\(\int e^{\frac {3}{2} i \arctan (a+b x)} x \, dx\) [236]

Optimal result

Integrand size = 16, antiderivative size = 314 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} x \, dx=\frac {(3-4 i a) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{7/4}}{2 b^2}-\frac {3 (3-4 i a) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {3 (3-4 i a) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {3 (3-4 i a) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x} \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}\right )}\right )}{4 \sqrt {2} b^2} \] Output:

1/4*(3-4*I*a)*(1-I*a-I*b*x)^(1/4)*(1+I*a+I*b*x)^(3/4)/b^2+1/2*(1-I*a-I*b*x 
)^(1/4)*(1+I*a+I*b*x)^(7/4)/b^2-3/8*(3-4*I*a)*arctan(1-2^(1/2)*(1-I*a-I*b* 
x)^(1/4)/(1+I*a+I*b*x)^(1/4))*2^(1/2)/b^2+3/8*(3-4*I*a)*arctan(1+2^(1/2)*( 
1-I*a-I*b*x)^(1/4)/(1+I*a+I*b*x)^(1/4))*2^(1/2)/b^2+3/8*(3-4*I*a)*arctanh( 
2^(1/2)*(1-I*a-I*b*x)^(1/4)/(1+I*a+I*b*x)^(1/4)/(1+(1-I*a-I*b*x)^(1/2)/(1+ 
I*a+I*b*x)^(1/2)))*2^(1/2)/b^2
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.25 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} x \, dx=\frac {\sqrt [4]{-i (i+a+b x)} \left ((1+i a+i b x)^{7/4}+2\ 2^{3/4} (3-4 i a) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{4},\frac {5}{4},-\frac {1}{2} i (i+a+b x)\right )\right )}{2 b^2} \] Input:

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

Output:

(((-I)*(I + a + b*x))^(1/4)*((1 + I*a + I*b*x)^(7/4) + 2*2^(3/4)*(3 - (4*I 
)*a)*Hypergeometric2F1[-3/4, 1/4, 5/4, (-1/2*I)*(I + a + b*x)]))/(2*b^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {5618, 90, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \int \frac {(i a+i b x+1)^{3/4}}{(-i a-i b x+1)^{3/4}}dx}{4 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {3}{2} \int \frac {1}{(-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}dx+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \int \frac {1}{\sqrt [4]{i a+i b x+1}}d\sqrt [4]{-i a-i b x+1}}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \int \frac {1}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \int \frac {\sqrt {-i a-i b x+1}+1}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{7/4}}{2 b^2}-\frac {(4 a+3 i) \left (\frac {6 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{b}+\frac {i \sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{b}\right )}{4 b}\)

Input:

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

Output:

((1 - I*a - I*b*x)^(1/4)*(1 + I*a + I*b*x)^(7/4))/(2*b^2) - ((3*I + 4*a)*( 
(I*(1 - I*a - I*b*x)^(1/4)*(1 + I*a + I*b*x)^(3/4))/b + ((6*I)*((-(ArcTan[ 
1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2]) + 
ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt 
[2])/2 + (-1/2*Log[1 + Sqrt[1 - I*a - I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^ 
(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2] + Log[1 + Sqrt[1 - I*a - I*b*x] + 
(Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt[2]))/2) 
)/b))/(4*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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

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 [F]

\[\int {\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}^{\frac {3}{2}} x d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (212) = 424\).

Time = 0.13 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.37 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} x \, dx=\frac {3 \, b^{2} \sqrt {-\frac {16 i \, a^{2} - 24 \, a - 9 i}{b^{4}}} \log \left (\frac {b^{2} \sqrt {-\frac {16 i \, a^{2} - 24 \, a - 9 i}{b^{4}}} + {\left (4 \, a + 3 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + 3 i}\right ) - 3 \, b^{2} \sqrt {-\frac {16 i \, a^{2} - 24 \, a - 9 i}{b^{4}}} \log \left (-\frac {b^{2} \sqrt {-\frac {16 i \, a^{2} - 24 \, a - 9 i}{b^{4}}} - {\left (4 \, a + 3 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + 3 i}\right ) - 3 \, b^{2} \sqrt {-\frac {-16 i \, a^{2} + 24 \, a + 9 i}{b^{4}}} \log \left (\frac {b^{2} \sqrt {-\frac {-16 i \, a^{2} + 24 \, a + 9 i}{b^{4}}} + {\left (4 \, a + 3 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + 3 i}\right ) + 3 \, b^{2} \sqrt {-\frac {-16 i \, a^{2} + 24 \, a + 9 i}{b^{4}}} \log \left (-\frac {b^{2} \sqrt {-\frac {-16 i \, a^{2} + 24 \, a + 9 i}{b^{4}}} - {\left (4 \, a + 3 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + 3 i}\right ) + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 i \, b x - 2 i \, a + 5\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, b^{2}} \] Input:

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

Output:

1/8*(3*b^2*sqrt(-(16*I*a^2 - 24*a - 9*I)/b^4)*log((b^2*sqrt(-(16*I*a^2 - 2 
4*a - 9*I)/b^4) + (4*a + 3*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b* 
x + a + I)))/(4*a + 3*I)) - 3*b^2*sqrt(-(16*I*a^2 - 24*a - 9*I)/b^4)*log(- 
(b^2*sqrt(-(16*I*a^2 - 24*a - 9*I)/b^4) - (4*a + 3*I)*sqrt(I*sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + 3*I)) - 3*b^2*sqrt(-(-16*I*a^2 
 + 24*a + 9*I)/b^4)*log((b^2*sqrt(-(-16*I*a^2 + 24*a + 9*I)/b^4) + (4*a + 
3*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + 3*I)) 
 + 3*b^2*sqrt(-(-16*I*a^2 + 24*a + 9*I)/b^4)*log(-(b^2*sqrt(-(-16*I*a^2 + 
24*a + 9*I)/b^4) - (4*a + 3*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b 
*x + a + I)))/(4*a + 3*I)) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*I*b*x 
- 2*I*a + 5)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b^2
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int e^{\frac {3}{2} i \arctan (a+b x)} x \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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