\(\int e^{4 i \arctan (a x)} x^2 \, dx\) [28]

Optimal result

Integrand size = 14, antiderivative size = 53 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=-\frac {8 x}{a^2}-\frac {2 i x^2}{a}+\frac {x^3}{3}-\frac {4}{a^3 (i+a x)}+\frac {12 i \log (i+a x)}{a^3} \]

[Out]

-8*x/a^2-2*I*x^2/a+1/3*x^3-4/a^3/(I+a*x)+12*I*ln(I+a*x)/a^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 90} \[ \int e^{4 i \arctan (a x)} x^2 \, dx=-\frac {4}{a^3 (a x+i)}+\frac {12 i \log (a x+i)}{a^3}-\frac {8 x}{a^2}-\frac {2 i x^2}{a}+\frac {x^3}{3} \]

[In]

Int[E^((4*I)*ArcTan[a*x])*x^2,x]

[Out]

(-8*x)/a^2 - ((2*I)*x^2)/a + x^3/3 - 4/(a^3*(I + a*x)) + ((12*I)*Log[I + a*x])/a^3

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 5170

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (1+i a x)^2}{(1-i a x)^2} \, dx \\ & = \int \left (-\frac {8}{a^2}-\frac {4 i x}{a}+x^2+\frac {4}{a^2 (i+a x)^2}+\frac {12 i}{a^2 (i+a x)}\right ) \, dx \\ & = -\frac {8 x}{a^2}-\frac {2 i x^2}{a}+\frac {x^3}{3}-\frac {4}{a^3 (i+a x)}+\frac {12 i \log (i+a x)}{a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=-\frac {8 x}{a^2}-\frac {2 i x^2}{a}+\frac {x^3}{3}-\frac {4}{a^3 (i+a x)}+\frac {12 i \log (i+a x)}{a^3} \]

[In]

Integrate[E^((4*I)*ArcTan[a*x])*x^2,x]

[Out]

(-8*x)/a^2 - ((2*I)*x^2)/a + x^3/3 - 4/(a^3*(I + a*x)) + ((12*I)*Log[I + a*x])/a^3

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09

method	result	size
default	\(-\frac {8 x -\frac {1}{3} a^{2} x^{3}+2 i a \,x^{2}}{a^{2}}+\frac {-\frac {4}{a \left (a x +i\right )}+\frac {12 i \ln \left (a x +i\right )}{a}}{a^{2}}\)	\(58\)
risch	\(-\frac {8 x}{a^{2}}+\frac {x^{3}}{3}-\frac {2 i x^{2}}{a}-\frac {4}{a^{3} \left (a x +i\right )}+\frac {6 i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}+\frac {12 \arctan \left (a x \right )}{a^{3}}\)	\(60\)
parallelrisch	\(\frac {a^{5} x^{5}-6 i a^{4} x^{4}+36 i \ln \left (a x +i\right ) x^{2} a^{2}-23 a^{3} x^{3}-18 i a^{2} x^{2}+36 i \ln \left (a x +i\right )-36 a x}{3 a^{3} \left (a^{2} x^{2}+1\right )}\)	\(81\)
meijerg	\(\frac {-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {a^{2}}}+\frac {2 i \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{a^{3}}-\frac {3 \left (\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {2 i \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{3 a^{2} x^{2}+3}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{a^{3}}+\frac {-\frac {x \left (a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}+70 a^{2} x^{2}+105\right )}{21 a^{6} \left (a^{2} x^{2}+1\right )}+\frac {5 \left (a^{2}\right )^{\frac {7}{2}} \arctan \left (a x \right )}{a^{7}}}{2 a^{2} \sqrt {a^{2}}}\)	\(254\)

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=\frac {a^{4} x^{4} - 5 i \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 24 i \, a x - 36 \, {\left (-i \, a x + 1\right )} \log \left (\frac {a x + i}{a}\right ) - 12}{3 \, {\left (a^{4} x + i \, a^{3}\right )}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="fricas")

[Out]

1/3*(a^4*x^4 - 5*I*a^3*x^3 - 18*a^2*x^2 - 24*I*a*x - 36*(-I*a*x + 1)*log((a*x + I)/a) - 12)/(a^4*x + I*a^3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=\frac {x^{3}}{3} - \frac {4}{a^{4} x + i a^{3}} - \frac {2 i x^{2}}{a} - \frac {8 x}{a^{2}} + \frac {12 i \log {\left (a x + i \right )}}{a^{3}} \]

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2*x**2,x)

[Out]

x**3/3 - 4/(a**4*x + I*a**3) - 2*I*x**2/a - 8*x/a**2 + 12*I*log(a*x + I)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=-\frac {4 \, {\left (a x - i\right )}}{a^{5} x^{2} + a^{3}} + \frac {a^{2} x^{3} - 6 i \, a x^{2} - 24 \, x}{3 \, a^{2}} + \frac {12 \, \arctan \left (a x\right )}{a^{3}} + \frac {6 i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="maxima")

[Out]

-4*(a*x - I)/(a^5*x^2 + a^3) + 1/3*(a^2*x^3 - 6*I*a*x^2 - 24*x)/a^2 + 12*arctan(a*x)/a^3 + 6*I*log(a^2*x^2 + 1
)/a^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=\frac {12 i \, \log \left (a x + i\right )}{a^{3}} - \frac {4}{{\left (a x + i\right )} a^{3}} + \frac {a^{6} x^{3} - 6 i \, a^{5} x^{2} - 24 \, a^{4} x}{3 \, a^{6}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2*x^2,x, algorithm="giac")

[Out]

12*I*log(a*x + I)/a^3 - 4/((a*x + I)*a^3) + 1/3*(a^6*x^3 - 6*I*a^5*x^2 - 24*a^4*x)/a^6

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int e^{4 i \arctan (a x)} x^2 \, dx=\frac {x^3}{3}+\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,12{}\mathrm {i}}{a^3}-\frac {8\,x}{a^2}-\frac {4}{a^4\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {x^2\,2{}\mathrm {i}}{a} \]

[In]

int((x^2*(a*x*1i + 1)^4)/(a^2*x^2 + 1)^2,x)

[Out]

(log(x + 1i/a)*12i)/a^3 - 4/(a^4*(x + 1i/a)) - (8*x)/a^2 + x^3/3 - (x^2*2i)/a