\(\int e^{3 i \arctan (a+b x)} x^4 \, dx\) [194]

Optimal result

Integrand size = 16, antiderivative size = 359 \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=-\frac {3 \left (19 i+68 a-88 i a^2-48 a^3+8 i a^4\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^5}-\frac {2 i x^4 (1+i a+i b x)^{3/2}}{b \sqrt {1-i a-i b x}}-\frac {3 \left (95 i+272 a-254 i a^2-72 a^3\right ) \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{40 b^5}+\frac {3 (17 i+16 a) x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{20 b^3}-\frac {11 x^3 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{5 b^2}+\frac {\left (61 i+118 a-52 i a^2\right ) (1-i a-i b x)^{3/2} (1+i a+i b x)^{3/2}}{20 b^5}-\frac {3 \left (19-68 i a-88 a^2+48 i a^3+8 a^4\right ) \text {arcsinh}(a+b x)}{8 b^5} \] Output:

-3/8*(19*I+68*a-88*I*a^2-48*a^3+8*I*a^4)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x) 
^(1/2)/b^5-2*I*x^4*(1+I*a+I*b*x)^(3/2)/b/(1-I*a-I*b*x)^(1/2)-3/40*(95*I+27 
2*a-254*I*a^2-72*a^3)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(3/2)/b^5+3/20*(17 
*I+16*a)*x^2*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(3/2)/b^3-11/5*x^3*(1-I*a-I 
*b*x)^(1/2)*(1+I*a+I*b*x)^(3/2)/b^2+1/20*(61*I+118*a-52*I*a^2)*(1-I*a-I*b* 
x)^(3/2)*(1+I*a+I*b*x)^(3/2)/b^5-3/8*(19-68*I*a-88*a^2+48*I*a^3+8*a^4)*arc 
sinh(b*x+a)/b^5
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.69 \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=-\frac {\sqrt {1+i a+i b x} \left (448 i+418 i a^4+8 a^5+163 b x+61 i b^2 x^2-34 b^3 x^3-22 i b^4 x^4+8 b^5 x^5+14 i a^3 (121 i+8 b x)-i a^2 \left (2599-422 i b x+52 b^2 x^2\right )+a \left (1763-458 i b x+118 b^2 x^2+32 i b^3 x^3\right )\right )}{40 b^5 \sqrt {-i (i+a+b x)}}+\frac {3 (-1)^{3/4} \left (19-68 i a-88 a^2+48 i a^3+8 a^4\right ) \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{4 \sqrt {-i b} b^{9/2}} \] Input:

Integrate[E^((3*I)*ArcTan[a + b*x])*x^4,x]
 

Output:

-1/40*(Sqrt[1 + I*a + I*b*x]*(448*I + (418*I)*a^4 + 8*a^5 + 163*b*x + (61* 
I)*b^2*x^2 - 34*b^3*x^3 - (22*I)*b^4*x^4 + 8*b^5*x^5 + (14*I)*a^3*(121*I + 
 8*b*x) - I*a^2*(2599 - (422*I)*b*x + 52*b^2*x^2) + a*(1763 - (458*I)*b*x 
+ 118*b^2*x^2 + (32*I)*b^3*x^3)))/(b^5*Sqrt[(-I)*(I + a + b*x)]) + (3*(-1) 
^(3/4)*(19 - (68*I)*a - 88*a^2 + (48*I)*a^3 + 8*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 = 0.88 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.95, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {5618, 108, 27, 170, 27, 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.

\(\displaystyle \int x^4 e^{3 i \arctan (a+b x)} \, dx\)

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {x^4 (i a+i b x+1)^{3/2}}{(-i a-i b x+1)^{3/2}}dx\)

\(\Big \downarrow \) 108

\(\displaystyle \frac {2 i \int \frac {x^3 \sqrt {i a+i b x+1} (8 (i a+1)+11 i b x)}{2 \sqrt {-i a-i b x+1}}dx}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \int \frac {x^3 \sqrt {i a+i b x+1} (8 (i a+1)+11 i b x)}{\sqrt {-i a-i b x+1}}dx}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {i \left (\frac {\int -\frac {3 b x^2 \sqrt {i a+i b x+1} (11 (i-a) (1-i a)-(17-16 i a) b x)}{\sqrt {-i a-i b x+1}}dx}{5 b^2}+\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \int \frac {x^2 \sqrt {i a+i b x+1} \left (11 i \left (a^2+1\right )-(17-16 i a) b x\right )}{\sqrt {-i a-i b x+1}}dx}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (\frac {\int -\frac {b x \sqrt {i a+i b x+1} \left (2 (i-a) (17-16 i a) (a+i)-\left (-52 i a^2+118 a+61 i\right ) b x\right )}{\sqrt {-i a-i b x+1}}dx}{4 b^2}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\int \frac {b x \sqrt {i a+i b x+1} \left (2 (i a+1) (a+i) (16 a+17 i)-\left (-52 i a^2+118 a+61 i\right ) b x\right )}{\sqrt {-i a-i b x+1}}dx}{4 b^2}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\int \frac {x \sqrt {i a+i b x+1} \left (2 (i a+1) (a+i) (16 a+17 i)-\left (-52 i a^2+118 a+61 i\right ) b x\right )}{\sqrt {-i a-i b x+1}}dx}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\frac {5 \left (8 i a^4-48 a^3-88 i a^2+68 a+19 i\right ) \int \frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}dx}{2 b}-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (112 i a^3+2 \left (-52 i a^2+118 a+61 i\right ) b x-422 a^2-458 i a+163\right )}{6 b^2}}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\frac {5 \left (8 i a^4-48 a^3-88 i a^2+68 a+19 i\right ) \left (\int \frac {1}{\sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (112 i a^3+2 \left (-52 i a^2+118 a+61 i\right ) b x-422 a^2-458 i a+163\right )}{6 b^2}}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\frac {5 \left (8 i a^4-48 a^3-88 i a^2+68 a+19 i\right ) \left (\int \frac {1}{\sqrt {b^2 x^2+2 a b x+(1-i a) (i a+1)}}dx+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (112 i a^3+2 \left (-52 i a^2+118 a+61 i\right ) b x-422 a^2-458 i a+163\right )}{6 b^2}}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\frac {5 \left (8 i a^4-48 a^3-88 i a^2+68 a+19 i\right ) \left (\frac {\int \frac {1}{\sqrt {\frac {\left (2 x b^2+2 a b\right )^2}{4 b^2}+1}}d\left (2 x b^2+2 a b\right )}{2 b^2}+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (112 i a^3+2 \left (-52 i a^2+118 a+61 i\right ) b x-422 a^2-458 i a+163\right )}{6 b^2}}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {i \left (\frac {11 i x^3 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{5 b}-\frac {3 \left (-\frac {\frac {5 \left (8 i a^4-48 a^3-88 i a^2+68 a+19 i\right ) \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}+\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (112 i a^3+2 \left (-52 i a^2+118 a+61 i\right ) b x-422 a^2-458 i a+163\right )}{6 b^2}}{4 b}-\frac {(17-16 i a) x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b}\right )}{5 b}\right )}{b}-\frac {2 i x^4 (i a+i b x+1)^{3/2}}{b \sqrt {-i a-i b x+1}}\)

Input:

Int[E^((3*I)*ArcTan[a + b*x])*x^4,x]
 

Output:

((-2*I)*x^4*(1 + I*a + I*b*x)^(3/2))/(b*Sqrt[1 - I*a - I*b*x]) + (I*((((11 
*I)/5)*x^3*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/b - (3*(-1/4*((1 
7 - (16*I)*a)*x^2*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/b - (-1/6 
*(Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(163 - (458*I)*a - 422*a^2 
 + (112*I)*a^3 + 2*(61*I + 118*a - (52*I)*a^2)*b*x))/b^2 + (5*(19*I + 68*a 
 - (88*I)*a^2 - 48*a^3 + (8*I)*a^4)*((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*b))/(4*b)))/(5*b))) 
/b
 

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 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 = 2.32 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.21

method	result	size
risch	\(-\frac {i \left (8 b^{4} x^{4}-8 a \,b^{3} x^{3}-30 i b^{3} x^{3}+8 a^{2} b^{2} x^{2}+70 i a \,b^{2} x^{2}-8 a^{3} b x -130 i a^{2} b x +8 a^{4}+250 i a^{3}-64 b^{2} x^{2}+252 a b x +125 i b x -804 a^{2}-835 i a +288\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{40 b^{5}}-\frac {-\frac {204 i a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {144 i a^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {i \left (-32 i a^{4}+128 a^{3}+192 i a^{2}-128 a -32 i\right ) \sqrt {\left (x +\frac {i+a}{b}\right )^{2} b^{2}-2 i b \left (x +\frac {i+a}{b}\right )}}{b^{2} \left (x +\frac {i+a}{b}\right )}+\frac {57 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\frac {264 a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}+\frac {24 a^{4} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}}{8 b^{4}}\)	\(436\)
default	\(\text {Expression too large to display}\)	\(4893\)

Input:

int((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/40*I*(8*b^4*x^4-30*I*b^3*x^3-8*a*b^3*x^3+70*I*a*b^2*x^2+8*a^2*b^2*x^2-1 
30*I*a^2*b*x-8*a^3*b*x+250*I*a^3+8*a^4-64*b^2*x^2+125*I*b*x+252*a*b*x-835* 
I*a-804*a^2+288)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)/b^5-1/8/b^4*(-204*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)+144*I*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)-I*( 
128*a^3-32*I*a^4-128*a+192*I*a^2-32*I)/b^2/(x+(I+a)/b)*((x+(I+a)/b)^2*b^2- 
2*I*b*(x+(I+a)/b))^(1/2)+57*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)-264*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)+24*a^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 = 0.15 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.74 \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=\frac {-62 i \, a^{6} + 2687 \, a^{5} + 11575 i \, a^{4} - 20350 \, a^{3} + {\left (-62 i \, a^{5} + 2625 \, a^{4} + 8950 i \, a^{3} - 11400 \, a^{2} - 6340 i \, a + 1280\right )} b x - 17740 i \, a^{2} + 120 \, {\left (8 \, a^{5} + 56 i \, a^{4} - 136 \, a^{3} + {\left (8 \, a^{4} + 48 i \, a^{3} - 88 \, a^{2} - 68 i \, a + 19\right )} b x - 156 i \, a^{2} + 87 \, a + 19 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 8 \, {\left (8 i \, b^{5} x^{5} + 22 \, b^{4} x^{4} - 2 \, {\left (16 \, a + 17 i\right )} b^{3} x^{3} + 8 i \, a^{5} + {\left (52 \, a^{2} + 118 i \, a - 61\right )} b^{2} x^{2} - 418 \, a^{4} - 1694 i \, a^{3} - {\left (112 \, a^{3} + 422 i \, a^{2} - 458 \, a - 163 i\right )} b x + 2599 \, a^{2} + 1763 i \, a - 448\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 7620 \, a + 1280 i}{320 \, {\left (b^{6} x + {\left (a + i\right )} b^{5}\right )}} \] Input:

integrate((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x, algorithm="fricas")
 

Output:

1/320*(-62*I*a^6 + 2687*a^5 + 11575*I*a^4 - 20350*a^3 + (-62*I*a^5 + 2625* 
a^4 + 8950*I*a^3 - 11400*a^2 - 6340*I*a + 1280)*b*x - 17740*I*a^2 + 120*(8 
*a^5 + 56*I*a^4 - 136*a^3 + (8*a^4 + 48*I*a^3 - 88*a^2 - 68*I*a + 19)*b*x 
- 156*I*a^2 + 87*a + 19*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1 
)) - 8*(8*I*b^5*x^5 + 22*b^4*x^4 - 2*(16*a + 17*I)*b^3*x^3 + 8*I*a^5 + (52 
*a^2 + 118*I*a - 61)*b^2*x^2 - 418*a^4 - 1694*I*a^3 - (112*a^3 + 422*I*a^2 
 - 458*a - 163*I)*b*x + 2599*a^2 + 1763*I*a - 448)*sqrt(b^2*x^2 + 2*a*b*x 
+ a^2 + 1) + 7620*a + 1280*I)/(b^6*x + (a + I)*b^5)
 

Sympy [F]

\[ \int e^{3 i \arctan (a+b x)} x^4 \, dx =\text {Too large to display} \] Input:

integrate((1+I*(b*x+a))**3/(1+(b*x+a)**2)**(3/2)*x**4,x)
 

Output:

-I*(Integral(I*x**4/(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** 
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(a**3*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*b*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(b**3*x**7/(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**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*I*b**2*x**6/(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**6/(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. 3081 vs. \(2 (253) = 506\).

Time = 0.08 (sec) , antiderivative size = 3081, normalized size of antiderivative = 8.58 \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=\text {Too large to display} \] Input:

integrate((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x, algorithm="maxima")
 

Output:

-1/5*I*b*x^6/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 11/20*I*a*x^5/sqrt(b^2*x^ 
2 + 2*a*b*x + a^2 + 1) - 693/4*I*a^7*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2 
*x^2 + 2*a*b*x + a^2 + 1)*b^2) - 33/20*I*a^2*x^4/(sqrt(b^2*x^2 + 2*a*b*x + 
 a^2 + 1)*b) + 2415/8*I*(a^2 + 1)*a^5*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1)*b^2) + 231/40*I*a^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x 
 + a^2 + 1)*b^2) - 2/5*(-I*a^2 - I)*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) 
*b) - 3/4*(I*a*b^2 + b^2)*x^5/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b^2) - 23 
1/8*I*(a^2 + 1)*a^6/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^ 
2 + 1)*b^3) + 945/4*(I*a*b^2 + b^2)*a^6*x/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt( 
b^2*x^2 + 2*a*b*x + a^2 + 1)*b^4) - 105*(I*a^2*b + 2*a*b - I*b)*a^5*x/((a^ 
2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b^3) - 2919/20*I* 
(a^2 + 1)^2*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^2) - 15*(-I*a^3 - 3*a^2 + 3*I*a + 1)*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) - 231/8*I*a^4*x^2/(sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)*b^3) - 111/40*I*(a^2 + 1)*a*x^3/(sqrt(b^2*x^2 + 2*a*b 
*x + a^2 + 1)*b^2) + 9/4*(I*a*b^2 + b^2)*a*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a 
^2 + 1)*b^3) - (I*a^2*b + 2*a*b - I*b)*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 
 1)*b^2) + 189/5*I*(a^2 + 1)^2*a^4/((a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 
 + 2*a*b*x + a^2 + 1)*b^3) - 2835/8*(I*a*b^2 + b^2)*(a^2 + 1)*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^4) + 265/2*(I*...
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.93 \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=-\frac {1}{40} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, {\left (x {\left (\frac {4 i \, x}{b} - \frac {4 i \, a b^{17} - 15 \, b^{17}}{b^{19}}\right )} - \frac {-4 i \, a^{2} b^{16} + 35 \, a b^{16} + 32 i \, b^{16}}{b^{19}}\right )} x - \frac {8 i \, a^{3} b^{15} - 130 \, a^{2} b^{15} - 252 i \, a b^{15} + 125 \, b^{15}}{b^{19}}\right )} x - \frac {-8 i \, a^{4} b^{14} + 250 \, a^{3} b^{14} + 804 i \, a^{2} b^{14} - 835 \, a b^{14} - 288 i \, b^{14}}{b^{19}}\right )} + \frac {{\left (8 \, a^{4} + 48 i \, a^{3} - 88 \, a^{2} - 68 i \, a + 19\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} + 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 i \, a^{2} b + 4 \, {\left (i \, x {\left | b \right |} - i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{4} {\left | b \right |}} \] Input:

integrate((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x, algorithm="giac")
 

Output:

-1/40*sqrt((b*x + a)^2 + 1)*((2*(x*(4*I*x/b - (4*I*a*b^17 - 15*b^17)/b^19) 
 - (-4*I*a^2*b^16 + 35*a*b^16 + 32*I*b^16)/b^19)*x - (8*I*a^3*b^15 - 130*a 
^2*b^15 - 252*I*a*b^15 + 125*b^15)/b^19)*x - (-8*I*a^4*b^14 + 250*a^3*b^14 
 + 804*I*a^2*b^14 - 835*a*b^14 - 288*I*b^14)/b^19) + 1/8*(8*a^4 + 48*I*a^3 
 - 88*a^2 - 68*I*a + 19)*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^4*abs(b))
 

Mupad [F(-1)]

Timed out. \[ \int e^{3 i \arctan (a+b x)} x^4 \, dx=\int \frac {x^4\,{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3}{{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}} \,d x \] Input:

int((x^4*(a*1i + b*x*1i + 1)^3)/((a + b*x)^2 + 1)^(3/2),x)
 

Output:

int((x^4*(a*1i + b*x*1i + 1)^3)/((a + b*x)^2 + 1)^(3/2), x)
 

Reduce [F]

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

int((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x)
 

Output:

int((1+I*(b*x+a))^3/(1+(b*x+a)^2)^(3/2)*x^4,x)