\(\int e^{-3 i \arctan (a x)} x \, dx\) [54]

Optimal result

Integrand size = 12, antiderivative size = 92 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {9 i \text {arcsinh}(a x)}{2 a^2} \]

[Out]

-3/2*(a^2*x^2+1)^(3/2)/a^2/(1+I*a*x)-(a^2*x^2+1)^(5/2)/a^2/(1+I*a*x)^3-9/2*I*arcsinh(a*x)/a^2-9/2*(a^2*x^2+1)^
(1/2)/a^2

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5168, 1647, 1607, 12, 807, 679, 221} \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {9 i \text {arcsinh}(a x)}{2 a^2}-\frac {\left (a^2 x^2+1\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {3 \left (a^2 x^2+1\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {9 \sqrt {a^2 x^2+1}}{2 a^2} \]

[In]

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

[Out]

(-9*Sqrt[1 + a^2*x^2])/(2*a^2) - (3*(1 + a^2*x^2)^(3/2))/(2*a^2*(1 + I*a*x)) - (1 + a^2*x^2)^(5/2)/(a^2*(1 + I
*a*x)^3) - (((9*I)/2)*ArcSinh[a*x])/a^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 221

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 679

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] - Dist[2*c*d*(p/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1647

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

Rule 5168

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] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1-i a x)^2}{(1+i a x) \sqrt {1+a^2 x^2}} \, dx \\ & = (i a) \int \frac {\left (-\frac {i x}{a}-x^2\right ) \sqrt {1+a^2 x^2}}{(1+i a x)^2} \, dx \\ & = (i a) \int \frac {\left (-\frac {i}{a}-x\right ) x \sqrt {1+a^2 x^2}}{(1+i a x)^2} \, dx \\ & = a^2 \int \frac {x \left (1+a^2 x^2\right )^{3/2}}{a^2 (1+i a x)^3} \, dx \\ & = \int \frac {x \left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^3} \, dx \\ & = -\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {(3 i) \int \frac {\left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^2} \, dx}{a} \\ & = -\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {(9 i) \int \frac {\sqrt {1+a^2 x^2}}{1+i a x} \, dx}{2 a} \\ & = -\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {(9 i) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{2 a} \\ & = -\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac {9 i \text {arcsinh}(a x)}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int e^{-3 i \arctan (a x)} x \, dx=\sqrt {1+a^2 x^2} \left (-\frac {3}{a^2}+\frac {i x}{2 a}+\frac {4 i}{a^2 (-i+a x)}\right )-\frac {9 i \text {arcsinh}(a x)}{2 a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {i \left (a x +6 i\right ) \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {i \left (\frac {9 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2}}}-\frac {8 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a^{2} \left (x -\frac {i}{a}\right )}\right )}{2 a}\)	\(106\)
default	\(-\frac {\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-2 i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{4}}+\frac {i \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{3}}\)	\(463\)

[In]

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

[Out]

1/2*I*(a*x+6*I)*(a^2*x^2+1)^(1/2)/a^2-1/2*I/a*(9*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)-8/a^2/(x-
I/a)*((x-I/a)^2*a^2+2*I*a*(x-I/a))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.78 \[ \int e^{-3 i \arctan (a x)} x \, dx=\frac {8 i \, a x - 9 \, {\left (-i \, a x - 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (i \, a^{2} x^{2} - 5 \, a x + 14 i\right )} + 8}{2 \, {\left (a^{3} x - i \, a^{2}\right )}} \]

[In]

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

[Out]

1/2*(8*I*a*x - 9*(-I*a*x - 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(I*a^2*x^2 - 5*a*x + 14*I) + 8
)/(a^3*x - I*a^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.22 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4} x^{2} - 2 i \, a^{3} x - a^{2}} - \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{2 i \, a^{3} x + 2 \, a^{2}} - \frac {6 \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{3} x + a^{2}} - \frac {9 i \, \operatorname {arsinh}\left (a x\right )}{2 \, a^{2}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1}}{2 \, a^{2}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

undef

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14 \[ \int e^{-3 i \arctan (a x)} x \, dx=-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {3\,\sqrt {a^2}}{a^2}-\frac {x\,\sqrt {a^2}\,1{}\mathrm {i}}{2\,a}\right )}{\sqrt {a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,9{}\mathrm {i}}{2\,a\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a\,\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]

[In]

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

[Out]

- ((a^2*x^2 + 1)^(1/2)*((3*(a^2)^(1/2))/a^2 - (x*(a^2)^(1/2)*1i)/(2*a)))/(a^2)^(1/2) - (asinh(x*(a^2)^(1/2))*9
i)/(2*a*(a^2)^(1/2)) - ((a^2*x^2 + 1)^(1/2)*4i)/(a*(((a^2)^(1/2)*1i)/a - x*(a^2)^(1/2))*(a^2)^(1/2))