Integrand size = 16, antiderivative size = 171 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \]
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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 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{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) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{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 (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) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{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 (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) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{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 (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) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{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 (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) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (-4 i+2 i a^2+3 b x+2 i b^2 x^2+a (-9-2 i b x)\right )}{6 b^3}+\frac {\sqrt [4]{-1} \left (-1+2 i a+2 a^2\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \]
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Time = 0.54 (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 | \(i b \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )+\left (i a +1\right ) \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )\) | \(436\) |
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Time = 0.26 (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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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (133) = 266\).
Time = 1.33 (sec) , antiderivative size = 585, normalized size of antiderivative = 3.42 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (- \frac {2 i a}{3} + 1\right )}{2 b} - \frac {i \left (2 a^{2} + 2\right )}{3 b}\right )}{b} - \frac {\left (a^{2} + 1\right ) \left (- \frac {2 i a}{3} + 1\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \left (\frac {i x^{2}}{3 b} + \frac {x \left (- \frac {2 i a}{3} + 1\right )}{2 b^{2}} + \frac {- \frac {3 a \left (- \frac {2 i a}{3} + 1\right )}{2 b} - \frac {i \left (2 a^{2} + 2\right )}{3 b}}{b^{2}}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {i \left (a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}\right )}{2 a b^{2}} + \frac {a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}}{2 a^{2} b^{2}} + \frac {i \left (- a^{6} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{4} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 3 a^{2} - 3\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \cdot \left (3 a^{4} + 6 a^{2} + 3\right )}{3} - \sqrt {a^{2} + 2 a b x + 1}\right )}{4 a^{3} b^{2}}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {i a x^{3}}{3} + \frac {i b x^{4}}{4} + \frac {x^{3}}{3}}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (119) = 238\).
Time = 0.17 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.05 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{2}}{3 \, b} - \frac {5 i \, a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {3 \, a^{2} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{6 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )} x}{2 \, b^{2}} + \frac {3 i \, {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {{\left (a^{2} + 1\right )} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a {\left (i \, a + 1\right )}}{2 \, b^{3}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a^{2} + i\right )}}{3 \, b^{3}} \]
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Time = 0.30 (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. \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\int \frac {x^2\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \]
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