\(\int e^{3 i \arctan (a+b x)} x^2 \, dx\) [196]

Optimal result

Integrand size = 16, antiderivative size = 227 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=\frac {\left (11 i+18 a-6 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {\left (11 i+18 a-6 i a^2\right ) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{6 b^3}-\frac {i (i+a)^2 (1+i a+i b x)^{5/2}}{b^3 \sqrt {1-i a-i b x}}+\frac {i \sqrt {1-i a-i b x} (1+i a+i b x)^{5/2}}{3 b^3}+\frac {\left (11-18 i a-6 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.70 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=\frac {\sqrt {1+i a+i b x} \left (52 i-53 i a^2-2 a^3+19 b x+7 i b^2 x^2-2 b^3 x^3+a (103-16 i b x)\right )}{6 b^3 \sqrt {-i (i+a+b x)}}+\frac {(-1)^{3/4} \left (-11+18 i a+6 a^2\right ) \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{\sqrt {-i b} b^{5/2}} \] Input:

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

Output:

(Sqrt[1 + I*a + I*b*x]*(52*I - (53*I)*a^2 - 2*a^3 + 19*b*x + (7*I)*b^2*x^2 
 - 2*b^3*x^3 + a*(103 - (16*I)*b*x)))/(6*b^3*Sqrt[(-I)*(I + a + b*x)]) + ( 
(-1)^(3/4)*(-11 + (18*I)*a + 6*a^2)*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I) 
*(I + a + b*x)])/Sqrt[(-I)*b]])/(Sqrt[(-I)*b]*b^(5/2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5618, 100, 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

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

((-I)*(I + a)^2*(1 + I*a + I*b*x)^(5/2))/(b^3*Sqrt[1 - I*a - I*b*x]) - (I* 
(-1/3*(Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(5/2))/b + ((11*I + 18*a - 
(6*I)*a^2)*(((I/2)*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/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 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 = 1.32 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.14

method	result	size
risch	\(-\frac {i \left (2 b^{2} x^{2}-2 a b x -9 i b x +2 a^{2}+27 i a -28\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}-\frac {\frac {18 i a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {i \left (-8 i a^{2}+16 a +8 i\right ) \sqrt {\left (x +\frac {i+a}{b}\right )^{2} b^{2}-2 i b \left (x +\frac {i+a}{b}\right )}}{b^{2} \left (x +\frac {i+a}{b}\right )}-\frac {11 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {6 a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}}{2 b^{2}}\)	\(259\)
default	\(\text {Expression too large to display}\)	\(1785\)

Input:

int((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^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 = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.77 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=\frac {-7 i \, a^{4} + 166 \, a^{3} + {\left (-7 i \, a^{3} + 159 \, a^{2} + 249 i \, a - 96\right )} b x + 408 i \, a^{2} + 12 \, {\left (6 \, a^{3} + {\left (6 \, a^{2} + 18 i \, a - 11\right )} b x + 24 i \, a^{2} - 29 \, a - 11 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, {\left (2 i \, b^{3} x^{3} + 7 \, b^{2} x^{2} + 2 i \, a^{3} - {\left (16 \, a + 19 i\right )} b x - 53 \, a^{2} - 103 i \, a + 52\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 345 \, a - 96 i}{24 \, {\left (b^{4} x + {\left (a + i\right )} b^{3}\right )}} \] Input:

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

\[ \int e^{3 i \arctan (a+b x)} x^2 \, dx =\text {Too large to display} \] Input:

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

Output:

-I*(Integral(I*x**2/(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x*s 
qrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2 
*x**2 + 1) + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x) + Integral(-3*a*x** 
2/(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x*sqrt(a**2 + 2*a*b*x 
 + b**2*x**2 + 1) + b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + sqrt( 
a**2 + 2*a*b*x + b**2*x**2 + 1)), x) + Integral(a**3*x**2/(a**2*sqrt(a**2 
+ 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) 
+ b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + sqrt(a**2 + 2*a*b*x + b 
**2*x**2 + 1)), x) + Integral(-3*b*x**3/(a**2*sqrt(a**2 + 2*a*b*x + b**2*x 
**2 + 1) + 2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + b**2*x**2*sqrt(a 
**2 + 2*a*b*x + b**2*x**2 + 1) + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x) 
 + Integral(b**3*x**5/(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x 
*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + b**2*x**2*sqrt(a**2 + 2*a*b*x + b* 
*2*x**2 + 1) + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x) + Integral(-3*I*a 
**2*x**2/(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x*sqrt(a**2 + 
2*a*b*x + b**2*x**2 + 1) + b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) 
+ sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x) + Integral(-3*I*b**2*x**4/(a** 
2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2*a*b*x*sqrt(a**2 + 2*a*b*x + b** 
2*x**2 + 1) + b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + sqrt(a**2 + 
 2*a*b*x + b**2*x**2 + 1)), x) + Integral(3*a*b**2*x**4/(a**2*sqrt(a**2...
 

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. 1608 vs. \(2 (155) = 310\).

Time = 0.05 (sec) , antiderivative size = 1608, normalized size of antiderivative = 7.08 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-35*I*a^5*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) 
- 1/3*I*b*x^4/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 265/6*I*(a^2 + 1)*a^3*x/ 
((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + 7/6*I*a*x^ 
3/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 35/6*I*(a^2 + 1)*a^4/((a^2*b^2 - (a^ 
2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b) - 61/6*I*(a^2 + 1)^2*a*x/ 
((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 2*(-I*a^3 
- 3*a^2 + 3*I*a + 1)*a^2*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b 
*x + a^2 + 1)) + 45*(I*a*b^2 + b^2)*a^4*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt( 
b^2*x^2 + 2*a*b*x + a^2 + 1)*b^2) - 18*(I*a^2*b + 2*a*b - I*b)*a^3*x/((a^2 
*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b) - 35/6*I*a^2*x^ 
2/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b) + 29/6*I*(a^2 + 1)^2*a^2/((a^2*b^2 
 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b) + (-I*a^3 - 3*a^2 + 
 3*I*a + 1)*(a^2 + 1)*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x 
+ a^2 + 1)) - 93/2*(I*a*b^2 + b^2)*(a^2 + 1)*a^2*x/((a^2*b^2 - (a^2 + 1)*b 
^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b^2) + 15*(I*a^2*b + 2*a*b - I*b)*(a 
^2 + 1)*a*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b 
) - 4/3*(-I*a^2 - I)*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b) - 3/2*(I*a* 
b^2 + b^2)*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b^2) + 35/2*I*a^3*arcsin 
h(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 + 15/2*(I*a*b^2 
+ b^2)*(a^2 + 1)*a^3/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x ...
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.07 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (\frac {2 i \, x}{b} - \frac {2 i \, a b^{6} - 9 \, b^{6}}{b^{8}}\right )} - \frac {-2 i \, a^{2} b^{5} + 27 \, a b^{5} + 28 i \, b^{5}}{b^{8}}\right )} + \frac {{\left (6 \, a^{2} + 18 i \, a - 11\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} + 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 i \, a^{2} b + 4 \, {\left (i \, x {\left | b \right |} - i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{6 \, b^{2} {\left | b \right |}} \] Input:

integrate((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^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*a 
bs(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*sqrt 
((b*x + a)^2 + 1))*a*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs 
(b))/(b^2*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 770, normalized size of antiderivative = 3.39 \[ \int e^{3 i \arctan (a+b x)} x^2 \, dx=\frac {-24+26 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{2} i \,x^{2}+50 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{2} i -33 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x -54 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a i +51 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{3}+9 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a \,b^{2} x^{2}-2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a \,b^{3} i \,x^{3}-108 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{2} b i x -54 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a \,b^{2} i \,x^{2}+33 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right )+48 a \,b^{2} i \,x^{2}+69 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{2} b x -2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{4} i \,x^{4}-36 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{3} b x -18 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{2} b^{2} x^{2}+66 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a b x -24 b^{2} x^{2}-2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{4} i -9 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{3} x^{3}-54 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{3} i +33 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) b^{2} x^{2}-18 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{4}+96 a^{2} b i x +24 a^{4}+106 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a b i x +48 a^{3} i +48 a i +51 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a +52 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, i +15 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{2}+48 a^{3} b x +24 a^{2} b^{2} x^{2}-48 a b x -2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{3} b i x}{6 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**4*i - 2*sqrt(a**2 + 2*a*b*x 
+ b**2*x**2 + 1)*a**3*b*i*x + 51*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**3 
 + 69*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2*b*x + 50*sqrt(a**2 + 2*a*b 
*x + b**2*x**2 + 1)*a**2*i - 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a*b**3 
*i*x**3 + 9*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a*b**2*x**2 + 106*sqrt(a* 
*2 + 2*a*b*x + b**2*x**2 + 1)*a*b*i*x + 51*sqrt(a**2 + 2*a*b*x + b**2*x**2 
 + 1)*a - 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*b**4*i*x**4 - 9*sqrt(a**2 
 + 2*a*b*x + b**2*x**2 + 1)*b**3*x**3 + 26*sqrt(a**2 + 2*a*b*x + b**2*x**2 
 + 1)*b**2*i*x**2 - 33*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*b*x + 52*sqrt( 
a**2 + 2*a*b*x + b**2*x**2 + 1)*i - 18*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 
 + 1) + a + b*x)*a**4 - 36*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + 
b*x)*a**3*b*x - 54*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a** 
3*i - 18*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a**2*b**2*x** 
2 - 108*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a**2*b*i*x + 1 
5*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a**2 - 54*log(sqrt(a 
**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a*b**2*i*x**2 + 66*log(sqrt(a**2 
 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a*b*x - 54*log(sqrt(a**2 + 2*a*b*x 
+ b**2*x**2 + 1) + a + b*x)*a*i + 33*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 
 1) + a + b*x)*b**2*x**2 + 33*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a 
 + b*x) + 24*a**4 + 48*a**3*b*x + 48*a**3*i + 24*a**2*b**2*x**2 + 96*a*...