\(\int e^{\frac {3}{2} i \arctan (a+b x)} \, dx\) [237]

Optimal result

Integrand size = 14, antiderivative size = 245 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} \, dx=\frac {i \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{b}-\frac {3 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {3 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {3 i \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x} \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}\right )}\right )}{\sqrt {2} b} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.18 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} \, dx=-\frac {8 i e^{\frac {7}{2} i \arctan (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},2,\frac {11}{4},-e^{2 i \arctan (a+b x)}\right )}{7 b} \] Input:

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

Output:

(((-8*I)/7)*E^(((7*I)/2)*ArcTan[a + b*x])*Hypergeometric2F1[7/4, 2, 11/4, 
-E^((2*I)*ArcTan[a + b*x])])/b
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5616, 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.

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

\(\Big \downarrow \) 5616

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(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
 

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 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 5616

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(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, n}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.09 \[ \int e^{\frac {3}{2} i \arctan (a+b x)} \, dx=\frac {b \sqrt {\frac {9 i}{b^{2}}} \log \left (\frac {1}{3} \, b \sqrt {\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {\frac {9 i}{b^{2}}} \log \left (-\frac {1}{3} \, b \sqrt {\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {-\frac {9 i}{b^{2}}} \log \left (\frac {1}{3} \, b \sqrt {-\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + b \sqrt {-\frac {9 i}{b^{2}}} \log \left (-\frac {1}{3} \, b \sqrt {-\frac {9 i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + 2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{2 \, b} \] Input:

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

Output:

1/2*(b*sqrt(9*I/b^2)*log(1/3*b*sqrt(9*I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b 
*x + a^2 + 1)/(b*x + a + I))) - b*sqrt(9*I/b^2)*log(-1/3*b*sqrt(9*I/b^2) + 
 sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - b*sqrt(-9*I/b^ 
2)*log(1/3*b*sqrt(-9*I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b* 
x + a + I))) + b*sqrt(-9*I/b^2)*log(-1/3*b*sqrt(-9*I/b^2) + sqrt(I*sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 2*I*sqrt(b^2*x^2 + 2*a*b*x + 
a^2 + 1)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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