Integrand size = 16, antiderivative size = 171 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=-\frac {\left (i+2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {(i+4 a) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}+\frac {x \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{3 b^2}-\frac {\left (1-2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \]
-1/2*(1-2*I*a-2*a^2)*arcsinh(b*x+a)/b^3-1/6*(I+4*a)*(1+I*a+I*b*x)^(3/2)*(1 -I*a-I*b*x)^(1/2)/b^3+1/3*x*(1+I*a+I*b*x)^(3/2)*(1-I*a-I*b*x)^(1/2)/b^2-1/ 2*(I+2*a-2*I*a^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (-4 i+2 i a^2+3 b x+2 i b^2 x^2+a (-9-2 i b x)\right )}{6 b^3}+\frac {\sqrt [4]{-1} \left (-1+2 i a+2 a^2\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \]
(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-4*I + (2*I)*a^2 + 3*b*x + (2*I)*b^2*x ^2 + a*(-9 - (2*I)*b*x)))/(6*b^3) + ((-1)^(1/4)*(-1 + (2*I)*a + 2*a^2)*Sqr t[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I) *b]])/b^(7/2)
Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5618, 101, 25, 90, 60, 62, 1090, 222}
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.
| \(\displaystyle \int x^2 e^{i \arctan (a+b x)} \, dx\) |
| \(\Big \downarrow \) 5618 |
| \(\displaystyle \int \frac {x^2 \sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {\int -\frac {\sqrt {i a+i b x+1} \left (a^2+(4 a+i) b x+1\right )}{\sqrt {-i a-i b x+1}}dx}{3 b^2}+\frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\int \frac {\sqrt {i a+i b x+1} \left (a^2+(4 a+i) b x+1\right )}{\sqrt {-i a-i b x+1}}dx}{3 b^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\frac {3}{2} \left (-2 a^2-2 i a+1\right ) \int \frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}dx+\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b}}{3 b^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\frac {3}{2} \left (-2 a^2-2 i a+1\right ) \left (\int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )+\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b}}{3 b^2}\) |
| \(\Big \downarrow \) 62 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\frac {3}{2} \left (-2 a^2-2 i a+1\right ) \left (\int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )+\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b}}{3 b^2}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\frac {3}{2} \left (-2 a^2-2 i a+1\right ) \left (\frac {\int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b^2}+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )+\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b}}{3 b^2}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {x \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{3 b^2}-\frac {\frac {3}{2} \left (-2 a^2-2 i a+1\right ) \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )+\frac {(4 a+i) \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 b}}{3 b^2}\) |
(x*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(3*b^2) - (((I + 4*a)*Sq rt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(2*b) + (3*(1 - (2*I)*a - 2*a ^2)*((I*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/b + ArcSinh[(2*a*b + 2*b^2*x)/(2*b)]/b))/2)/(3*b^2)
3.2.64.3.1 Defintions of rubi rules used
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[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
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]
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 | \(i b \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )+\left (i a +1\right ) \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )\) | \(436\) |
1/6*I*(2*b^2*x^2-3*I*b*x-2*a*b*x+9*I*a+2*a^2-4)*(b^2*x^2+2*a*b*x+a^2+1)^(1 /2)/b^3+1/2*(2*I*a+2*a^2-1)/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b* x+a^2+1)^(1/2))/(b^2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\frac {7 i \, a^{3} - 21 \, a^{2} - 12 \, {\left (2 \, a^{2} + 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-2 i \, b^{2} x^{2} + {\left (2 i \, a - 3\right )} b x - 2 i \, a^{2} + 9 \, a + 4 i\right )} - 9 i \, a}{24 \, b^{3}} \]
1/24*(7*I*a^3 - 21*a^2 - 12*(2*a^2 + 2*I*a - 1)*log(-b*x - a + sqrt(b^2*x^ 2 + 2*a*b*x + a^2 + 1)) - 4*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(-2*I*b^2*x^ 2 + (2*I*a - 3)*b*x - 2*I*a^2 + 9*a + 4*I) - 9*I*a)/b^3
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 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (- \frac {2 i a}{3} + 1\right )}{2 b} - \frac {i \left (2 a^{2} + 2\right )}{3 b}\right )}{b} - \frac {\left (a^{2} + 1\right ) \left (- \frac {2 i a}{3} + 1\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \left (\frac {i x^{2}}{3 b} + \frac {x \left (- \frac {2 i a}{3} + 1\right )}{2 b^{2}} + \frac {- \frac {3 a \left (- \frac {2 i a}{3} + 1\right )}{2 b} - \frac {i \left (2 a^{2} + 2\right )}{3 b}}{b^{2}}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {i \left (a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}\right )}{2 a b^{2}} + \frac {a^{4} \sqrt {a^{2} + 2 a b x + 1} + 2 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {a^{2} + 2 a b x + 1}}{2 a^{2} b^{2}} + \frac {i \left (- a^{6} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{4} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 3 a^{2} - 3\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \cdot \left (3 a^{4} + 6 a^{2} + 3\right )}{3} - \sqrt {a^{2} + 2 a b x + 1}\right )}{4 a^{3} b^{2}}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {i a x^{3}}{3} + \frac {i b x^{4}}{4} + \frac {x^{3}}{3}}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]
Piecewise(((-a*(-3*a*(-2*I*a/3 + 1)/(2*b) - I*(2*a**2 + 2)/(3*b))/b - (a** 2 + 1)*(-2*I*a/3 + 1)/(2*b**2))*log(2*a*b + 2*b**2*x + 2*sqrt(a**2 + 2*a*b *x + b**2*x**2 + 1)*sqrt(b**2))/sqrt(b**2) + (I*x**2/(3*b) + x*(-2*I*a/3 + 1)/(2*b**2) + (-3*a*(-2*I*a/3 + 1)/(2*b) - I*(2*a**2 + 2)/(3*b))/b**2)*sq rt(a**2 + 2*a*b*x + b**2*x**2 + 1), Ne(b**2, 0)), ((I*(a**4*sqrt(a**2 + 2* a*b*x + 1) + 2*a**2*sqrt(a**2 + 2*a*b*x + 1) + (-2*a**2 - 2)*(a**2 + 2*a*b *x + 1)**(3/2)/3 + (a**2 + 2*a*b*x + 1)**(5/2)/5 + sqrt(a**2 + 2*a*b*x + 1 ))/(2*a*b**2) + (a**4*sqrt(a**2 + 2*a*b*x + 1) + 2*a**2*sqrt(a**2 + 2*a*b* x + 1) + (-2*a**2 - 2)*(a**2 + 2*a*b*x + 1)**(3/2)/3 + (a**2 + 2*a*b*x + 1 )**(5/2)/5 + sqrt(a**2 + 2*a*b*x + 1))/(2*a**2*b**2) + I*(-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) + (-3*a**2 - 3)*(a**2 + 2*a*b*x + 1)**(5/2)/5 + (a**2 + 2*a*b*x + 1)**(7/2)/7 + (a**2 + 2*a*b*x + 1)**(3/2)*(3*a**4 + 6*a**2 + 3)/3 - sqrt( a**2 + 2*a*b*x + 1))/(4*a**3*b**2))/(2*a*b), Ne(a*b, 0)), ((I*a*x**3/3 + I *b*x**4/4 + x**3/3)/sqrt(a**2 + 1), True))
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 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{2}}{3 \, b} - \frac {5 i \, a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {3 \, a^{2} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{6 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )} x}{2 \, b^{2}} + \frac {3 i \, {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} - \frac {{\left (a^{2} + 1\right )} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{3}} + \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a {\left (i \, a + 1\right )}}{2 \, b^{3}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a^{2} + i\right )}}{3 \, b^{3}} \]
1/3*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x^2/b - 5/2*I*a^3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 + 3/2*a^2*(I*a + 1)*arcsin h(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 5/6*I*sqrt(b^2 *x^2 + 2*a*b*x + a^2 + 1)*a*x/b^2 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)* (-I*a - 1)*x/b^2 + 3/2*I*(a^2 + 1)*a*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b ^2 + 4*(a^2 + 1)*b^2))/b^3 - 1/2*(a^2 + 1)*(I*a + 1)*arcsinh(2*(b^2*x + a* b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 + 5/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^3 - 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*(I*a + 1)/b^3 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(I*a^2 + I)/b^3
Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int e^{i \arctan (a+b x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (-\frac {2 i \, x}{b} - \frac {-2 i \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} - \frac {2 i \, a^{2} b^{2} - 9 \, a b^{2} - 4 i \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} + 2 i \, a - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \]
-1/6*sqrt((b*x + a)^2 + 1)*(x*(-2*I*x/b - (-2*I*a*b^3 + 3*b^3)/b^5) - (2*I *a^2*b^2 - 9*a*b^2 - 4*I*b^2)/b^5) - 1/2*(2*a^2 + 2*I*a - 1)*log(-a*b - (x *abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^2*abs(b))
Timed out. \[ \int e^{i \arctan (a+b x)} x^2 \, dx=\int \frac {x^2\,\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}{\sqrt {{\left (a+b\,x\right )}^2+1}} \,d x \]