\(\int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx\) [148]

Optimal result

Integrand size = 16, antiderivative size = 295 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=-\frac {i a (1-i a x)^{7/8} \sqrt [8]{1+i a x}}{8 x}-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}+\frac {1}{16} a^2 \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{16 \sqrt {2}}+\frac {1}{16} a^2 \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )+\frac {a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x} \left (1+\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )}\right )}{16 \sqrt {2}} \] Output:

-1/8*I*a*(1-I*a*x)^(7/8)*(1+I*a*x)^(1/8)/x-1/2*(1-I*a*x)^(7/8)*(1+I*a*x)^( 
9/8)/x^2+1/16*a^2*arctan((1+I*a*x)^(1/8)/(1-I*a*x)^(1/8))-1/32*a^2*arctan( 
1-2^(1/2)*(1+I*a*x)^(1/8)/(1-I*a*x)^(1/8))*2^(1/2)+1/32*a^2*arctan(1+2^(1/ 
2)*(1+I*a*x)^(1/8)/(1-I*a*x)^(1/8))*2^(1/2)+1/16*a^2*arctanh((1+I*a*x)^(1/ 
8)/(1-I*a*x)^(1/8))+1/32*a^2*arctanh(2^(1/2)*(1+I*a*x)^(1/8)/(1-I*a*x)^(1/ 
8)/(1+(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4)))*2^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.28 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\frac {(1-i a x)^{7/8} \left (7 \left (-4-9 i a x+5 a^2 x^2\right )+2 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{8},1,\frac {15}{8},\frac {i+a x}{i-a x}\right )\right )}{56 x^2 (1+i a x)^{7/8}} \] Input:

Integrate[E^((I/4)*ArcTan[a*x])/x^3,x]
 

Output:

((1 - I*a*x)^(7/8)*(7*(-4 - (9*I)*a*x + 5*a^2*x^2) + 2*a^2*x^2*Hypergeomet 
ric2F1[7/8, 1, 15/8, (I + a*x)/(I - a*x)]))/(56*x^2*(1 + I*a*x)^(7/8))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.25, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5585, 107, 105, 104, 758, 755, 756, 216, 219, 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.

\(\displaystyle \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {\sqrt [8]{1+i a x}}{x^3 \sqrt [8]{1-i a x}}dx\)

\(\Big \downarrow \) 107

\(\displaystyle \frac {1}{8} i a \int \frac {\sqrt [8]{i a x+1}}{x^2 \sqrt [8]{1-i a x}}dx-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {1}{8} i a \left (\frac {1}{4} i a \int \frac {1}{x \sqrt [8]{1-i a x} (i a x+1)^{7/8}}dx-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{8} i a \left (2 i a \int \frac {1}{\frac {i a x+1}{1-i a x}-1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 758

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-1}d\left (\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-1}d\left (1-\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1\right )}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1\right )}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}{\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}+1}d\frac {\sqrt [8]{i a x+1}}{\sqrt [8]{1-i a x}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{8} i a \left (2 i a \left (\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )\right )+\frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}+\frac {\sqrt {2} \sqrt [8]{1+i a x}}{\sqrt [8]{1-i a x}}+1\right )}{2 \sqrt {2}}\right )\right )\right )-\frac {(1-i a x)^{7/8} \sqrt [8]{1+i a x}}{x}\right )-\frac {(1-i a x)^{7/8} (1+i a x)^{9/8}}{2 x^2}\)

Input:

Int[E^((I/4)*ArcTan[a*x])/x^3,x]
 

Output:

-1/2*((1 - I*a*x)^(7/8)*(1 + I*a*x)^(9/8))/x^2 + (I/8)*a*(-(((1 - I*a*x)^( 
7/8)*(1 + I*a*x)^(1/8))/x) + (2*I)*a*((-1/2*ArcTan[(1 + I*a*x)^(1/8)/(1 - 
I*a*x)^(1/8)] - ArcTanh[(1 + I*a*x)^(1/8)/(1 - I*a*x)^(1/8)]/2)/2 + ((ArcT 
an[1 - (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)]/Sqrt[2] - ArcTan[1 + 
 (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8)]/Sqrt[2])/2 + (Log[1 - (Sqr 
t[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) + (1 + I*a*x)^(1/4)/(1 - I*a*x)^ 
(1/4)]/(2*Sqrt[2]) - Log[1 + (Sqrt[2]*(1 + I*a*x)^(1/8))/(1 - I*a*x)^(1/8) 
 + (1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/(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 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 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 216

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

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 219

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

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 758

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), 
 x], x] + Simp[r/(2*a)   Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] 
 && IGtQ[n/4, 1] &&  !GtQ[a/b, 0]
 

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 5585

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a 
*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))), x] /; FreeQ[{a, m, n}, x] &&  !Intege 
rQ[(I*n - 1)/2]
 

Maple [F]

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

Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + 1\right ) + i \, a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + i\right ) - i \, a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - i\right ) - a^{2} x^{2} \log \left (\left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - 1\right ) + \sqrt {i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + \sqrt {i \, a^{4}}}{a^{2}}\right ) - \sqrt {i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - \sqrt {i \, a^{4}}}{a^{2}}\right ) + \sqrt {-i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} + \sqrt {-i \, a^{4}}}{a^{2}}\right ) - \sqrt {-i \, a^{4}} x^{2} \log \left (\frac {a^{2} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}} - \sqrt {-i \, a^{4}}}{a^{2}}\right ) - 4 \, {\left (5 \, a^{2} x^{2} + i \, a x + 4\right )} \left (\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}\right )^{\frac {1}{4}}}{32 \, x^{2}} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="fricas")
 

Output:

1/32*(a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + 1) + I*a^2*x^2*l 
og((I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) + I) - I*a^2*x^2*log((I*sqrt(a^2* 
x^2 + 1)/(a*x + I))^(1/4) - I) - a^2*x^2*log((I*sqrt(a^2*x^2 + 1)/(a*x + I 
))^(1/4) - 1) + sqrt(I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^( 
1/4) + sqrt(I*a^4))/a^2) - sqrt(I*a^4)*x^2*log((a^2*(I*sqrt(a^2*x^2 + 1)/( 
a*x + I))^(1/4) - sqrt(I*a^4))/a^2) + sqrt(-I*a^4)*x^2*log((a^2*(I*sqrt(a^ 
2*x^2 + 1)/(a*x + I))^(1/4) + sqrt(-I*a^4))/a^2) - sqrt(-I*a^4)*x^2*log((a 
^2*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4) - sqrt(-I*a^4))/a^2) - 4*(5*a^2*x 
^2 + I*a*x + 4)*(I*sqrt(a^2*x^2 + 1)/(a*x + I))^(1/4))/x^2
 

Sympy [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int \frac {\sqrt [4]{\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}{x^{3}}\, dx \] Input:

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/4)/x**3,x)
 

Output:

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(1/4)/x**3, x)
 

Maxima [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {1}{4}}}{x^{3}} \,d x } \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x, algorithm="maxima")
 

Output:

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(1/4)/x^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,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. \[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{1/4}}{x^3} \,d x \] Input:

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^3,x)
 

Output:

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/4)/x^3, x)
 

Reduce [F]

\[ \int \frac {e^{\frac {1}{4} i \arctan (a x)}}{x^3} \, dx=\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {1}{4}}}{x^{3}}d x \] Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/4)/x^3,x)