\(\int e^{-3 i \arctan (a+b x)} x^2 \, dx\) [223]

Optimal result

Integrand size = 16, antiderivative size = 229 \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\frac {i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt {1+i a+i b x}}-\frac {\left (11 i-18 a-6 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}-\frac {\left (11 i-18 a-6 i a^2\right ) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}-\frac {i (1-i a-i b x)^{5/2} \sqrt {1+i a+i b x}}{3 b^3}+\frac {\left (11+18 i a-6 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \] Output:

I*(I-a)^2*(1-I*a-I*b*x)^(5/2)/b^3/(1+I*a+I*b*x)^(1/2)-1/2*(11*I-18*a-6*I*a 
^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3-1/6*(11*I-18*a-6*I*a^2)*(1 
-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3-1/3*I*(1-I*a-I*b*x)^(5/2)*(1+I*a 
+I*b*x)^(1/2)/b^3+1/2*(11+18*I*a-6*a^2)*arcsinh(b*x+a)/b^3
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.86 \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\frac {2 i a^4+a^3 (51+2 i b x)+a^2 (-50 i+69 b x)+a \left (51-106 i b x+9 b^2 x^2+2 i b^3 x^3\right )+i \left (-52+33 i b x-26 b^2 x^2+9 i b^3 x^3+2 b^4 x^4\right )}{6 b^3 \sqrt {1+a^2+2 a b x+b^2 x^2}}+\frac {\sqrt [4]{-1} \left (11+18 i a-6 a^2\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \] Input:

Integrate[x^2/E^((3*I)*ArcTan[a + b*x]),x]
 

Output:

((2*I)*a^4 + a^3*(51 + (2*I)*b*x) + a^2*(-50*I + 69*b*x) + a*(51 - (106*I) 
*b*x + 9*b^2*x^2 + (2*I)*b^3*x^3) + I*(-52 + (33*I)*b*x - 26*b^2*x^2 + (9* 
I)*b^3*x^3 + 2*b^4*x^4))/(6*b^3*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]) + ((-1) 
^(1/4)*(11 + (18*I)*a - 6*a^2)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*S 
qrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5618, 100, 25, 27, 90, 60, 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.

\(\displaystyle \int x^2 e^{-3 i \arctan (a+b x)} \, dx\)

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {x^2 (-i a-i b x+1)^{3/2}}{(i a+i b x+1)^{3/2}}dx\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {i \int -\frac {b (-i a-i b x+1)^{3/2} ((i-a) (2 i a+3)+b x)}{\sqrt {i a+i b x+1}}dx}{b^3}+\frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \int \frac {b (-i a-i b x+1)^{3/2} ((i-a) (2 i a+3)+b x)}{\sqrt {i a+i b x+1}}dx}{b^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \int \frac {(-i a-i b x+1)^{3/2} ((i-a) (2 i a+3)+b x)}{\sqrt {i a+i b x+1}}dx}{b^2}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \int \frac {(-i a-i b x+1)^{3/2}}{\sqrt {i a+i b x+1}}dx\right )}{b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \left (\frac {3}{2} \int \frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}dx-\frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 b}\right )\right )}{b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )-\frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 b}\right )\right )}{b^2}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )-\frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 b}\right )\right )}{b^2}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \left (\frac {3}{2} \left (\frac {\int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b^2}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )-\frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 b}\right )\right )}{b^2}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt {i a+i b x+1}}-\frac {i \left (\frac {(-i a-i b x+1)^{5/2} \sqrt {i a+i b x+1}}{3 b}-\frac {1}{3} \left (18 a-i \left (11-6 a^2\right )\right ) \left (\frac {3}{2} \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )-\frac {i (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{2 b}\right )\right )}{b^2}\)

Input:

Int[x^2/E^((3*I)*ArcTan[a + b*x]),x]
 

Output:

(I*(I - a)^2*(1 - I*a - I*b*x)^(5/2))/(b^3*Sqrt[1 + I*a + I*b*x]) - (I*((( 
1 - I*a - I*b*x)^(5/2)*Sqrt[1 + I*a + I*b*x])/(3*b) - ((18*a - I*(11 - 6*a 
^2))*(((-1/2*I)*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/b + (3*(((- 
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
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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]
 

rule 62

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]
 

rule 90

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]
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 222

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]
 

rule 1090

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]
 

rule 5618

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]
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.17

method	result	size
risch	\(\frac {i \left (2 b^{2} x^{2}-2 a b x +9 i b x +2 a^{2}-27 i a -28\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}-\frac {-\frac {18 i a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {i \left (-8 i a^{2}-16 a +8 i\right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{2} \left (x -\frac {i-a}{b}\right )}-\frac {11 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {6 a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}}{2 b^{2}}\)	\(268\)
default	\(\frac {i \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )}{b^{3}}+\frac {i \left (i-a \right )^{2} \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{5}}+\frac {2 i \left (i-a \right ) \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )}{b^{4}}\)	\(799\)

Input:

int(x^2/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*I*(2*b^2*x^2+9*I*b*x-2*a*b*x-27*I*a+2*a^2-28)*(b^2*x^2+2*a*b*x+a^2+1)^ 
(1/2)/b^3-1/2/b^2*(-18*I*a*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)-I*(-16*a-8*I*a^2+8*I)/b^2/(x-(I-a)/b)*((x-(I-a)/b)^ 
2*b^2+2*I*b*(x-(I-a)/b))^(1/2)-11*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)+6*a^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))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.76 \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\frac {7 i \, a^{4} + 166 \, a^{3} + {\left (7 i \, a^{3} + 159 \, a^{2} - 249 i \, a - 96\right )} b x - 408 i \, a^{2} + 12 \, {\left (6 \, a^{3} + {\left (6 \, a^{2} - 18 i \, a - 11\right )} b x - 24 i \, a^{2} - 29 \, a + 11 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, {\left (-2 i \, b^{3} x^{3} + 7 \, b^{2} x^{2} - 2 i \, a^{3} - {\left (16 \, a - 19 i\right )} b x - 53 \, a^{2} + 103 i \, a + 52\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 345 \, a + 96 i}{24 \, {\left (b^{4} x + {\left (a - i\right )} b^{3}\right )}} \] Input:

integrate(x^2/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="fricas")
 

Output:

1/24*(7*I*a^4 + 166*a^3 + (7*I*a^3 + 159*a^2 - 249*I*a - 96)*b*x - 408*I*a 
^2 + 12*(6*a^3 + (6*a^2 - 18*I*a - 11)*b*x - 24*I*a^2 - 29*a + 11*I)*log(- 
b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 4*(-2*I*b^3*x^3 + 7*b^2*x^2 
 - 2*I*a^3 - (16*a - 19*I)*b*x - 53*a^2 + 103*I*a + 52)*sqrt(b^2*x^2 + 2*a 
*b*x + a^2 + 1) - 345*a + 96*I)/(b^4*x + (a - I)*b^3)
 

Sympy [F(-1)]

Timed out. \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\text {Timed out} \] Input:

integrate(x**2/(1+I*(b*x+a))**3*(1+(b*x+a)**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (155) = 310\).

Time = 0.12 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.72 \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{2 i \, b^{4} x + 2 i \, a b^{3} + 2 \, b^{3}} + \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} - \frac {2 i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{2 i \, b^{4} x + 2 i \, a b^{3} + 2 \, b^{3}} + \frac {12 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac {3 \, a^{2} \operatorname {arsinh}\left (b x + a\right )}{b^{3}} - \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3} x}{2 \, b^{2}} + \frac {9 i \, a \operatorname {arsinh}\left (b x + a\right )}{b^{3}} + \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3} a}{2 \, b^{3}} + \frac {\arcsin \left (i \, b x + i \, a + 2\right )}{2 \, b^{3}} + \frac {6 \, \operatorname {arsinh}\left (b x + a\right )}{b^{3}} - \frac {3 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{3}} + \frac {i \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3}}{b^{3}} \] Input:

integrate(x^2/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="maxima")
 

Output:

I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 - 
 2*I*b^4*x - 2*I*a*b^3 - b^3) + 2*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(b 
^5*x^2 + 2*a*b^4*x + a^2*b^3 - 2*I*b^4*x - 2*I*a*b^3 - b^3) + 2*(b^2*x^2 + 
 2*a*b*x + a^2 + 1)^(3/2)*a/(2*I*b^4*x + 2*I*a*b^3 + 2*b^3) + 6*I*sqrt(b^2 
*x^2 + 2*a*b*x + a^2 + 1)*a^2/(I*b^4*x + I*a*b^3 + b^3) - I*(b^2*x^2 + 2*a 
*b*x + a^2 + 1)^(3/2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 - 2*I*b^4*x - 2*I*a*b 
^3 - b^3) - 2*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(2*I*b^4*x + 2*I*a*b^3 
 + 2*b^3) + 12*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/(I*b^4*x + I*a*b^3 + b^ 
3) - 3*a^2*arcsinh(b*x + a)/b^3 - 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(I 
*b^4*x + I*a*b^3 + b^3) - 1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4* 
I*a + 3)*x/b^2 + 9*I*a*arcsinh(b*x + a)/b^3 + 1/3*I*(b^2*x^2 + 2*a*b*x + a 
^2 + 1)^(3/2)/b^3 + 3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^3 - 1/2*sqrt(- 
b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a/b^3 + 1/2*arcsin(I*b*x + 
I*a + 2)/b^3 + 6*arcsinh(b*x + a)/b^3 - 3*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 
 1)/b^3 + I*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)/b^3
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.05 \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (-\frac {2 i \, x}{b} + \frac {2 i \, a b^{6} + 9 \, b^{6}}{b^{8}}\right )} + \frac {-2 i \, a^{2} b^{5} - 27 \, a b^{5} + 28 i \, b^{5}}{b^{8}}\right )} + \frac {{\left (6 \, a^{2} - 18 i \, a - 11\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 i \, a^{2} b + 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{6 \, b^{2} {\left | b \right |}} \] Input:

integrate(x^2/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="giac")
 

Output:

-1/6*sqrt((b*x + a)^2 + 1)*(x*(-2*I*x/b + (2*I*a*b^6 + 9*b^6)/b^8) + (-2*I 
*a^2*b^5 - 27*a*b^5 + 28*I*b^5)/b^8) + 1/6*(6*a^2 - 18*I*a - 11)*log(3*(x* 
abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b + a^3*b + (x*abs(b) - sqrt((b*x + a) 
^2 + 1))^3*abs(b) + 3*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^2*abs(b) - 2*I* 
(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b - 2*I*a^2*b + 4*(-I*x*abs(b) + I*sq 
rt((b*x + a)^2 + 1))*a*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*a 
bs(b))/(b^2*abs(b))
 

Mupad [F(-1)]

Timed out. \[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=\int \frac {x^2\,{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((x^2*((a + b*x)^2 + 1)^(3/2))/(a*1i + b*x*1i + 1)^3,x)
 

Output:

int((x^2*((a + b*x)^2 + 1)^(3/2))/(a*1i + b*x*1i + 1)^3, x)
 

Reduce [F]

\[ \int e^{-3 i \arctan (a+b x)} x^2 \, dx=-\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{4}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) b^{2}-2 \left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{3}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) a b -\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{2}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) a^{2}-\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{2}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) \] Input:

int(x^2/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2),x)
 

Output:

 - int((sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*x**4)/(a**3*i + 3*a**2*b*i*x 
+ 3*a**2 + 3*a*b**2*i*x**2 + 6*a*b*x - 3*a*i + b**3*i*x**3 + 3*b**2*x**2 - 
 3*b*i*x - 1),x)*b**2 - 2*int((sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*x**3)/ 
(a**3*i + 3*a**2*b*i*x + 3*a**2 + 3*a*b**2*i*x**2 + 6*a*b*x - 3*a*i + b**3 
*i*x**3 + 3*b**2*x**2 - 3*b*i*x - 1),x)*a*b - int((sqrt(a**2 + 2*a*b*x + b 
**2*x**2 + 1)*x**2)/(a**3*i + 3*a**2*b*i*x + 3*a**2 + 3*a*b**2*i*x**2 + 6* 
a*b*x - 3*a*i + b**3*i*x**3 + 3*b**2*x**2 - 3*b*i*x - 1),x)*a**2 - int((sq 
rt(a**2 + 2*a*b*x + b**2*x**2 + 1)*x**2)/(a**3*i + 3*a**2*b*i*x + 3*a**2 + 
 3*a*b**2*i*x**2 + 6*a*b*x - 3*a*i + b**3*i*x**3 + 3*b**2*x**2 - 3*b*i*x - 
 1),x)