3.2.100 \(\int e^{-2 i \text {ArcTan}(a+b x)} x^2 \, dx\) [200]

Optimal. Leaf size=59 \[ \frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {2 i (i-a)^2 \log (i-a-b x)}{b^3} \]

[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

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Rubi [A]
time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5203, 78} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/E^((2*I)*ArcTan[a + b*x]),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*} \int e^{-2 i \tan ^{-1}(a+b x)} x^2 \, dx &=\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*}

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Mathematica [A]
time = 0.03, size = 55, normalized size = 0.93 \begin {gather*} \frac {b x \left (6+6 i a-3 i b x-b^2 x^2\right )-6 i (-i+a)^2 \log (i-a-b x)}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*(6 + (6*I)*a - (3*I)*b*x - b^2*x^2) - (6*I)*(-I + a)^2*Log[I - a - b*x])/(3*b^3)

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Maple [A]
time = 0.09, size = 59, normalized size = 1.00

method	result	size
default	\(\frac {i \left (\frac {1}{3} i b^{2} x^{3}-x^{2} b -2 i x +2 a x \right )}{b^{2}}+\frac {\left (-2 i a^{2}-4 a +2 i\right ) \ln \left (-b x -a +i\right )}{b^{3}}\)	\(59\)
risch	\(-\frac {x^{3}}{3}-\frac {i x^{2}}{b}+\frac {2 x}{b^{2}}+\frac {2 i a x}{b^{2}}-\frac {2 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{3}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{3}}-\frac {4 i \arctan \left (b x +a \right ) a}{b^{3}}+\frac {2 \arctan \left (b x +a \right ) a^{2}}{b^{3}}-\frac {2 \arctan \left (b x +a \right )}{b^{3}}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 53, normalized size = 0.90 \begin {gather*} -\frac {b^{2} x^{3} + 3 i \, b x^{2} + 6 \, {\left (-i \, a - 1\right )} x}{3 \, b^{2}} - \frac {2 \, {\left (i \, a^{2} + 2 \, a - i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.32, size = 53, normalized size = 0.90 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]
time = 0.16, size = 49, normalized size = 0.83 \begin {gather*} - \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (45) = 90\).
time = 0.44, size = 109, normalized size = 1.85 \begin {gather*} -\frac {i \, {\left (i \, b x + i \, a + 1\right )}^{3} {\left (-\frac {3 i \, {\left (a b - 2 i \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} - \frac {3 \, {\left (a^{2} b^{2} - 6 i \, a b^{2} - 5 \, b^{2}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{2} b^{2}} + 1\right )}}{3 \, b^{3}} - \frac {2 \, {\left (-i \, a^{2} - 2 \, a + i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*(I*b*x + I*a + 1)^3*(-3*I*(a*b - 2*I*b)/((I*b*x + I*a + 1)*b) - 3*(a^2*b^2 - 6*I*a*b^2 - 5*b^2)/((I*b*x
+ I*a + 1)^2*b^2) + 1)/b^3 - 2*(-I*a^2 - 2*a + I)*log(1/(sqrt((b*x + a)^2 + 1)*abs(b)))/b^3

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Mupad [B]
time = 0.54, size = 90, normalized size = 1.53 \begin {gather*} -\ln \left (x+\frac {a-\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 {a-\mathrm {i}}{2\,b}-\frac {a+1{}\mathrm {i}}{2\,b}\right )-\frac {x^3}{3}-\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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