\(\int e^{2 i \arctan (a+b x)} x^2 \, dx\) [173]

Optimal result

Integrand size = 16, antiderivative size = 54 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=\frac {2 (1-i a) x}{b^2}+\frac {i x^2}{b}-\frac {x^3}{3}+\frac {2 i (i+a)^2 \log (i+a+b x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5203, 78} \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=\frac {2 i (a+i)^2 \log (a+b x+i)}{b^3}+\frac {2 (1-i a) x}{b^2}+\frac {i x^2}{b}-\frac {x^3}{3} \]

[In]

Int[E^((2*I)*ArcTan[a + b*x])*x^2,x]

[Out]

(2*(1 - I*a)*x)/b^2 + (I*x^2)/b - x^3/3 + ((2*I)*(I + a)^2*Log[I + a + b*x])/b^3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5203

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((1 -
I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c + I*b*c*x)^(I*(n/2))), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=\frac {2 (1-i a) x}{b^2}+\frac {i x^2}{b}-\frac {x^3}{3}+\frac {2 i (i+a)^2 \log (i+a+b x)}{b^3} \]

[In]

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

[Out]

(2*(1 - I*a)*x)/b^2 + (I*x^2)/b - x^3/3 + ((2*I)*(I + a)^2*Log[I + a + b*x])/b^3

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(\frac {-b^{3} x^{3}+3 i x^{2} b^{2}-12 \ln \left (b x +a +i\right ) a +6 i \ln \left (b x +a +i\right ) a^{2}-6 i a b x -6 i \ln \left (b x +a +i\right )+6 b x}{3 b^{3}}\)	\(70\)
default	\(\frac {i \left (\frac {1}{3} i b^{2} x^{3}+x^{2} b -2 i x -2 a x \right )}{b^{2}}+\frac {\frac {\left (2 i a^{2} b -4 a b -2 i b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (2 i a^{3}+2 i a -2 a^{2}-2-\frac {\left (2 i a^{2} b -4 a b -2 i b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{2}}\)	\(135\)
risch	\(-\frac {x^{3}}{3}+\frac {i x^{2}}{b}+\frac {2 x}{b^{2}}-\frac {2 i a x}{b^{2}}-\frac {2 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{3}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{3}}+\frac {4 i \arctan \left (b x +a \right ) a}{b^{3}}+\frac {2 \arctan \left (b x +a \right ) a^{2}}{b^{3}}-\frac {2 \arctan \left (b x +a \right )}{b^{3}}\)	\(143\)

[In]

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

[Out]

1/3*(-b^3*x^3+3*I*x^2*b^2-12*ln(I+a+b*x)*a+6*I*ln(I+a+b*x)*a^2-6*I*x*a*b-6*I*ln(I+a+b*x)+6*b*x)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=-\frac {b^{3} x^{3} - 3 i \, b^{2} x^{2} + 6 \, {\left (i \, a - 1\right )} b x + 6 \, {\left (-i \, a^{2} + 2 \, a + i\right )} \log \left (\frac {b x + a + i}{b}\right )}{3 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=- \frac {x^{3}}{3} - x \left (\frac {2 i a}{b^{2}} - \frac {2}{b^{2}}\right ) + \frac {i x^{2}}{b} + \frac {2 i \left (a + i\right )^{2} \log {\left (a + b x + i \right )}}{b^{3}} \]

[In]

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

[Out]

-x**3/3 - x*(2*I*a/b**2 - 2/b**2) + I*x**2/b + 2*I*(a + I)**2*log(a + b*x + I)/b**3

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (42) = 84\).

Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=-\frac {b^{2} x^{3} - 3 i \, b x^{2} + 6 \, {\left (i \, a - 1\right )} x}{3 \, b^{2}} + \frac {2 \, {\left (a^{2} + 2 i \, a - 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{3}} + \frac {{\left (i \, a^{2} - 2 \, a - i\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}} \]

[In]

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

[Out]

-1/3*(b^2*x^3 - 3*I*b*x^2 + 6*(I*a - 1)*x)/b^2 + 2*(a^2 + 2*I*a - 1)*arctan((b^2*x + a*b)/b)/b^3 + (I*a^2 - 2*
a - I)*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=-\frac {2 \, {\left (-i \, a^{2} + 2 \, a + i\right )} \log \left (b x + a + i\right )}{b^{3}} - \frac {b^{3} x^{3} - 3 i \, b^{2} x^{2} + 6 i \, a b x - 6 \, b x}{3 \, b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.98 \[ \int e^{2 i \arctan (a+b x)} x^2 \, dx=-\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (\frac {4\,a}{b^3}-\frac {\left (2\,a^2-2\right )\,1{}\mathrm {i}}{b^3}\right )-x^2\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}\right )-\frac {x^3}{3}-\frac {x\,\left (-1+a\,1{}\mathrm {i}\right )\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{b} \]

[In]

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

[Out]

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