3.199 \(\int e^{-2 i \tan ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=77 \[ -\frac {2 (1+i a)^3 \log (-a-b x+i)}{b^4}-\frac {2 i (-a+i)^2 x}{b^3}+\frac {(1+i a) x^2}{b^2}-\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 77, 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} \[ \frac {(1+i a) x^2}{b^2}-\frac {2 i (-a+i)^2 x}{b^3}-\frac {2 (1+i a)^3 \log (-a-b x+i)}{b^4}-\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

((-2*I)*(I - a)^2*x)/b^3 + ((1 + I*a)*x^2)/b^2 - (((2*I)/3)*x^3)/b - x^4/4 - (2*(1 + I*a)^3*Log[I - a - b*x])/
b^4

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^3 \, dx &=\int \frac {x^3 (1-i a-i b x)}{1+i a+i b x} \, dx\\ &=\int \left (-\frac {2 i (-i+a)^2}{b^3}+\frac {2 (1+i a) x}{b^2}-\frac {2 i x^2}{b}-x^3+\frac {2 (-1-i a)^3}{b^3 (-i+a+b x)}\right ) \, dx\\ &=-\frac {2 i (i-a)^2 x}{b^3}+\frac {(1+i a) x^2}{b^2}-\frac {2 i x^3}{3 b}-\frac {x^4}{4}-\frac {2 (1+i a)^3 \log (i-a-b x)}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 1.00 \[ -\frac {2 (1+i a)^3 \log (-a-b x+i)}{b^4}-\frac {2 i (-a+i)^2 x}{b^3}+\frac {(1+i a) x^2}{b^2}-\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

((-2*I)*(I - a)^2*x)/b^3 + ((1 + I*a)*x^2)/b^2 - (((2*I)/3)*x^3)/b - x^4/4 - (2*(1 + I*a)^3*Log[I - a - b*x])/
b^4

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fricas [A]  time = 0.44, size = 77, normalized size = 1.00 \[ -\frac {3 \, b^{4} x^{4} + 8 i \, b^{3} x^{3} + 12 \, {\left (-i \, a - 1\right )} b^{2} x^{2} - {\left (-24 i \, a^{2} - 48 \, a + 24 i\right )} b x - {\left (24 i \, a^{3} + 72 \, a^{2} - 72 i \, a - 24\right )} \log \left (\frac {b x + a - i}{b}\right )}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^4*x^4 + 8*I*b^3*x^3 + 12*(-I*a - 1)*b^2*x^2 - (-24*I*a^2 - 48*a + 24*I)*b*x - (24*I*a^3 + 72*a^2 -
72*I*a - 24)*log((b*x + a - I)/b))/b^4

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giac [B]  time = 0.13, size = 173, normalized size = 2.25 \[ -\frac {2 \, {\left (a^{3} i + 3 \, a^{2} - 3 \, a i - 1\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{4}} + \frac {{\left (b i x + a i + 1\right )}^{4} {\left (\frac {4 \, {\left (3 \, a b - 5 \, b i\right )} i}{{\left (b i x + a i + 1\right )} b} - \frac {18 \, {\left (a^{2} b^{2} - 4 \, a b^{2} i - 3 \, b^{2}\right )} i^{2}}{{\left (b i x + a i + 1\right )}^{2} b^{2}} + \frac {12 \, {\left (a^{3} b^{3} - 9 \, a^{2} b^{3} i - 15 \, a b^{3} + 7 \, b^{3} i\right )} i^{3}}{{\left (b i x + a i + 1\right )}^{3} b^{3}} - 3\right )}}{12 \, b^{4} i^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a^3*i + 3*a^2 - 3*a*i - 1)*log(1/(sqrt((b*x + a)^2 + 1)*abs(b)))/b^4 + 1/12*(b*i*x + a*i + 1)^4*(4*(3*a*b
- 5*b*i)*i/((b*i*x + a*i + 1)*b) - 18*(a^2*b^2 - 4*a*b^2*i - 3*b^2)*i^2/((b*i*x + a*i + 1)^2*b^2) + 12*(a^3*b^
3 - 9*a^2*b^3*i - 15*a*b^3 + 7*b^3*i)*i^3/((b*i*x + a*i + 1)^3*b^3) - 3)/(b^4*i^4)

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maple [B]  time = 0.06, size = 211, normalized size = 2.74 \[ -\frac {x^{4}}{4}-\frac {2 i x^{3}}{3 b}+\frac {i x^{2} a}{b^{2}}-\frac {2 i a^{2} x}{b^{3}}+\frac {x^{2}}{b^{2}}+\frac {2 i x}{b^{3}}-\frac {4 a x}{b^{3}}-\frac {2 \arctan \left (b x +a \right ) a^{3}}{b^{4}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{4}}+\frac {6 \arctan \left (b x +a \right ) a}{b^{4}}+\frac {6 i \arctan \left (b x +a \right ) a^{2}}{b^{4}}-\frac {3 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{4}}+\frac {3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{4}}-\frac {2 i \arctan \left (b x +a \right )}{b^{4}}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x)

[Out]

-1/4*x^4-2/3*I*x^3/b+I/b^2*x^2*a-2*I/b^3*a^2*x+1/b^2*x^2+2*I/b^3*x-4*a*x/b^3-2/b^4*arctan(b*x+a)*a^3+I/b^4*ln(
b^2*x^2+2*a*b*x+a^2+1)*a^3+6/b^4*arctan(b*x+a)*a+6*I/b^4*arctan(b*x+a)*a^2-3*I/b^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a
+3/b^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a^2-2*I/b^4*arctan(b*x+a)-1/b^4*ln(b^2*x^2+2*a*b*x+a^2+1)

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maxima [A]  time = 0.32, size = 74, normalized size = 0.96 \[ -\frac {i \, {\left (-3 i \, b^{3} x^{4} + 8 \, b^{2} x^{3} - {\left (12 \, a - 12 i\right )} b x^{2} + 24 \, {\left (a^{2} - 2 i \, a - 1\right )} x\right )}}{12 \, b^{3}} + \frac {{\left (2 i \, a^{3} + 6 \, a^{2} - 6 i \, a - 2\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*I*(-3*I*b^3*x^4 + 8*b^2*x^3 - (12*a - 12*I)*b*x^2 + 24*(a^2 - 2*I*a - 1)*x)/b^3 + (2*I*a^3 + 6*a^2 - 6*I
*a - 2)*log(I*b*x + I*a + 1)/b^4

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mupad [B]  time = 0.54, size = 129, normalized size = 1.68 \[ x^3\,\left (\frac {a-\mathrm {i}}{3\,b}-\frac {a+1{}\mathrm {i}}{3\,b}\right )-\frac {x^4}{4}-\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (-\frac {6\,a^2-2}{b^4}+\frac {\left (6\,a-2\,a^3\right )\,1{}\mathrm {i}}{b^4}\right )-\frac {x^2\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{2\,b}+\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,{\left (a-\mathrm {i}\right )}^2}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*((a + b*x)^2 + 1))/(a*1i + b*x*1i + 1)^2,x)

[Out]

x^3*((a - 1i)/(3*b) - (a + 1i)/(3*b)) - x^4/4 - log(x + (a - 1i)/b)*(((6*a - 2*a^3)*1i)/b^4 - (6*a^2 - 2)/b^4)
- (x^2*((a - 1i)/b - (a + 1i)/b)*(a - 1i))/(2*b) + (x*((a - 1i)/b - (a + 1i)/b)*(a - 1i)^2)/b^2

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sympy [A]  time = 0.49, size = 80, normalized size = 1.04 \[ - \frac {x^{4}}{4} - x^{2} \left (- \frac {i a}{b^{2}} - \frac {1}{b^{2}}\right ) - x \left (\frac {2 i a^{2}}{b^{3}} + \frac {4 a}{b^{3}} - \frac {2 i}{b^{3}}\right ) - \frac {2 i x^{3}}{3 b} + \frac {2 i \left (a - i\right )^{3} \log {\left (i a + i b x + 1 \right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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