\(\int e^{-3 i \arctan (a x)} x^2 \, dx\) [68]

Optimal result

Integrand size = 14, antiderivative size = 103 \[ \int e^{-3 i \arctan (a x)} x^2 \, dx=-\frac {4 i (1-i a x)}{a^3 \sqrt {1+a^2 x^2}}-\frac {5 i \sqrt {1+a^2 x^2}}{a^3}-\frac {3 x \sqrt {1+a^2 x^2}}{2 a^2}+\frac {i \left (1+a^2 x^2\right )^{3/2}}{3 a^3}+\frac {11 \text {arcsinh}(a x)}{2 a^3} \] Output:

-4*I*(1-I*a*x)/a^3/(a^2*x^2+1)^(1/2)-5*I*(a^2*x^2+1)^(1/2)/a^3-3/2*x*(a^2* 
x^2+1)^(1/2)/a^2+1/3*I*(a^2*x^2+1)^(3/2)/a^3+11/2*arcsinh(a*x)/a^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61 \[ \int e^{-3 i \arctan (a x)} x^2 \, dx=\frac {\frac {\sqrt {1+a^2 x^2} \left (-52-19 i a x-7 a^2 x^2+2 i a^3 x^3\right )}{-i+a x}+33 \text {arcsinh}(a x)}{6 a^3} \] Input:

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

Output:

((Sqrt[1 + a^2*x^2]*(-52 - (19*I)*a*x - 7*a^2*x^2 + (2*I)*a^3*x^3))/(-I + 
a*x) + 33*ArcSinh[a*x])/(6*a^3)
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5583, 2164, 25, 2027, 2164, 27, 563, 2346, 2346, 27, 455, 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.

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

\(\Big \downarrow \) 5583

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

\(\Big \downarrow \) 2164

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2027

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

\(\Big \downarrow \) 2164

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 563

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {\int \frac {-9 x^2 a^4-14 i x a^3+12 a^2}{\sqrt {a^2 x^2+1}}dx}{3 a^2}+\frac {1}{3} i a x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {4 i \sqrt {a^2 x^2+1}}{a^3 (1+i a x)}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {-\frac {9}{2} a^2 x \sqrt {a^2 x^2+1}+\frac {\int \frac {a^4 (33-28 i a x)}{\sqrt {a^2 x^2+1}}dx}{2 a^2}}{3 a^2}+\frac {1}{3} i a x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {4 i \sqrt {a^2 x^2+1}}{a^3 (1+i a x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {9}{2} a^2 x \sqrt {a^2 x^2+1}+\frac {1}{2} a^2 \int \frac {33-28 i a x}{\sqrt {a^2 x^2+1}}dx}{3 a^2}+\frac {1}{3} i a x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {4 i \sqrt {a^2 x^2+1}}{a^3 (1+i a x)}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {-\frac {9}{2} a^2 x \sqrt {a^2 x^2+1}+\frac {1}{2} a^2 \left (33 \int \frac {1}{\sqrt {a^2 x^2+1}}dx-\frac {28 i \sqrt {a^2 x^2+1}}{a}\right )}{3 a^2}+\frac {1}{3} i a x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {4 i \sqrt {a^2 x^2+1}}{a^3 (1+i a x)}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {-\frac {9}{2} a^2 x \sqrt {a^2 x^2+1}+\frac {1}{2} a^2 \left (\frac {33 \text {arcsinh}(a x)}{a}-\frac {28 i \sqrt {a^2 x^2+1}}{a}\right )}{3 a^2}+\frac {1}{3} i a x^2 \sqrt {a^2 x^2+1}}{a^2}-\frac {4 i \sqrt {a^2 x^2+1}}{a^3 (1+i a x)}\)

Input:

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

Output:

((-4*I)*Sqrt[1 + a^2*x^2])/(a^3*(1 + I*a*x)) + ((I/3)*a*x^2*Sqrt[1 + a^2*x 
^2] + ((-9*a^2*x*Sqrt[1 + a^2*x^2])/2 + (a^2*(((-28*I)*Sqrt[1 + a^2*x^2])/ 
a + (33*ArcSinh[a*x])/a))/2)/(3*a^2))/a^2
 

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 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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 563

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* 
b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1)   Int[(1/Sqrt[a 
+ b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n 
 - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 
, 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2164

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Simp[d*e   Int[(d + e*x)^(m - 1)*PolynomialQuotient[Pq, a*e + b*d*x, x]* 
(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] 
 && EqQ[b*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + b*d*x, x], 0 
]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

rule 5583

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a* 
x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1 + a^2*x^2])), x] /; Free 
Q[{a, m}, x] && IntegerQ[(I*n - 1)/2]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {i \left (2 a^{2} x^{2}+9 i a x -28\right ) \sqrt {a^{2} x^{2}+1}}{6 a^{3}}-\frac {4 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a^{4} \left (x -\frac {i}{a}\right )}+\frac {11 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\)	\(111\)
default	\(-\frac {i \left (\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-2 i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )\right )}{a^{5}}+\frac {i \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )}{a^{3}}-\frac {2 \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{4}}\)	\(619\)

Input:

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

Output:

1/6*I*(2*a^2*x^2+9*I*a*x-28)*(a^2*x^2+1)^(1/2)/a^3-4/a^4/(x-I/a)*((x-I/a)^ 
2*a^2+2*I*a*(x-I/a))^(1/2)+11/2/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2) 
)/(a^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int e^{-3 i \arctan (a x)} x^2 \, dx=-\frac {24 \, a x + 33 \, {\left (a x - i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) - {\left (2 i \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 19 i \, a x - 52\right )} \sqrt {a^{2} x^{2} + 1} - 24 i}{6 \, {\left (a^{4} x - i \, a^{3}\right )}} \] Input:

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

Output:

-1/6*(24*a*x + 33*(a*x - I)*log(-a*x + sqrt(a^2*x^2 + 1)) - (2*I*a^3*x^3 - 
 7*a^2*x^2 - 19*I*a*x - 52)*sqrt(a^2*x^2 + 1) - 24*I)/(a^4*x - I*a^3)
 

Sympy [F]

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

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

Output:

I*(Integral(x**2*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + 
I), x) + Integral(a**2*x**4*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 
 - 3*a*x + I), x))
 

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. 181 vs. \(2 (80) = 160\).

Time = 0.11 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.76 \[ \int e^{-3 i \arctan (a x)} x^2 \, dx=-\frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{5} x^{2} - 2 i \, a^{4} x - a^{3}} - \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{i \, a^{4} x + a^{3}} - \frac {6 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{4} x + a^{3}} - \frac {\sqrt {-a^{2} x^{2} + 4 i \, a x + 3} x}{2 \, a^{2}} + \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{3}} + \frac {\arcsin \left (i \, a x + 2\right )}{2 \, a^{3}} + \frac {6 \, \operatorname {arsinh}\left (a x\right )}{a^{3}} - \frac {3 i \, \sqrt {a^{2} x^{2} + 1}}{a^{3}} + \frac {i \, \sqrt {-a^{2} x^{2} + 4 i \, a x + 3}}{a^{3}} \] Input:

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

Output:

-I*(a^2*x^2 + 1)^(3/2)/(a^5*x^2 - 2*I*a^4*x - a^3) - I*(a^2*x^2 + 1)^(3/2) 
/(I*a^4*x + a^3) - 6*I*sqrt(a^2*x^2 + 1)/(I*a^4*x + a^3) - 1/2*sqrt(-a^2*x 
^2 + 4*I*a*x + 3)*x/a^2 + 1/3*I*(a^2*x^2 + 1)^(3/2)/a^3 + 1/2*arcsin(I*a*x 
 + 2)/a^3 + 6*arcsinh(a*x)/a^3 - 3*I*sqrt(a^2*x^2 + 1)/a^3 + I*sqrt(-a^2*x 
^2 + 4*I*a*x + 3)/a^3
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 22.78 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12 \[ \int e^{-3 i \arctan (a x)} x^2 \, dx=\frac {11\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{2\,a^2\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {3\,x\,\sqrt {a^2}}{2\,a^2}+\frac {a\,14{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}-\frac {a^3\,x^2\,1{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}\right )}{\sqrt {a^2}}+\frac {4\,\sqrt {a^2\,x^2+1}}{a^2\,\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \] Input:

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

Output:

(11*asinh(x*(a^2)^(1/2)))/(2*a^2*(a^2)^(1/2)) - ((a^2*x^2 + 1)^(1/2)*((a*1 
4i)/(3*(a^2)^(3/2)) - (a^3*x^2*1i)/(3*(a^2)^(3/2)) + (3*x*(a^2)^(1/2))/(2* 
a^2)))/(a^2)^(1/2) + (4*(a^2*x^2 + 1)^(1/2))/(a^2*(((a^2)^(1/2)*1i)/a - x* 
(a^2)^(1/2))*(a^2)^(1/2))
 

Reduce [F]

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

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

Output:

 - (int((sqrt(a**2*x**2 + 1)*x**4)/(a**3*i*x**3 + 3*a**2*x**2 - 3*a*i*x - 
1),x)*a**2 + int((sqrt(a**2*x**2 + 1)*x**2)/(a**3*i*x**3 + 3*a**2*x**2 - 3 
*a*i*x - 1),x))