Integrand size = 16, antiderivative size = 178 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {i+a} b \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{5/2}} \]
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Time = 0.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5203, 96, 95, 214} \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=-\frac {6 i \sqrt {a+i} b \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{5/2}}-\frac {(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt {i a+i b x+1}}+\frac {6 i b \sqrt {-i a-i b x+1}}{(-a+i)^2 \sqrt {i a+i b x+1}} \]
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Rule 95
Rule 96
Rule 214
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a-i b x)^{3/2}}{x^2 (1+i a+i b x)^{3/2}} \, dx \\ & = -\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}+\frac {(3 b) \int \frac {\sqrt {1-i a-i b x}}{x (1+i a+i b x)^{3/2}} \, dx}{i-a} \\ & = \frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}+\frac {(3 (i+a) b) \int \frac {1}{x \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{(i-a)^2} \\ & = \frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}+\frac {(6 (i+a) b) \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 )}{(i-a)^2} \\ & = \frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {i+a} b \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 )}{(i-a)^{5/2}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\frac {\frac {\sqrt {-i (i+a+b x)} \left (1+a^2+5 i b x+a b x\right )}{x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {-1+i a} b \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a}}}{(-i+a)^2} \]
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Time = 1.34 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \left (i+a \right )}{\left (a -i\right )^{2} x}+\frac {b \left (-\frac {\left (-3 a^{2}-3\right ) \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (i-a \right ) \sqrt {a^{2}+1}}+\frac {4 i \left (i a +1\right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b \left (i-a \right ) \left (x -\frac {i-a}{b}\right )}\right )}{a^{2}-2 i a -1}\) | \(194\) |
default | \(\text {Expression too large to display}\) | \(1543\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (116) = 232\).
Time = 0.27 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.19 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=-\frac {{\left (i \, a - 5\right )} b^{2} x^{2} + {\left (i \, a^{2} - 4 \, a + 5 i\right )} b x - 3 \, {\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + 3 \, {\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x - {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (i \, a - 5\right )} b x + i \, a^{2} + i\right )}}{{\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x} \]
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Timed out. \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{2}} \,d x } \]
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\[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\int \frac {{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{x^2\,{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \]
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