\(\int e^{-3 i \arctan (a+b x)} x^3 \, dx\) [222]

Optimal result

Integrand size = 16, antiderivative size = 282 \[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx=\frac {2 i x^3 (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 \left (17+44 i a-36 a^2-8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {7 \left (1+8 i a-6 a^2\right ) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{8 b^4}-\frac {9 x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{4 b^2}+\frac {(11+10 i a) (1-i a-i b x)^{5/2} \sqrt {1+i a+i b x}}{4 b^4}+\frac {3 \left (17 i-44 a-36 i a^2+8 a^3\right ) \text {arcsinh}(a+b x)}{8 b^4} \] Output:

2*I*x^3*(1-I*a-I*b*x)^(3/2)/b/(1+I*a+I*b*x)^(1/2)+3/8*(17+44*I*a-36*a^2-8* 
I*a^3)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^4+7/8*(1+8*I*a-6*a^2)*(1- 
I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^4-9/4*x^2*(1-I*a-I*b*x)^(3/2)*(1+I* 
a+I*b*x)^(1/2)/b^2+1/4*(11+10*I*a)*(1-I*a-I*b*x)^(5/2)*(1+I*a+I*b*x)^(1/2) 
/b^4+3/8*(17*I-44*a-36*I*a^2+8*a^3)*arcsinh(b*x+a)/b^4
 

Mathematica [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.87 \[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx=\frac {80-2 i a^5-51 i b x+40 b^2 x^2-17 i b^3 x^3-8 b^4 x^4+2 i b^5 x^5+a^4 (-76-2 i b x)-5 a^3 (-31 i+20 b x)+a^2 \left (4+265 i b x-12 b^2 x^2\right )+a \left (157 i+212 b x+53 i b^2 x^2+4 b^3 x^3+2 i b^4 x^4\right )}{8 b^4 \sqrt {1+a^2+2 a b x+b^2 x^2}}+\frac {3 \sqrt [4]{-1} \left (17 i-44 a-36 i a^2+8 a^3\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{4 b^{9/2}} \] Input:

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

Output:

(80 - (2*I)*a^5 - (51*I)*b*x + 40*b^2*x^2 - (17*I)*b^3*x^3 - 8*b^4*x^4 + ( 
2*I)*b^5*x^5 + a^4*(-76 - (2*I)*b*x) - 5*a^3*(-31*I + 20*b*x) + a^2*(4 + ( 
265*I)*b*x - 12*b^2*x^2) + a*(157*I + 212*b*x + (53*I)*b^2*x^2 + 4*b^3*x^3 
 + (2*I)*b^4*x^4))/(8*b^4*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]) + (3*(-1)^(1/ 
4)*(17*I - 44*a - (36*I)*a^2 + 8*a^3)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sq 
rt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(4*b^(9/2))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.94, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5618, 108, 27, 170, 27, 164, 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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 170

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\int \frac {b x \sqrt {-i a-i b x+1} \left (6 i \left (a^2+1\right )+(10 i a+11) b x\right )}{\sqrt {i a+i b x+1}}dx}{4 b^2}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\int \frac {x \sqrt {-i a-i b x+1} \left (6 i \left (a^2+1\right )+(10 i a+11) b x\right )}{\sqrt {i a+i b x+1}}dx}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-22 i a^2+2 (11+10 i a) b x-54 a+29 i\right )}{6 b^2}-\frac {\left (-8 i a^3-36 a^2+44 i a+17\right ) \int \frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}dx}{2 b}}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-22 i a^2+2 (11+10 i a) b x-54 a+29 i\right )}{6 b^2}-\frac {\left (-8 i a^3-36 a^2+44 i a+17\right ) \left (\int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-22 i a^2+2 (11+10 i a) b x-54 a+29 i\right )}{6 b^2}-\frac {\left (-8 i a^3-36 a^2+44 i a+17\right ) \left (\int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-22 i a^2+2 (11+10 i a) b x-54 a+29 i\right )}{6 b^2}-\frac {\left (-8 i a^3-36 a^2+44 i a+17\right ) \left (\frac {\int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b^2}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {2 i x^3 (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-\frac {3 i \left (\frac {\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (-22 i a^2+2 (11+10 i a) b x-54 a+29 i\right )}{6 b^2}-\frac {\left (-8 i a^3-36 a^2+44 i a+17\right ) \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}}{4 b}-\frac {3 i x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}\right )}{b}\)

Input:

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

Output:

((2*I)*x^3*(1 - I*a - I*b*x)^(3/2))/(b*Sqrt[1 + I*a + I*b*x]) - ((3*I)*((( 
(-3*I)/4)*x^2*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/b + (((1 - I* 
a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x]*(29*I - 54*a - (22*I)*a^2 + 2*(11 + 
 (10*I)*a)*b*x))/(6*b^2) - ((17 + (44*I)*a - 36*a^2 - (8*I)*a^3)*(((-I)*Sq 
rt[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*b))/(4*b)))/b
 

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 108

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

rule 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
 

rule 170

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegerQ[m]
 

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 = 0.85 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.24

method	result	size
risch	\(-\frac {i \left (-2 b^{3} x^{3}+2 a \,b^{2} x^{2}-8 i b^{2} x^{2}-2 a^{2} b x +20 i a b x +2 a^{3}-44 i a^{2}+19 b x -93 a +48 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{8 b^{4}}+\frac {\frac {51 i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {108 i a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {i \left (-32 i a^{3}-96 a^{2}+96 i a +32\right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{2} \left (x -\frac {i-a}{b}\right )}-\frac {132 a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {24 a^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}}{8 b^{3}}\)	\(351\)
default	\(\frac {i \left (\frac {\left (2 b^{2} x +2 a b \right ) \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{8 b^{2}}+\frac {3 \left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\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 )}{8 b^{2} \sqrt {b^{2}}}\right )}{16 b^{2}}\right )}{b^{3}}-\frac {3 \left (i a +1\right ) \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^{4}}+\frac {3 \left (i a^{2}+2 a -i\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^{5}}-\frac {\left (i a^{3}+3 a^{2}-3 i a -1\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 )^{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^{6}}\)	\(984\)

Input:

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

Output:

-1/8*I*(-2*b^3*x^3-8*I*b^2*x^2+2*a*b^2*x^2+20*I*a*b*x-2*a^2*b*x-44*I*a^2+2 
*a^3+19*b*x+48*I-93*a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b^4+1/8/b^3*(51*I*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)-108*I*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)-I 
*(-96*a^2-32*I*a^3+32+96*I*a)/b^2/(x-(I-a)/b)*((x-(I-a)/b)^2*b^2+2*I*b*(x- 
(I-a)/b))^(1/2)-132*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)+24*a^3*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 = 216, normalized size of antiderivative = 0.77 \[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx=\frac {-15 i \, a^{5} - 495 \, a^{4} + 1664 i \, a^{3} + {\left (-15 i \, a^{4} - 480 \, a^{3} + 1184 i \, a^{2} + 968 \, a - 256 i\right )} b x + 2152 \, a^{2} - 24 \, {\left (8 \, a^{4} - 44 i \, a^{3} + {\left (8 \, a^{3} - 36 i \, a^{2} - 44 \, a + 17 i\right )} b x - 80 \, a^{2} + 61 i \, a + 17\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 8 \, {\left (-2 i \, b^{4} x^{4} + 6 \, b^{3} x^{3} - {\left (10 \, a - 11 i\right )} b^{2} x^{2} + 2 i \, a^{4} + 78 \, a^{3} + {\left (22 \, a^{2} - 54 i \, a - 29\right )} b x - 233 i \, a^{2} - 237 \, a + 80 i\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 1224 i \, a - 256}{64 \, {\left (b^{5} x + {\left (a - i\right )} b^{4}\right )}} \] Input:

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

Output:

1/64*(-15*I*a^5 - 495*a^4 + 1664*I*a^3 + (-15*I*a^4 - 480*a^3 + 1184*I*a^2 
 + 968*a - 256*I)*b*x + 2152*a^2 - 24*(8*a^4 - 44*I*a^3 + (8*a^3 - 36*I*a^ 
2 - 44*a + 17*I)*b*x - 80*a^2 + 61*I*a + 17)*log(-b*x - a + sqrt(b^2*x^2 + 
 2*a*b*x + a^2 + 1)) - 8*(-2*I*b^4*x^4 + 6*b^3*x^3 - (10*a - 11*I)*b^2*x^2 
 + 2*I*a^4 + 78*a^3 + (22*a^2 - 54*I*a - 29)*b*x - 233*I*a^2 - 237*a + 80* 
I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 1224*I*a - 256)/(b^5*x + (a - I)*b^ 
4)
 

Sympy [F(-1)]

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

integrate(x**3/(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. 979 vs. \(2 (198) = 396\).

Time = 0.14 (sec) , antiderivative size = 979, normalized size of antiderivative = 3.47 \[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx =\text {Too large to display} \] Input:

integrate(x^3/(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^3/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 
- 2*I*b^5*x - 2*I*a*b^4 - b^4) - 3*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2 
/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 - 2*I*b^5*x - 2*I*a*b^4 - b^4) - 3*(b^2*x^ 
2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/(2*I*b^5*x + 2*I*a*b^4 + 2*b^4) - 6*I*sqr 
t(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^3/(I*b^5*x + I*a*b^4 + b^4) + 3*I*(b^2*x^ 
2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(b^6*x^2 + 2*a*b^5*x + a^2*b^4 - 2*I*b^5*x 
- 2*I*a*b^4 - b^4) + 6*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(2*I*b^5*x 
+ 2*I*a*b^4 + 2*b^4) - 18*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/(I*b^5*x + 
 I*a*b^4 + b^4) + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(b^6*x^2 + 2*a*b^5*x 
 + a^2*b^4 - 2*I*b^5*x - 2*I*a*b^4 - b^4) + 3*(b^2*x^2 + 2*a*b*x + a^2 + 1 
)^(3/2)/(2*I*b^5*x + 2*I*a*b^4 + 2*b^4) + 18*I*sqrt(b^2*x^2 + 2*a*b*x + a^ 
2 + 1)*a/(I*b^5*x + I*a*b^4 + b^4) + 3*a^3*arcsinh(b*x + a)/b^4 + 6*sqrt(b 
^2*x^2 + 2*a*b*x + a^2 + 1)/(I*b^5*x + I*a*b^4 + b^4) + 1/4*I*(b^2*x^2 + 2 
*a*b*x + a^2 + 1)^(3/2)*x/b^3 + 3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b* 
x + 4*I*a + 3)*a*x/b^3 - 27/2*I*a^2*arcsinh(b*x + a)/b^4 - 3/4*I*(b^2*x^2 
+ 2*a*b*x + a^2 + 1)^(3/2)*a/b^4 - 9/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a 
^2/b^4 + 3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a^2/b^4 
+ 3/8*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x/b^3 - 3/2*I*sqrt(-b^2*x^2 - 2* 
a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*x/b^3 - 3/2*a*arcsin(I*b*x + I*a + 2)/b 
^4 - 18*a*arcsinh(b*x + a)/b^4 - (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/b^...
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.01 \[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx=-\frac {1}{8} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, x {\left (-\frac {i \, x}{b} - \frac {-i \, a b^{11} - 4 \, b^{11}}{b^{13}}\right )} - \frac {2 i \, a^{2} b^{10} + 20 \, a b^{10} - 19 i \, b^{10}}{b^{13}}\right )} x - \frac {-2 i \, a^{3} b^{9} - 44 \, a^{2} b^{9} + 93 i \, a b^{9} + 48 \, b^{9}}{b^{13}}\right )} - \frac {{\left (8 \, a^{3} - 36 i \, a^{2} - 44 \, a + 17 i\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 i \, a^{2} b + 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{3} {\left | b \right |}} \] Input:

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

Output:

-1/8*sqrt((b*x + a)^2 + 1)*((2*x*(-I*x/b - (-I*a*b^11 - 4*b^11)/b^13) - (2 
*I*a^2*b^10 + 20*a*b^10 - 19*I*b^10)/b^13)*x - (-2*I*a^3*b^9 - 44*a^2*b^9 
+ 93*I*a*b^9 + 48*b^9)/b^13) - 1/8*(8*a^3 - 36*I*a^2 - 44*a + 17*I)*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 
*sqrt((b*x + a)^2 + 1))*a*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1) 
)*abs(b))/(b^3*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int e^{-3 i \arctan (a+b x)} x^3 \, dx=-\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{5}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) b^{2}-2 \left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{4}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) a b -\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{3}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) a^{2}-\left (\int \frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x^{3}}{b^{3} i \,x^{3}+3 a \,b^{2} i \,x^{2}+3 a^{2} b i x +a^{3} i +3 b^{2} x^{2}+6 a b x -3 b i x +3 a^{2}-3 a i -1}d x \right ) \] Input:

int(x^3/(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**5)/(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**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)*a*b - 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**2 - int((sq 
rt(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)