\(\int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx\) [248]

Optimal result

Integrand size = 18, antiderivative size = 344 \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=-\frac {2 (i+a)^{3/4} \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{3/4}}-\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )+\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )-\frac {2 (i+a)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{3/4}}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x} \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}\right )}\right ) \] Output:

-2*(I+a)^(3/4)*arctan((I+a)^(1/4)*(1+I*a+I*b*x)^(1/4)/(I-a)^(1/4)/(1-I*a-I 
*b*x)^(1/4))/(I-a)^(3/4)-2^(1/2)*arctan(1-2^(1/2)*(1-I*a-I*b*x)^(1/4)/(1+I 
*a+I*b*x)^(1/4))+2^(1/2)*arctan(1+2^(1/2)*(1-I*a-I*b*x)^(1/4)/(1+I*a+I*b*x 
)^(1/4))-2*(I+a)^(3/4)*arctanh((I+a)^(1/4)*(1+I*a+I*b*x)^(1/4)/(I-a)^(1/4) 
/(1-I*a-I*b*x)^(1/4))/(I-a)^(3/4)-2^(1/2)*arctanh(2^(1/2)*(1-I*a-I*b*x)^(1 
/4)/(1+I*a+I*b*x)^(1/4)/(1+(1-I*a-I*b*x)^(1/2)/(1+I*a+I*b*x)^(1/2)))
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.37 \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\frac {2 (-i (i+a+b x))^{3/4} \left (\sqrt [4]{2} (1+i a+i b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )-2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{3 (1+i a+i b x)^{3/4}} \] Input:

Integrate[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x),x]
 

Output:

(2*((-I)*(I + a + b*x))^(3/4)*(2^(1/4)*(1 + I*a + I*b*x)^(3/4)*Hypergeomet 
ric2F1[3/4, 3/4, 7/4, (-1/2*I)*(I + a + b*x)] - 2*Hypergeometric2F1[3/4, 1 
, 7/4, (1 + a^2 - I*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/(3*(1 + I*a 
+ I*b*x)^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.96 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.24, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.944, Rules used = {5618, 140, 27, 73, 104, 756, 218, 221, 854, 826, 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.

\(\displaystyle \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx\)

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {(-i a-i b x+1)^{3/4}}{x (i a+i b x+1)^{3/4}}dx\)

\(\Big \downarrow \) 140

\(\displaystyle \int \frac {1-i a}{x \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx-i b \int \frac {1}{\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle (1-i a) \int \frac {1}{x \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx-i b \int \frac {1}{\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx\)

\(\Big \downarrow \) 73

\(\displaystyle (1-i a) \int \frac {1}{x \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx+4 \int \frac {\sqrt {-i a-i b x+1}}{(i a+i b x+1)^{3/4}}d\sqrt [4]{-i a-i b x+1}\)

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x),x]
 

Output:

4*(1 - I*a)*(((-1/2*I)*ArcTan[((I + a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I 
- a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(3/4)*(I + a)^(1/4)) - ((I/ 
2)*ArcTanh[((I + a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a 
 - I*b*x)^(1/4))])/((I - a)^(3/4)*(I + a)^(1/4))) + 4*((-(ArcTan[1 - (Sqrt 
[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2]) + ArcTan[1 
+ (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2])/2 + 
(Log[1 + Sqrt[1 - I*a - I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I* 
a + I*b*x)^(1/4)]/(2*Sqrt[2]) - Log[1 + Sqrt[1 - I*a - I*b*x] + (Sqrt[2]*( 
1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt[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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*d^(m + n)*f^p   Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] 
, x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x 
)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 
0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n, -1]))
 

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 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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

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 5618

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

Maple [F]

\[\int \frac {1}{{\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}^{\frac {3}{2}} x}d x\]

Input:

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)
 

Output:

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (228) = 456\).

Time = 0.17 (sec) , antiderivative size = 629, normalized size of antiderivative = 1.83 \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx =\text {Too large to display} \] Input:

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x, algorithm="fric 
as")
 

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^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log( 
((a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (a - I) 
*(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4))/(a + I)) - 
(-(a^3 + 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a + I)* 
sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (a - I)*(-(a^3 + 
 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4))/(a + I)) + I*(-(a^3 
+ 3*I*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a + I)*sqrt(I* 
sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (I*a + 1)*(-(a^3 + 3*I* 
a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4))/(a + I)) - I*(-(a^3 + 3*I 
*a^2 - 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4)*log(((a + I)*sqrt(I*sqrt( 
b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-I*a - 1)*(-(a^3 + 3*I*a^2 
- 3*a - I)/(a^3 - 3*I*a^2 - 3*a + I))^(1/4))/(a + I))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\text {Timed out} \] Input:

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(3/2)/x,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\int { \frac {1}{x \left (\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x, algorithm="maxi 
ma")
 

Output:

integrate(1/(x*((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))^(3/2)), x)
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/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. \[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {1}{x\,{\left (\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}\right )}^{3/2}} \,d x \] Input:

int(1/(x*((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(3/2)),x)
 

Output:

int(1/(x*((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {e^{-\frac {3}{2} i \arctan (a+b x)}}{x} \, dx=\int \frac {1}{{\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}^{\frac {3}{2}} x}d x \] Input:

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)
 

Output:

int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x,x)