Integrand size = 16, antiderivative size = 171 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {\left (i-2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \]
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Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5203, 92, 81, 52, 55, 633, 221} \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=-\frac {\left (-2 a^2+2 i a+1\right ) \text {arcsinh}(a+b x)}{2 b^3}+\frac {\left (-2 i a^2-2 a+i\right ) \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{2 b^3}+\frac {(-4 a+i) \sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{6 b^3}+\frac {x \sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{3 b^2} \]
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Rule 52
Rule 55
Rule 81
Rule 92
Rule 221
Rule 633
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx \\ & = \frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}+\frac {\int \frac {\sqrt {1-i a-i b x} \left (-1-a^2+(i-4 a) b x\right )}{\sqrt {1+i a+i b x}} \, dx}{3 b^2} \\ & = \frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx}{2 b^2} \\ & = -\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b^2} \\ & = -\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b^2} \\ & = -\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{4 b^4} \\ & = -\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.95 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {i \sqrt {1+i a+i b x} \left (4+7 a^2+2 i a^3-7 i b x-5 b^2 x^2+2 i b^3 x^3+a (5 i+8 b x)\right )}{6 b^3 \sqrt {-i (i+a+b x)}}+\frac {\sqrt [4]{-1} \left (-1-2 i a+2 a^2\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \]
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Time = 0.60 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66
method | result | size |
risch | \(-\frac {i \left (2 b^{2} x^{2}-2 a b x +3 b x i+2 a^{2}-9 i a -4\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}+\frac {\left (2 a^{2}-2 i a -1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\) | \(113\) |
default | \(-\frac {i \left (i \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )-a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )+b \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )\right )}{b^{2}}+\frac {\left (-i a^{2}-2 a +i\right ) \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b^{3}}\) | \(485\) |
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Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {-7 i \, a^{3} - 21 \, a^{2} - 12 \, {\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 i \, b^{2} x^{2} + {\left (-2 i \, a - 3\right )} b x + 2 i \, a^{2} + 9 \, a - 4 i\right )} + 9 i \, a}{24 \, b^{3}} \]
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\[ \int e^{-i \arctan (a+b x)} x^2 \, dx=- i \int \frac {x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a + b x - i}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{b^{2}} + \frac {a^{2} \operatorname {arsinh}\left (b x + a\right )}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b^{2}} - \frac {i \, a \operatorname {arsinh}\left (b x + a\right )}{b^{3}} - \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{3}} - \frac {\operatorname {arsinh}\left (b x + a\right )}{2 \, b^{3}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (\frac {2 i \, x}{b} - \frac {2 i \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} - \frac {-2 i \, a^{2} b^{2} - 9 \, a b^{2} + 4 i \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \]
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Timed out. \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\int \frac {x^2\,\sqrt {{\left (a+b\,x\right )}^2+1}}{1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}} \,d x \]
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