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

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[E^(I*ArcTan[a*x])*x^3,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[E^(I*ArcTan[a*x])*x^3,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 = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {i \left (6 a^{3} x^{3}-8 i a^{2} x^{2}-9 a x +16 i\right ) \sqrt {a^{2} x^{2}+1}}{24 a^{4}}+\frac {3 i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 a^{3} \sqrt {a^{2}}}\)	\(77\)
meijerg	\(\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right ) \sqrt {a^{2} x^{2}+1}}{6}}{2 a^{4} \sqrt {\pi }}+\frac {i \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right ) \sqrt {a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \operatorname {arcsinh}\left (a x \right )}{4 a^{5}}\right )}{2 a^{3} \sqrt {\pi }\, \sqrt {a^{2}}}\)	\(109\)
default	\(\frac {x^{2} \sqrt {a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {a^{2} x^{2}+1}}{3 a^{4}}+i a \left (\frac {x^{3} \sqrt {a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 \left (\frac {x \sqrt {a^{2} x^{2}+1}}{2 a^{2}}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}\right )\)	\(117\)

[In]

int((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^3,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 = 0.26 (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((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^3,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 [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.16 \[ \int e^{i \arctan (a x)} x^3 \, dx=\begin {cases} \sqrt {a^{2} x^{2} + 1} \left (\frac {i x^{3}}{4 a} + \frac {x^{2}}{3 a^{2}} - \frac {3 i x}{8 a^{3}} - \frac {2}{3 a^{4}}\right ) + \frac {3 i \log {\left (2 a^{2} x + 2 \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2}} \right )}}{8 a^{3} \sqrt {a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {i a x^{5}}{5} + \frac {x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2*x**2 + 1)*(I*x**3/(4*a) + x**2/(3*a**2) - 3*I*x/(8*a**3) - 2/(3*a**4)) + 3*I*log(2*a**2*x
+ 2*sqrt(a**2*x**2 + 1)*sqrt(a**2))/(8*a**3*sqrt(a**2)), Ne(a**2, 0)), (I*a*x**5/5 + x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int e^{i \arctan (a x)} x^3 \, dx=\frac {i \, \sqrt {a^{2} x^{2} + 1} x^{3}}{4 \, a} + \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac {3 i \, \sqrt {a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {3 i \, \operatorname {arsinh}\left (a x\right )}{8 \, a^{4}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{3 \, a^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int e^{i \arctan (a x)} x^3 \, dx=-\frac {1}{24} \, \sqrt {a^{2} x^{2} + 1} {\left ({\left (2 \, x {\left (-\frac {3 i \, x}{a} - \frac {4}{a^{2}}\right )} + \frac {9 i}{a^{3}}\right )} x + \frac {16}{a^{4}}\right )} - \frac {3 i \, \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{8 \, a^{3} {\left | a \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.50 (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*x*1i + 1))/(a^2*x^2 + 1)^(1/2),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)