\(\int e^{-3 i \arctan (a+b x)} \, dx\) [225]

Optimal result

Integrand size = 12, antiderivative size = 94 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {2 i (1-i a-i b x)^{3/2}}{b \sqrt {1+i a+i b x}}+\frac {3 i \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {3 \text {arcsinh}(a+b x)}{b} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.48 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {\sqrt {1+(a+b x)^2} \left (i+\frac {4}{-i+a+b x}\right )}{b}-\frac {3 \text {arcsinh}(a+b x)}{b} \] Input:

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

Output:

(Sqrt[1 + (a + b*x)^2]*(I + 4/(-I + a + b*x)))/b - (3*ArcSinh[a + b*x])/b
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5616, 57, 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.

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

\(\Big \downarrow \) 5616

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {2 i (-i a-i b x+1)^{3/2}}{b \sqrt {i a+i b x+1}}-3 \left (\frac {\text {arcsinh}\left (\frac {2 a b+2 b^2 x}{2 b}\right )}{b}-\frac {i \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{b}\right )\)

Input:

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

Output:

((2*I)*(1 - I*a - I*b*x)^(3/2))/(b*Sqrt[1 + I*a + I*b*x]) - 3*(((-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)
 

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 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 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 [A] (verified)

Time = 0.80 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b}+\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{2} \left (x -\frac {i-a}{b}\right )}-\frac {3 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}\)	\(129\)
default	\(\frac {i \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{3}}\)	\(327\)

Input:

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

Output:

I/b*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+4/b^2/(x-(I-a)/b)*((x-(I-a)/b)^2*b^2+2*I 
*b*(x-(I-a)/b))^(1/2)-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 = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {{\left (i \, a + 8\right )} b x + i \, a^{2} + 6 \, {\left (b x + a - i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, b x - i \, a - 5\right )} + 9 \, a - 8 i}{2 \, {\left (b^{2} x + {\left (a - i\right )} b\right )}} \] Input:

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

Output:

1/2*((I*a + 8)*b*x + I*a^2 + 6*(b*x + a - I)*log(-b*x - a + sqrt(b^2*x^2 + 
 2*a*b*x + a^2 + 1)) - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(-I*b*x - I*a - 
 5) + 9*a - 8*I)/(b^2*x + (a - I)*b)
 

Sympy [F]

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

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

Output:

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I* 
a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x 
 + I), x) + Integral(a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a 
**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b** 
2*x**2 - 3*b*x + I), x) + Integral(b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x* 
*2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + 
b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(2*a*b*x*sqrt(a**2 + 
2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6 
*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b - 2 i \, b^{2} x - 2 i \, a b - b} - \frac {3 \, \operatorname {arsinh}\left (b x + a\right )}{b} + \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{2} x + i \, a b + b} \] Input:

integrate(1/(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)/(b^3*x^2 + 2*a*b^2*x + a^2*b - 2*I*b 
^2*x - 2*I*a*b - b) - 3*arcsinh(b*x + a)/b + 6*I*sqrt(b^2*x^2 + 2*a*b*x + 
a^2 + 1)/(I*b^2*x + I*a*b + b)
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (66) = 132\).

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.91 \[ \int e^{-3 i \arctan (a+b x)} \, dx=\frac {\log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 i \, a^{2} b + 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + \frac {i \, \sqrt {{\left (b x + a\right )}^{2} + 1}}{b} \] Input:

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

Output:

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))/abs(b) + I*sqrt((b*x + a)^2 + 1)/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/(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)/(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(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(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) - int((sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*x**2)/ 
(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)/(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