/(1 - I*(a + b*x))^(1/4)]/Sqrt[2])/2 
 + (-1/2*Log[1 - (Sqrt[2]*(1 + I*(a + b*x))^(1/4))/(1 - I*(a + b*x))^(1/4) 
 + Sqrt[1 + I*(a + b*x)]/Sqrt[1 - I*(a + b*x)]]/Sqrt[2] + Log[1 + (Sqrt[2] 
*(1 + I*(a + b*x))^(1/4))/(1 - I*(a + b*x))^(1/4) + Sqrt[1 + I*(a + b*x)]/ 
Sqrt[1 - I*(a + b*x)]]/(2*Sqrt[2]))/2)/2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 947

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Int[x^(m + n*(p + q))*(b + a/x^n)^p*(d + c/x^n)^q, x] /; Fr 
eeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[p, q] && NegQ[ 
n]
 

rule 981

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), 
 x_Symbol] :> Simp[(-a)*(e^n/(b*c - a*d))   Int[(e*x)^(m - n)/(a + b*x^n), 
x], x] + Simp[c*(e^n/(b*c - a*d))   Int[(e*x)^(m - n)/(c + d*x^n), x], x] / 
; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, 
 m, 2*n - 1]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 5617

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Simp 
[4/(I^m*n*b^(m + 1)*c^(m + 1))   Subst[Int[x^(2/(I*n))*((1 - I*a*c - (1 + I 
*a*c)*x^(2/(I*n)))^m/(1 + x^(2/(I*n)))^(m + 2)), x], x, (1 - I*c*(a + b*x)) 
^(I*(n/2))/(1 + I*c*(a + b*x))^(I*(n/2))], x] /; FreeQ[{a, b, c}, x] && ILt 
Q[m, 0] && LtQ[-1, I*n, 1]
 

Maple [F]

\[\int \frac {\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}{x}d x\]

Input:

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x)
 

Output:

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\frac {1}{2} \, \sqrt {4 i} \log \left (\frac {1}{2} \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac {1}{2} \, \sqrt {4 i} \log \left (-\frac {1}{2} \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + \frac {1}{2} \, \sqrt {-4 i} \log \left (\frac {1}{2} \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \frac {1}{2} \, \sqrt {-4 i} \log \left (-\frac {1}{2} \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) - i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) + i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - i \, \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) + \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}} \log \left (\sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - \left (-\frac {a - i}{a + i}\right )^{\frac {1}{4}}\right ) \] Input:

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x, algorithm="fricas 
")
 

Output:

1/2*sqrt(4*I)*log(1/2*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) 
/(b*x + a + I))) - 1/2*sqrt(4*I)*log(-1/2*sqrt(4*I) + sqrt(I*sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 1/2*sqrt(-4*I)*log(1/2*sqrt(-4*I) + 
 sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - 1/2*sqrt(-4*I) 
*log(-1/2*sqrt(-4*I) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + 
 I))) - (-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 
 1)/(b*x + a + I)) + (-(a - I)/(a + I))^(1/4)) - I*(-(a - I)/(a + I))^(1/4 
)*log(sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + I*(-(a - I 
)/(a + I))^(1/4)) + I*(-(a - I)/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x^2 + 2 
*a*b*x + a^2 + 1)/(b*x + a + I)) - I*(-(a - I)/(a + I))^(1/4)) + (-(a - I) 
/(a + I))^(1/4)*log(sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I) 
) - (-(a - I)/(a + I))^(1/4))
 

Sympy [F]

\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {\sqrt {\frac {i \left (a + b x - i\right )}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}}}{x}\, dx \] Input:

integrate(((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(1/2)/x,x)
 

Output:

Integral(sqrt(I*(a + b*x - I)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1))/x, x)
 

Maxima [F]

\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int { \frac {\sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}}{x} \,d x } \] Input:

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x, algorithm="maxima 
")
 

Output:

integrate(sqrt((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))/x, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}}{x} \,d x \] Input:

int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(1/2)/x,x)
 

Output:

int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(1/2)/x, x)
 

Reduce [F]

\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}{x}d x \] Input:

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x)
 

Output:

int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x,x)