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

Optimal result

Integrand size = 14, antiderivative size = 75 \[ \int e^{i \arctan (a x)} x^2 \, dx=-\frac {i \sqrt {1+a^2 x^2}}{a^3}+\frac {x \sqrt {1+a^2 x^2}}{2 a^2}+\frac {i \left (1+a^2 x^2\right )^{3/2}}{3 a^3}-\frac {\text {arcsinh}(a x)}{2 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5168, 811, 655, 201, 221} \[ \int e^{i \arctan (a x)} x^2 \, dx=-\frac {\text {arcsinh}(a x)}{2 a^3}+\frac {x \sqrt {a^2 x^2+1}}{2 a^2}+\frac {i \left (a^2 x^2+1\right )^{3/2}}{3 a^3}-\frac {i \sqrt {a^2 x^2+1}}{a^3} \]

[In]

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

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 655

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

Rule 811

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 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^2 (1+i a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {\int \frac {1+i a x}{\sqrt {1+a^2 x^2}} \, dx}{a^2}+\frac {\int (1+i a x) \sqrt {1+a^2 x^2} \, dx}{a^2} \\ & = -\frac {i \sqrt {1+a^2 x^2}}{a^3}+\frac {i \left (1+a^2 x^2\right )^{3/2}}{3 a^3}-\frac {\int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{a^2}+\frac {\int \sqrt {1+a^2 x^2} \, dx}{a^2} \\ & = -\frac {i \sqrt {1+a^2 x^2}}{a^3}+\frac {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 {\text {arcsinh}(a x)}{a^3}+\frac {\int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{2 a^2} \\ & = -\frac {i \sqrt {1+a^2 x^2}}{a^3}+\frac {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 {\text {arcsinh}(a x)}{2 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61 \[ \int e^{i \arctan (a x)} x^2 \, dx=\frac {\left (-4 i+3 a x+2 i a^2 x^2\right ) \sqrt {1+a^2 x^2}-3 \text {arcsinh}(a x)}{6 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {i \left (2 a^{2} x^{2}-3 i a x -4\right ) \sqrt {a^{2} x^{2}+1}}{6 a^{3}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\)	\(67\)
default	\(\frac {x \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}+i a \left (\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}}\right )\)	\(92\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {3}{2}} \sqrt {a^{2} x^{2}+1}}{a^{2}}-\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {3}{2}} \operatorname {arcsinh}\left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {\pi }\, \sqrt {a^{2}}}+\frac {i \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right ) \sqrt {a^{2} x^{2}+1}}{6}\right )}{2 a^{3} \sqrt {\pi }}\)	\(98\)

[In]

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

[Out]

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

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68 \[ \int e^{i \arctan (a x)} x^2 \, dx=\frac {\sqrt {a^{2} x^{2} + 1} {\left (2 i \, a^{2} x^{2} + 3 \, a x - 4 i\right )} + 3 \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{6 \, a^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int e^{i \arctan (a x)} x^2 \, dx=\begin {cases} \sqrt {a^{2} x^{2} + 1} \left (\frac {i x^{2}}{3 a} + \frac {x}{2 a^{2}} - \frac {2 i}{3 a^{3}}\right ) - \frac {\log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{2 a^{2} \sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{4}}{4} + \frac {x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2*x**2 + 1)*(I*x**2/(3*a) + x/(2*a**2) - 2*I/(3*a**3)) - log(2*a**2*x + 2*sqrt(a**2*x**2 +
1)*sqrt(a**2))/(2*a**2*sqrt(a**2)), Ne(a**2, 0)), (I*a*x**4/4 + x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int e^{i \arctan (a x)} x^2 \, dx=\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{2}}{3 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x}{2 \, a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{2 \, a^{3}} - \frac {2 i \, \sqrt {a^{2} x^{2} + 1}}{3 \, a^{3}} \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int e^{i \arctan (a x)} x^2 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^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 = 0.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int e^{i \arctan (a x)} x^2 \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {x\,\sqrt {a^2}}{2\,a^2}-\frac {a\,2{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}+\frac {a^3\,x^2\,1{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}\right )}{\sqrt {a^2}}-\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{2\,a^2\,\sqrt {a^2}} \]

[In]

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

[Out]

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