3.173 \(\int e^{2 i \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=54 \[ \frac {2 i (a+i)^2 \log (a+b x+i)}{b^3}+\frac {2 (1-i a) x}{b^2}+\frac {i x^2}{b}-\frac {x^3}{3} \]

[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 = 0.05, antiderivative size = 54, 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 = {5095, 77} \[ \frac {2 (1-i a) x}{b^2}+\frac {2 i (a+i)^2 \log (a+b x+i)}{b^3}+\frac {i x^2}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[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 77

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 5095

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 = 0.04, size = 54, normalized size = 1.00 \[ \frac {2 i (a+i)^2 \log (a+b x+i)}{b^3}+\frac {2 (1-i a) x}{b^2}+\frac {i x^2}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.41, size = 53, normalized size = 0.98 \[ -\frac {b^{3} x^{3} - 3 i \, b^{2} x^{2} + 6 \, {\left (i \, a - 1\right )} b x - {\left (6 i \, a^{2} - 12 \, a - 6 i\right )} \log \left (\frac {b x + a + i}{b}\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 - 12*a - 6*I)*log((b*x + a + I)/b))/b^3

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giac [A]  time = 0.14, size = 57, normalized size = 1.06 \[ \frac {2 \, {\left (a^{2} i - 2 \, a - i\right )} \log \left (b x + a + i\right )}{b^{3}} - \frac {b^{3} x^{3} - 3 \, b^{2} i x^{2} + 6 \, a b i x - 6 \, b x}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 176, normalized size = 3.26 \[ -\frac {x^{3}}{3}+\frac {i x^{2}}{b}-\frac {2 i a x}{b^{2}}+\frac {2 x}{b^{2}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{3}}-\frac {2 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{3}}+\frac {4 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{b^{3}}+\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{b^{3}}-\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 87, normalized size = 1.61 \[ -\frac {b^{2} x^{3} - 3 i \, b x^{2} + 6 \, {\left (i \, a - 1\right )} x}{3 \, b^{2}} + \frac {2 \, {\left (a^{2} + 2 i \, a - 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{3}} + \frac {{\left (i \, a^{2} - 2 \, a - i\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.51, size = 107, normalized size = 1.98 \[ -\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (\frac {4\,a}{b^3}-\frac {\left (2\,a^2-2\right )\,1{}\mathrm {i}}{b^3}\right )-x^2\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,b}\right )-\frac {x^3}{3}-\frac {x\,\left (-1+a\,1{}\mathrm {i}\right )\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.31, size = 49, normalized size = 0.91 \[ - \frac {x^{3}}{3} - x \left (\frac {2 i a}{b^{2}} - \frac {2}{b^{2}}\right ) + \frac {i x^{2}}{b} + \frac {2 i \left (a + i\right )^{2} \log {\left (i a + i b x - 1 \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(I*a + I*b*x - 1)/b**3

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