\(\int e^{-i \arctan (a+b x)} x^2 \, dx\) [205]

Optimal result

Integrand size = 16, antiderivative size = 171 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {\left (i-2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \text {arcsinh}(a+b x)}{2 b^3} \] Output:

1/2*(I-2*a-2*I*a^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3+1/6*(I-4*a 
)*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3+1/3*x*(1-I*a-I*b*x)^(3/2)*(1 
+I*a+I*b*x)^(1/2)/b^2-1/2*(1+2*I*a-2*a^2)*arcsinh(b*x+a)/b^3
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.95 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {i \sqrt {1+i a+i b x} \left (4+7 a^2+2 i a^3-7 i b x-5 b^2 x^2+2 i b^3 x^3+a (5 i+8 b x)\right )}{6 b^3 \sqrt {-i (i+a+b x)}}+\frac {\sqrt [4]{-1} \left (-1-2 i a+2 a^2\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \] Input:

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

Output:

((I/6)*Sqrt[1 + I*a + I*b*x]*(4 + 7*a^2 + (2*I)*a^3 - (7*I)*b*x - 5*b^2*x^ 
2 + (2*I)*b^3*x^3 + a*(5*I + 8*b*x)))/(b^3*Sqrt[(-I)*(I + a + b*x)]) + ((- 
1)^(1/4)*(-1 - (2*I)*a + 2*a^2)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]* 
Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5618, 101, 25, 90, 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.

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

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 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 = 0.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {i \left (2 b^{2} x^{2}-2 a b x +3 i b x +2 a^{2}-9 i a -4\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}+\frac {\left (2 a^{2}-2 i a -1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(113\)
default	\(-\frac {i \left (i \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )+b \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )-a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )\right )}{b^{2}}-\frac {\left (i a^{2}+2 a -i\right ) \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b^{3}}\)	\(486\)

Input:

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

Output:

-1/6*I*(2*b^2*x^2+3*I*b*x-2*a*b*x-9*I*a+2*a^2-4)*(b^2*x^2+2*a*b*x+a^2+1)^( 
1/2)/b^3+1/2*(-2*I*a+2*a^2-1)/b^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)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {-7 i \, a^{3} - 21 \, a^{2} - 12 \, {\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 i \, b^{2} x^{2} + {\left (-2 i \, a - 3\right )} b x + 2 i \, a^{2} + 9 \, a - 4 i\right )} + 9 i \, a}{24 \, b^{3}} \] Input:

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

Output:

1/24*(-7*I*a^3 - 21*a^2 - 12*(2*a^2 - 2*I*a - 1)*log(-b*x - a + sqrt(b^2*x 
^2 + 2*a*b*x + a^2 + 1)) - 4*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*I*b^2*x^ 
2 + (-2*I*a - 3)*b*x + 2*I*a^2 + 9*a - 4*I) + 9*I*a)/b^3
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{b^{2}} + \frac {a^{2} \operatorname {arsinh}\left (b x + a\right )}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b^{2}} - \frac {i \, a \operatorname {arsinh}\left (b x + a\right )}{b^{3}} - \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{3}} - \frac {\operatorname {arsinh}\left (b x + a\right )}{2 \, b^{3}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.67 \[ \int e^{-i \arctan (a+b x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (\frac {2 i \, x}{b} - \frac {2 i \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} - \frac {-2 i \, a^{2} b^{2} - 9 \, a b^{2} + 4 i \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left ({\left | -a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |} \right |}\right )}{2 \, b^{2} {\left | b \right |}} \] Input:

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

Output:

-1/6*sqrt((b*x + a)^2 + 1)*(x*(2*I*x/b - (2*I*a*b^3 + 3*b^3)/b^5) - (-2*I* 
a^2*b^2 - 9*a*b^2 + 4*I*b^2)/b^5) - 1/2*(2*a^2 - 2*I*a - 1)*log(abs(-a*b - 
 (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b)))/(b^2*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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