Integrand size = 14, antiderivative size = 338 \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=-\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b} \]
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Time = 0.14 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5201, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=-\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b}+\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2} b}-\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}+\frac {3 i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b}-\frac {3 i \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2} b} \]
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Rule 52
Rule 65
Rule 210
Rule 303
Rule 338
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 5201
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a-i b x)^{3/4}}{(1+i a+i b x)^{3/4}} \, dx \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {3}{2} \int \frac {1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {(6 i) \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{b} \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {(6 i) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b} \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {(3 i) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b}+\frac {(3 i) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{b} \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b}+\frac {(3 i) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}+\frac {(3 i) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b} \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b} \\ & = -\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {3 i \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} b}+\frac {3 i \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} b}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.13 \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=-\frac {8 i e^{\frac {1}{2} i \arctan (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},2,\frac {5}{4},-e^{2 i \arctan (a+b x)}\right )}{b} \]
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\[\int \frac {1}{{\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}^{\frac {3}{2}}}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.75 \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=\frac {b \sqrt {\frac {9 i}{b^{2}}} \log \left (\frac {1}{3} i \, b \sqrt {\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {\frac {9 i}{b^{2}}} \log \left (-\frac {1}{3} i \, b \sqrt {\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + b \sqrt {-\frac {9 i}{b^{2}}} \log \left (\frac {1}{3} i \, b \sqrt {-\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {-\frac {9 i}{b^{2}}} \log \left (-\frac {1}{3} i \, b \sqrt {-\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - 2 \, {\left (b x + a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{2 \, b} \]
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\[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=\int \frac {1}{\left (\frac {i \left (a + b x\right ) + 1}{\sqrt {\left (a + b x\right )^{2} + 1}}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=\int { \frac {1}{\left (\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}\right )^{\frac {3}{2}}} \,d x } \]
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Exception generated. \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{-\frac {3}{2} i \arctan (a+b x)} \, dx=\int \frac {1}{{\left (\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}\right )}^{3/2}} \,d x \]
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