\(\int e^{-i \arctan (a x)} x^3 \, dx\) [35]

Optimal result

Integrand size = 14, antiderivative size = 90 \[ \int e^{-i \arctan (a x)} x^3 \, dx=\frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16-9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}-\frac {3 i \text {arcsinh}(a x)}{8 a^4} \]

[Out]

-3/8*I*arcsinh(a*x)/a^4+1/3*x^2*(a^2*x^2+1)^(1/2)/a^2-1/4*I*x^3*(a^2*x^2+1)^(1/2)/a-1/24*(16-9*I*a*x)*(a^2*x^2
+1)^(1/2)/a^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5168, 847, 794, 221} \[ \int e^{-i \arctan (a x)} x^3 \, dx=-\frac {3 i \text {arcsinh}(a x)}{8 a^4}+\frac {x^2 \sqrt {a^2 x^2+1}}{3 a^2}-\frac {i x^3 \sqrt {a^2 x^2+1}}{4 a}-\frac {(16-9 i a x) \sqrt {a^2 x^2+1}}{24 a^4} \]

[In]

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

[Out]

(x^2*Sqrt[1 + a^2*x^2])/(3*a^2) - ((I/4)*x^3*Sqrt[1 + a^2*x^2])/a - ((16 - (9*I)*a*x)*Sqrt[1 + a^2*x^2])/(24*a
^4) - (((3*I)/8)*ArcSinh[a*x])/a^4

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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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^3 (1-i a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}+\frac {\int \frac {x^2 \left (3 i a+4 a^2 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2} \\ & = \frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}+\frac {\int \frac {x \left (-8 a^2+9 i a^3 x\right )}{\sqrt {1+a^2 x^2}} \, dx}{12 a^4} \\ & = \frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16-9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}-\frac {(3 i) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{8 a^3} \\ & = \frac {x^2 \sqrt {1+a^2 x^2}}{3 a^2}-\frac {i x^3 \sqrt {1+a^2 x^2}}{4 a}-\frac {(16-9 i a x) \sqrt {1+a^2 x^2}}{24 a^4}-\frac {3 i \text {arcsinh}(a x)}{8 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int e^{-i \arctan (a x)} x^3 \, dx=\frac {\sqrt {1+a^2 x^2} \left (-16+9 i a x+8 a^2 x^2-6 i a^3 x^3\right )-9 i \text {arcsinh}(a x)}{24 a^4} \]

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*(-16 + (9*I)*a*x + 8*a^2*x^2 - (6*I)*a^3*x^3) - (9*I)*ArcSinh[a*x])/(24*a^4)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {i \left (6 a^{3} x^{3}+8 i a^{2} x^{2}-9 a x -16 i\right ) \sqrt {a^{2} x^{2}+1}}{24 a^{4}}-\frac {3 i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{3} \sqrt {a^{2}}}\)	\(77\)
default	\(\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a^{4}}+\frac {i \left (\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}\right )}{a^{3}}-\frac {i \left (\frac {x \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 a^{2}}-\frac {\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}}{4 a^{2}}\right )}{a}-\frac {\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}}{a^{4}}\)	\(241\)

[In]

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

[Out]

-1/24*I*(6*a^3*x^3+8*I*a^2*x^2-9*a*x-16*I)*(a^2*x^2+1)^(1/2)/a^4-3/8*I/a^3*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1
/2))/(a^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int e^{-i \arctan (a x)} x^3 \, dx=\frac {{\left (-6 i \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 9 i \, a x - 16\right )} \sqrt {a^{2} x^{2} + 1} + 9 i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{24 \, a^{4}} \]

[In]

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

[Out]

1/24*((-6*I*a^3*x^3 + 8*a^2*x^2 + 9*I*a*x - 16)*sqrt(a^2*x^2 + 1) + 9*I*log(-a*x + sqrt(a^2*x^2 + 1)))/a^4

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int e^{-i \arctan (a x)} x^3 \, dx=-\frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, a^{3}} + \frac {5 i \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{4}} - \frac {3 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} - \frac {\sqrt {a^{2} x^{2} + 1}}{a^{4}} \]

[In]

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

[Out]

-1/4*I*(a^2*x^2 + 1)^(3/2)*x/a^3 + 5/8*I*sqrt(a^2*x^2 + 1)*x/a^3 + 1/3*(a^2*x^2 + 1)^(3/2)/a^4 - 3/8*I*arcsinh
(a*x)/a^4 - sqrt(a^2*x^2 + 1)/a^4

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int e^{-i \arctan (a x)} x^3 \, dx=-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,3{}\mathrm {i}}{8\,a^3\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {2}{3\,{\left (a^2\right )}^{3/2}}-\frac {a^2\,x^2}{3\,{\left (a^2\right )}^{3/2}}+\frac {x^3\,{\left (a^2\right )}^{3/2}\,1{}\mathrm {i}}{4\,a^3}-\frac {x\,\sqrt {a^2}\,3{}\mathrm {i}}{8\,a^3}\right )}{\sqrt {a^2}} \]

[In]

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

[Out]

- (asinh(x*(a^2)^(1/2))*3i)/(8*a^3*(a^2)^(1/2)) - ((a^2*x^2 + 1)^(1/2)*(2/(3*(a^2)^(3/2)) - (a^2*x^2)/(3*(a^2)
^(3/2)) + (x^3*(a^2)^(3/2)*1i)/(4*a^3) - (x*(a^2)^(1/2)*3i)/(8*a^3)))/(a^2)^(1/2)