3.199 \(\int e^{-2 i \tan ^{-1}(a+b x)} x^3 \, dx\)
Optimal. Leaf size=77 \[ \frac{(1+i a) x^2}{b^2}-\frac{2 i (-a+i)^2 x}{b^3}-\frac{2 (1+i a)^3 \log (-a-b x+i)}{b^4}-\frac{2 i x^3}{3 b}-\frac{x^4}{4} \]
[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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Rubi [A] time = 0.0592766, antiderivative size = 77, normalized size of antiderivative = 1.,
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 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]
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]))))
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*}
Mathematica [A] time = 0.0583609, size = 77, normalized size = 1. \[ \frac{(1+i a) x^2}{b^2}-\frac{2 i (-a+i)^2 x}{b^3}-\frac{2 (1+i a)^3 \log (-a-b x+i)}{b^4}-\frac{2 i x^3}{3 b}-\frac{x^4}{4} \]
Antiderivative was successfully verified.
[In]
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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Maple [B] time = 0.043, size = 211, normalized size = 2.7 \begin{align*} -{\frac{{x}^{4}}{4}}-{\frac{{\frac{2\,i}{3}}{x}^{3}}{b}}+{\frac{i{x}^{2}a}{{b}^{2}}}-{\frac{2\,i{a}^{2}x}{{b}^{3}}}+{\frac{{x}^{2}}{{b}^{2}}}+{\frac{2\,ix}{{b}^{3}}}-4\,{\frac{ax}{{b}^{3}}}-2\,{\frac{\arctan \left ( bx+a \right ){a}^{3}}{{b}^{4}}}+{\frac{i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ){a}^{3}}{{b}^{4}}}+6\,{\frac{\arctan \left ( bx+a \right ) a}{{b}^{4}}}+{\frac{6\,i\arctan \left ( bx+a \right ){a}^{2}}{{b}^{4}}}-{\frac{3\,i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{{b}^{4}}}+3\,{\frac{\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ){a}^{2}}{{b}^{4}}}-{\frac{2\,i\arctan \left ( bx+a \right ) }{{b}^{4}}}-{\frac{\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{{b}^{4}}} \end{align*}
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/b^3*a*x-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)
________________________________________________________________________________________
Maxima [A] time = 1.05035, size = 100, normalized size = 1.3 \begin{align*} -\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}} \end{align*}
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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Fricas [A] time = 2.12639, size = 203, normalized size = 2.64 \begin{align*} -\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}} \end{align*}
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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Sympy [B] time = 6.41207, size = 1212, normalized size = 15.74 \begin{align*} \text{result too large to display} \end{align*}
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**3*(2*I*a**6 + 12*a**5 - 30*I*a**4 - 40*a**3 + 30*I*a**2 + 12*a - 2*I)/(3*a**6*b - 18*I*a**5*b - 4
5*a**4*b + 60*I*a**3*b + 45*a**2*b - 18*I*a*b - 3*b) + x**2*(I*a**13 + 13*a**12 - 78*I*a**11 - 286*a**10 + 715
*I*a**9 + 1287*a**8 - 1716*I*a**7 - 1716*a**6 + 1287*I*a**5 + 715*a**4 - 286*I*a**3 - 78*a**2 + 13*I*a + 1)/(a
**12*b**2 - 12*I*a**11*b**2 - 66*a**10*b**2 + 220*I*a**9*b**2 + 495*a**8*b**2 - 792*I*a**7*b**2 - 924*a**6*b**
2 + 792*I*a**5*b**2 + 495*a**4*b**2 - 220*I*a**3*b**2 - 66*a**2*b**2 + 12*I*a*b**2 + b**2) - x*(2*I*a**20 + 40
*a**19 - 380*I*a**18 - 2280*a**17 + 9690*I*a**16 + 31008*a**15 - 77520*I*a**14 - 155040*a**13 + 251940*I*a**12
+ 335920*a**11 - 369512*I*a**10 - 335920*a**9 + 251940*I*a**8 + 155040*a**7 - 77520*I*a**6 - 31008*a**5 + 969
0*I*a**4 + 2280*a**3 - 380*I*a**2 - 40*a + 2*I)/(a**18*b**3 - 18*I*a**17*b**3 - 153*a**16*b**3 + 816*I*a**15*b
**3 + 3060*a**14*b**3 - 8568*I*a**13*b**3 - 18564*a**12*b**3 + 31824*I*a**11*b**3 + 43758*a**10*b**3 - 48620*I
*a**9*b**3 - 43758*a**8*b**3 + 31824*I*a**7*b**3 + 18564*a**6*b**3 - 8568*I*a**5*b**3 - 3060*a**4*b**3 + 816*I
*a**3*b**3 + 153*a**2*b**3 - 18*I*a*b**3 - b**3) + 2*(I*a**27 + 27*a**26 - 351*I*a**25 - 2925*a**24 + 17550*I*
a**23 + 80730*a**22 - 296010*I*a**21 - 888030*a**20 + 2220075*I*a**19 + 4686825*a**18 - 8436285*I*a**17 - 1303
7895*a**16 + 17383860*I*a**15 + 20058300*a**14 - 20058300*I*a**13 - 17383860*a**12 + 13037895*I*a**11 + 843628
5*a**10 - 4686825*I*a**9 - 2220075*a**8 + 888030*I*a**7 + 296010*a**6 - 80730*I*a**5 - 17550*a**4 + 2925*I*a**
3 + 351*a**2 - 27*I*a - 1)*log(a**25 - 25*I*a**24 - 300*a**23 + 2300*I*a**22 + 12650*a**21 - 53130*I*a**20 - 1
77100*a**19 + 480700*I*a**18 + 1081575*a**17 - 2042975*I*a**16 - 3268760*a**15 + 4457400*I*a**14 + 5200300*a**
13 - 5200300*I*a**12 - 4457400*a**11 + 3268760*I*a**10 + 2042975*a**9 - 1081575*I*a**8 - 480700*a**7 + 177100*
I*a**6 + 53130*a**5 - 12650*I*a**4 - 2300*a**3 + 300*I*a**2 + 25*a + x*(a**24*b - 24*I*a**23*b - 276*a**22*b +
2024*I*a**21*b + 10626*a**20*b - 42504*I*a**19*b - 134596*a**18*b + 346104*I*a**17*b + 735471*a**16*b - 13075
04*I*a**15*b - 1961256*a**14*b + 2496144*I*a**13*b + 2704156*a**12*b - 2496144*I*a**11*b - 1961256*a**10*b + 1
307504*I*a**9*b + 735471*a**8*b - 346104*I*a**7*b - 134596*a**6*b + 42504*I*a**5*b + 10626*a**4*b - 2024*I*a**
3*b - 276*a**2*b + 24*I*a*b + b) - I)/(b**4*(a**24 - 24*I*a**23 - 276*a**22 + 2024*I*a**21 + 10626*a**20 - 425
04*I*a**19 - 134596*a**18 + 346104*I*a**17 + 735471*a**16 - 1307504*I*a**15 - 1961256*a**14 + 2496144*I*a**13
+ 2704156*a**12 - 2496144*I*a**11 - 1961256*a**10 + 1307504*I*a**9 + 735471*a**8 - 346104*I*a**7 - 134596*a**6
+ 42504*I*a**5 + 10626*a**4 - 2024*I*a**3 - 276*a**2 + 24*I*a + 1))
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Giac [B] time = 1.09381, size = 234, normalized size = 3.04 \begin{align*} -\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}} \end{align*}
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)