Integrand size = 16, antiderivative size = 339 \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=-\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}+\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \]
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Time = 0.17 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5170, 92, 81, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}-\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}+\frac {3 i \log \left (\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{16 \sqrt {2} a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2} \]
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Rule 52
Rule 65
Rule 81
Rule 92
Rule 210
Rule 303
Rule 338
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx \\ & = \frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}+\frac {\int \frac {\left (-1-\frac {i a x}{2}\right ) \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx}{3 a^2} \\ & = -\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {3 \int \frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx}{8 a^2} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {3 \int \frac {1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{16 a^2} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {(3 i) \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{4 a^3} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {(3 i) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^3} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}+\frac {(3 i) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^3}-\frac {(3 i) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^3} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 a^3}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 a^3}-\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 x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}-\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 x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3} \\ & = -\frac {3 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{8 a^3}-\frac {i (1-i a x)^{3/4} (1+i a x)^{5/4}}{12 a^3}+\frac {x (1-i a x)^{3/4} (1+i a x)^{5/4}}{3 a^2}+\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3}+\frac {3 i \log \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{16 \sqrt {2} a^3} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.24 \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\frac {(1-i a x)^{3/4} \left (\sqrt [4]{1+i a x} \left (-i+5 a x+4 i a^2 x^2\right )-6 i \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-i a x)\right )\right )}{12 a^3} \]
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\[\int \sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{2}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.72 \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=-\frac {12 \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} \log \left (\frac {8}{3} i \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} \log \left (-\frac {8}{3} i \, a^{3} \sqrt {\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 12 \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} \log \left (\frac {8}{3} i \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 12 \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} \log \left (-\frac {8}{3} i \, a^{3} \sqrt {-\frac {9 i}{64 \, a^{6}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (8 \, a^{3} x^{3} - 2 i \, a^{2} x^{2} - a x - 11 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{24 \, a^{3}} \]
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\[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\int x^{2} \sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}\, dx \]
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\[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\int { x^{2} \sqrt {\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}} \,d x } \]
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Exception generated. \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{\frac {1}{2} i \arctan (a x)} x^2 \, dx=\int x^2\,\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}} \,d x \]
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