Integrand size = 18, antiderivative size = 494 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=-\frac {\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac {(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac {x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (3 i+4 a-8 i a^2\right ) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{16 \sqrt {2} b^3}+\frac {\left (3 i+4 a-8 i a^2\right ) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{16 \sqrt {2} b^3} \]
-1/8*(3*I+4*a-8*I*a^2)*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(1/4)/b^3-1/12*(I +8*a)*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(5/4)/b^3+1/3*x*(1-I*a-I*b*x)^(3/4 )*(1+I*a+I*b*x)^(5/4)/b^2+1/16*(3*I+4*a-8*I*a^2)*arctan(1-(1-I*a-I*b*x)^(1 /4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b^3*2^(1/2)-1/16*(3*I+4*a-8*I*a^2)*arctan (1+(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b^3*2^(1/2)-1/32*(3*I+ 4*a-8*I*a^2)*ln(1-(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I *b*x)^(1/2)/(1+I*a+I*b*x)^(1/2))/b^3*2^(1/2)+1/32*(3*I+4*a-8*I*a^2)*ln(1+( 1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I*b*x)^(1/2)/(1+I*a+ I*b*x)^(1/2))/b^3*2^(1/2)
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 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\frac {(-i (i+a+b x))^{3/4} \left (-i \sqrt [4]{1+i a+i b x} \left (1+8 a^2+5 i b x-4 b^2 x^2+a (-7 i+4 b x)\right )+2 i \sqrt [4]{2} \left (-3+4 i a+8 a^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )\right )}{12 b^3} \]
(((-I)*(I + a + b*x))^(3/4)*((-I)*(1 + I*a + I*b*x)^(1/4)*(1 + 8*a^2 + (5* I)*b*x - 4*b^2*x^2 + a*(-7*I + 4*b*x)) + (2*I)*2^(1/4)*(-3 + (4*I)*a + 8*a ^2)*Hypergeometric2F1[-1/4, 3/4, 7/4, (-1/2*I)*(I + a + b*x)]))/(12*b^3)
Time = 0.51 (sec) , antiderivative size = 414, normalized size of antiderivative = 0.84, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5618, 101, 27, 90, 60, 73, 854, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^2 e^{\frac {1}{2} i \arctan (a+b x)} \, dx\) |
| \(\Big \downarrow \) 5618 |
| \(\displaystyle \int \frac {x^2 \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {\int -\frac {\sqrt [4]{i a+i b x+1} \left (2 a^2+(8 a+i) b x+2\right )}{2 \sqrt [4]{-i a-i b x+1}}dx}{3 b^2}+\frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\int \frac {\sqrt [4]{i a+i b x+1} \left (2 \left (a^2+1\right )+(8 a+i) b x\right )}{\sqrt [4]{-i a-i b x+1}}dx}{6 b^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \int \frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}dx+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {1}{2} \int \frac {1}{\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \int \frac {\sqrt {-i a-i b x+1}}{(i a+i b x+1)^{3/4}}d\sqrt [4]{-i a-i b x+1}}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \int \frac {\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \int \frac {\sqrt {-i a-i b x+1}+1}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac {\frac {3}{4} \left (-8 a^2-4 i a+3\right ) \left (\frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\right )+\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b}}{6 b^2}\) |
(x*(1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(5/4))/(3*b^2) - (((I + 8*a)* (1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(5/4))/(2*b) + (3*(3 - (4*I)*a - 8*a^2)*((I*(1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/b + ((2*I)*(( -(ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sq rt[2]) + ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1 /4)]/Sqrt[2])/2 + (Log[1 + Sqrt[1 - I*a - I*b*x] - (Sqrt[2]*(1 - I*a - I*b *x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt[2]) - Log[1 + Sqrt[1 - I*a - I *b*x] + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt [2]))/2))/b))/4)/(6*b^2)
3.3.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((1 - I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c + I*b*c*x)^(I*(n/2))), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
\[\int \sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, x^{2}d x\]
Time = 0.29 (sec) , antiderivative size = 554, normalized size of antiderivative = 1.12 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\frac {3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 2 \, {\left (8 \, b^{3} x^{3} - 2 i \, b^{2} x^{2} + 8 \, a^{3} + {\left (8 i \, a - 1\right )} b x + 34 i \, a^{2} - 37 \, a - 11 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{48 \, b^{3}} \]
1/48*(3*b^3*sqrt((64*I*a^4 - 64*a^3 - 64*I*a^2 + 24*a + 9*I)/b^6)*log((I*b ^3*sqrt((64*I*a^4 - 64*a^3 - 64*I*a^2 + 24*a + 9*I)/b^6) + (8*a^2 + 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(8*a^2 + 4*I *a - 3)) - 3*b^3*sqrt((64*I*a^4 - 64*a^3 - 64*I*a^2 + 24*a + 9*I)/b^6)*log ((-I*b^3*sqrt((64*I*a^4 - 64*a^3 - 64*I*a^2 + 24*a + 9*I)/b^6) + (8*a^2 + 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(8*a^2 + 4*I*a - 3)) + 3*b^3*sqrt((-64*I*a^4 + 64*a^3 + 64*I*a^2 - 24*a - 9*I)/b ^6)*log((I*b^3*sqrt((-64*I*a^4 + 64*a^3 + 64*I*a^2 - 24*a - 9*I)/b^6) + (8 *a^2 + 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) /(8*a^2 + 4*I*a - 3)) - 3*b^3*sqrt((-64*I*a^4 + 64*a^3 + 64*I*a^2 - 24*a - 9*I)/b^6)*log((-I*b^3*sqrt((-64*I*a^4 + 64*a^3 + 64*I*a^2 - 24*a - 9*I)/b ^6) + (8*a^2 + 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(8*a^2 + 4*I*a - 3)) + 2*(8*b^3*x^3 - 2*I*b^2*x^2 + 8*a^3 + (8*I* a - 1)*b*x + 34*I*a^2 - 37*a - 11*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b^3
Timed out. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\text {Timed out} \]
\[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\int { x^{2} \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}} \,d x } \]
Exception generated. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ,0]Warning, replacing 0 by -27, a substitution variable should perhaps be purged.Wa
Timed out. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x^2 \, dx=\int x^2\,\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}} \,d x \]