\(\int e^{-i \arctan (a+b x)} x^4 \, dx\) [203]

Optimal result

Integrand size = 16, antiderivative size = 315 \[ \int e^{-i \arctan (a+b x)} x^4 \, dx=-\frac {\left (3 i-12 a-24 i a^2+16 a^3+8 i a^4\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^5}+\frac {7 \left (i+16 a+6 i a^2-24 a^3\right ) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{120 b^5}+\frac {(i-8 a) x^2 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{20 b^3}+\frac {x^3 (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{5 b^2}-\frac {\left (13 i-14 a-36 i a^2\right ) (1-i a-i b x)^{5/2} \sqrt {1+i a+i b x}}{60 b^5}+\frac {\left (3+12 i a-24 a^2-16 i a^3+8 a^4\right ) \text {arcsinh}(a+b x)}{8 b^5} \] Output:

-1/8*(3*I-12*a-24*I*a^2+16*a^3+8*I*a^4)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^ 
(1/2)/b^5+7/120*(I+16*a+6*I*a^2-24*a^3)*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^ 
(1/2)/b^5+1/20*(I-8*a)*x^2*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3+1/5 
*x^3*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^2-1/60*(13*I-14*a-36*I*a^2) 
*(1-I*a-I*b*x)^(5/2)*(1+I*a+I*b*x)^(1/2)/b^5+1/8*(3+12*I*a-24*a^2-16*I*a^3 
+8*a^4)*arcsinh(b*x+a)/b^5
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.79 \[ \int e^{-i \arctan (a+b x)} x^4 \, dx=\frac {i \sqrt {1+i a+i b x} \left (-64+226 a^4+24 i a^5+109 i b x+77 b^2 x^2-62 i b^3 x^3-54 b^4 x^4+24 i b^5 x^5+2 a^3 (-41 i+72 b x)+a^2 \left (57-346 i b x-84 b^2 x^2\right )+a \left (-211 i-346 b x+154 i b^2 x^2+64 b^3 x^3\right )\right )}{120 b^5 \sqrt {-i (i+a+b x)}}+\frac {\sqrt [4]{-1} \left (-3 i+12 a+24 i a^2-16 a^3-8 i a^4\right ) \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{4 \sqrt {-i b} b^{9/2}} \] Input:

Integrate[x^4/E^(I*ArcTan[a + b*x]),x]
 

Output:

((I/120)*Sqrt[1 + I*a + I*b*x]*(-64 + 226*a^4 + (24*I)*a^5 + (109*I)*b*x + 
 77*b^2*x^2 - (62*I)*b^3*x^3 - 54*b^4*x^4 + (24*I)*b^5*x^5 + 2*a^3*(-41*I 
+ 72*b*x) + a^2*(57 - (346*I)*b*x - 84*b^2*x^2) + a*(-211*I - 346*b*x + (1 
54*I)*b^2*x^2 + 64*b^3*x^3)))/(b^5*Sqrt[(-I)*(I + a + b*x)]) + ((-1)^(1/4) 
*(-3*I + 12*a + (24*I)*a^2 - 16*a^3 - (8*I)*a^4)*ArcSinh[((1/2 + I/2)*Sqrt 
[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(4*Sqrt[(-I)*b]*b^(9/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.90, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {5618, 111, 25, 170, 25, 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.

\(\displaystyle \int x^4 e^{-i \arctan (a+b x)} \, dx\)

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {x^4 \sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}dx\)

\(\Big \downarrow \) 111

\(\displaystyle \frac {\int -\frac {x^2 \sqrt {-i a-i b x+1} \left (3 \left (a^2+1\right )-(i-8 a) b x\right )}{\sqrt {i a+i b x+1}}dx}{5 b^2}+\frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {\int \frac {x^2 \sqrt {-i a-i b x+1} \left (3 \left (a^2+1\right )-(i-8 a) b x\right )}{\sqrt {i a+i b x+1}}dx}{5 b^2}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {\frac {\int -\frac {b x \sqrt {-i a-i b x+1} \left (2 (i-8 a) (i-a) (a+i)-\left (-36 a^2+14 i a+13\right ) b x\right )}{\sqrt {i a+i b x+1}}dx}{4 b^2}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\int \frac {b x \sqrt {-i a-i b x+1} \left (2 (i-8 a) (i-a) (a+i)-\left (-36 a^2+14 i a+13\right ) b x\right )}{\sqrt {i a+i b x+1}}dx}{4 b^2}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\int \frac {x \sqrt {-i a-i b x+1} \left (2 (i-8 a) (i-a) (a+i)-\left (-36 a^2+14 i a+13\right ) b x\right )}{\sqrt {i a+i b x+1}}dx}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\frac {5 \left (8 a^4-16 i a^3-24 a^2+12 i a+3\right ) \int \frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}dx}{2 b}-\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (96 a^3+2 \left (-36 a^2+14 i a+13\right ) b x-86 i a^2-114 a+19 i\right )}{6 b^2}}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\frac {5 \left (8 a^4-16 i a^3-24 a^2+12 i a+3\right ) \left (\int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (96 a^3+2 \left (-36 a^2+14 i a+13\right ) b x-86 i a^2-114 a+19 i\right )}{6 b^2}}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\frac {5 \left (8 a^4-16 i a^3-24 a^2+12 i a+3\right ) \left (\int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (96 a^3+2 \left (-36 a^2+14 i a+13\right ) b x-86 i a^2-114 a+19 i\right )}{6 b^2}}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\frac {5 \left (8 a^4-16 i a^3-24 a^2+12 i a+3\right ) \left (\frac {\int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b^2}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (96 a^3+2 \left (-36 a^2+14 i a+13\right ) b x-86 i a^2-114 a+19 i\right )}{6 b^2}}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {x^3 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{5 b^2}-\frac {-\frac {\frac {5 \left (8 a^4-16 i a^3-24 a^2+12 i a+3\right ) \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {(-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1} \left (96 a^3+2 \left (-36 a^2+14 i a+13\right ) b x-86 i a^2-114 a+19 i\right )}{6 b^2}}{4 b}-\frac {(-8 a+i) x^2 (-i a-i b x+1)^{3/2} \sqrt {i a+i b x+1}}{4 b}}{5 b^2}\)

Input:

Int[x^4/E^(I*ArcTan[a + b*x]),x]
 

Output:

(x^3*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(5*b^2) - (-1/4*((I - 
8*a)*x^2*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/b - (-1/6*((1 - I* 
a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x]*(19*I - 114*a - (86*I)*a^2 + 96*a^3 
 + 2*(13 + (14*I)*a - 36*a^2)*b*x))/b^2 + (5*(3 + (12*I)*a - 24*a^2 - (16* 
I)*a^3 + 8*a^4)*(((-I)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/b + Ar 
cSinh[(2*a*b + 2*b^2*x)/(2*b)]/b))/(2*b))/(4*b))/(5*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 111

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

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 = 0.96 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {i \left (24 b^{4} x^{4}-24 a \,b^{3} x^{3}+30 i b^{3} x^{3}+24 a^{2} b^{2} x^{2}-70 i a \,b^{2} x^{2}-24 a^{3} b x +130 i a^{2} b x +24 a^{4}-250 i a^{3}-32 b^{2} x^{2}+116 a b x -45 i b x -332 a^{2}+275 i a +64\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{120 b^{5}}+\frac {\left (8 a^{4}-16 i a^{3}-24 a^{2}+12 i a +3\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{4} \sqrt {b^{2}}}\)	\(197\)
default	\(\text {Expression too large to display}\)	\(1153\)

Input:

int(x^4/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/120*I*(24*b^4*x^4+30*I*b^3*x^3-24*a*b^3*x^3-70*I*a*b^2*x^2+24*a^2*b^2*x 
^2+130*I*a^2*b*x-24*a^3*b*x-250*I*a^3+24*a^4-32*b^2*x^2-45*I*b*x+116*a*b*x 
+275*I*a-332*a^2+64)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b^5+1/8*(3+12*I*a-24*a^ 
2-16*I*a^3+8*a^4)/b^4*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 = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.56 \[ \int e^{-i \arctan (a+b x)} x^4 \, dx=\frac {-186 i \, a^{5} - 1345 \, a^{4} + 1730 i \, a^{3} + 1320 \, a^{2} - 120 \, {\left (8 \, a^{4} - 16 i \, a^{3} - 24 \, a^{2} + 12 i \, a + 3\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 8 \, {\left (24 i \, b^{4} x^{4} + 6 \, {\left (-4 i \, a - 5\right )} b^{3} x^{3} + 2 \, {\left (12 i \, a^{2} + 35 \, a - 16 i\right )} b^{2} x^{2} + 24 i \, a^{4} + 250 \, a^{3} + {\left (-24 i \, a^{3} - 130 \, a^{2} + 116 i \, a + 45\right )} b x - 332 i \, a^{2} - 275 \, a + 64 i\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 300 i \, a}{960 \, b^{5}} \] Input:

integrate(x^4/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

1/960*(-186*I*a^5 - 1345*a^4 + 1730*I*a^3 + 1320*a^2 - 120*(8*a^4 - 16*I*a 
^3 - 24*a^2 + 12*I*a + 3)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) 
) - 8*(24*I*b^4*x^4 + 6*(-4*I*a - 5)*b^3*x^3 + 2*(12*I*a^2 + 35*a - 16*I)* 
b^2*x^2 + 24*I*a^4 + 250*a^3 + (-24*I*a^3 - 130*a^2 + 116*I*a + 45)*b*x - 
332*I*a^2 - 275*a + 64*I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 300*I*a)/b^5
 

Sympy [F]

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

integrate(x**4/(1+I*(b*x+a))*(1+(b*x+a)**2)**(1/2),x)
 

Output:

-I*Integral(x**4*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a + b*x - I), x)
 

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. 456 vs. \(2 (223) = 446\).

Time = 0.14 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.45 \[ \int e^{-i \arctan (a+b x)} x^4 \, dx=\frac {2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{3} x}{b^{4}} - \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{5 \, b^{3}} + \frac {a^{4} \operatorname {arsinh}\left (b x + a\right )}{b^{5}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{4}}{b^{5}} + \frac {3 i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a x}{5 \, b^{4}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} x}{b^{4}} - \frac {2 i \, a^{3} \operatorname {arsinh}\left (b x + a\right )}{b^{5}} - \frac {6 i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{5 \, b^{5}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{3}}{b^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{4}} - \frac {5 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{2 \, b^{4}} - \frac {3 \, a^{2} \operatorname {arsinh}\left (b x + a\right )}{b^{5}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{12 \, b^{5}} + \frac {7 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{5}} - \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{8 \, b^{4}} + \frac {3 i \, a \operatorname {arsinh}\left (b x + a\right )}{2 \, b^{5}} + \frac {7 i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{15 \, b^{5}} + \frac {27 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{8 \, b^{5}} + \frac {3 \, \operatorname {arsinh}\left (b x + a\right )}{8 \, b^{5}} - \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{5}} \] Input:

integrate(x^4/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^3*x/b^4 - 1/5*I*(b^2*x^2 + 2*a*b*x 
 + a^2 + 1)^(3/2)*x^2/b^3 + a^4*arcsinh(b*x + a)/b^5 + I*sqrt(b^2*x^2 + 2* 
a*b*x + a^2 + 1)*a^4/b^5 + 3/5*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a*x/b 
^4 + 3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*x/b^4 - 2*I*a^3*arcsinh(b*x + 
 a)/b^5 - 6/5*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/b^5 - sqrt(b^2*x^2 
 + 2*a*b*x + a^2 + 1)*a^3/b^5 + 1/4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x/ 
b^4 - 5/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^4 - 3*a^2*arcsinh(b*x 
+ a)/b^5 - 13/12*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/b^5 + 7/2*I*sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^5 - 5/8*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) 
*x/b^4 + 3/2*I*a*arcsinh(b*x + a)/b^5 + 7/15*I*(b^2*x^2 + 2*a*b*x + a^2 + 
1)^(3/2)/b^5 + 27/8*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^5 + 3/8*arcsinh( 
b*x + a)/b^5 - I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^5
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.65 \[ \int e^{-i \arctan (a+b x)} x^4 \, dx=-\frac {1}{120} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, {\left (3 \, x {\left (\frac {4 i \, x}{b} - \frac {4 i \, a b^{7} + 5 \, b^{7}}{b^{9}}\right )} - \frac {-12 i \, a^{2} b^{6} - 35 \, a b^{6} + 16 i \, b^{6}}{b^{9}}\right )} x - \frac {24 i \, a^{3} b^{5} + 130 \, a^{2} b^{5} - 116 i \, a b^{5} - 45 \, b^{5}}{b^{9}}\right )} x - \frac {-24 i \, a^{4} b^{4} - 250 \, a^{3} b^{4} + 332 i \, a^{2} b^{4} + 275 \, a b^{4} - 64 i \, b^{4}}{b^{9}}\right )} - \frac {{\left (8 \, a^{4} - 16 i \, a^{3} - 24 \, a^{2} + 12 i \, a + 3\right )} \log \left ({\left | -a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |} \right |}\right )}{8 \, b^{4} {\left | b \right |}} \] Input:

integrate(x^4/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

-1/120*sqrt((b*x + a)^2 + 1)*((2*(3*x*(4*I*x/b - (4*I*a*b^7 + 5*b^7)/b^9) 
- (-12*I*a^2*b^6 - 35*a*b^6 + 16*I*b^6)/b^9)*x - (24*I*a^3*b^5 + 130*a^2*b 
^5 - 116*I*a*b^5 - 45*b^5)/b^9)*x - (-24*I*a^4*b^4 - 250*a^3*b^4 + 332*I*a 
^2*b^4 + 275*a*b^4 - 64*I*b^4)/b^9) - 1/8*(8*a^4 - 16*I*a^3 - 24*a^2 + 12* 
I*a + 3)*log(abs(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b)))/(b^4*a 
bs(b))
 

Mupad [F(-1)]

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

int((x^4*((a + b*x)^2 + 1)^(1/2))/(a*1i + b*x*1i + 1),x)
 

Output:

int((x^4*((a + b*x)^2 + 1)^(1/2))/(a*1i + b*x*1i + 1), x)
 

Reduce [F]

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

int(x^4/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x)
 

Output:

int((sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*x**4)/(a*i + b*i*x + 1),x)