Integrand size = 18, antiderivative size = 494 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=-\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{16 \sqrt {2} b^3}+\frac {\left (3 i+4 a-8 i a^2\right ) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{16 \sqrt {2} b^3} \]
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Time = 0.30 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5203, 92, 81, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\frac {\left (-8 i a^2+4 a+3 i\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {\left (-8 i a^2+4 a+3 i\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {\left (-8 i a^2+4 a+3 i\right ) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{8 b^3}-\frac {\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}+\frac {\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}-\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{12 b^3}+\frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2} \]
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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 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx \\ & = \frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac {\int \frac {\sqrt [4]{1+i a+i b x} \left (-1-a^2-\frac {1}{2} (i+8 a) b x\right )}{\sqrt [4]{1-i a-i b x}} \, dx}{3 b^2} \\ & = -\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3-4 i a-8 a^2\right ) \int \frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx}{8 b^2} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3-4 i a-8 a^2\right ) \int \frac {1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx}{16 b^2} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{4 b^3} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{4 b^3} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{8 b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{8 b^3} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3}+\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{8 \sqrt {2} b^3}+\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{8 \sqrt {2} b^3} \\ & = -\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3}+\frac {\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt {2} b^3} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.24 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\frac {(-i (i+a+b x))^{3/4} \left (-i \sqrt [4]{1+i a+i b x} \left (1+8 a^2+5 i b x-4 b^2 x^2+a (-7 i+4 b x)\right )+2 i \sqrt [4]{2} \left (-3+4 i a+8 a^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )\right )}{12 b^3} \]
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\[\int \sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, x^{2}d x\]
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none
Time = 0.29 (sec) , antiderivative size = 554, normalized size of antiderivative = 1.12 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\frac {3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 2 \, {\left (8 \, b^{3} x^{3} - 2 i \, b^{2} x^{2} + 8 \, a^{3} + {\left (8 i \, a - 1\right )} b x + 34 i \, a^{2} - 37 \, a - 11 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{48 \, b^{3}} \]
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Timed out. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\text {Timed out} \]
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\[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\int { x^{2} \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}} \,d x } \]
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Exception generated. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\int x^2\,\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}} \,d x \]
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