3.2.99 \(\int e^{-2 i \text {ArcTan}(a+b x)} x^3 \, dx\) [199]

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

[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 = 0.05, 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 = {5203, 78} \begin {gather*} -\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} \end {gather*}

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 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^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 = 0.05, size = 77, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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 [A]
time = 0.13, size = 85, normalized size = 1.10

method	result	size
default	\(\frac {i \left (\frac {1}{4} i b^{3} x^{4}-\frac {2}{3} b^{2} x^{3}-i b \,x^{2}+a b \,x^{2}+4 i a x -2 a^{2} x +2 x \right )}{b^{3}}+\frac {\left (2 i a^{3}+6 a^{2}-6 i a -2\right ) \ln \left (-b x -a +i\right )}{b^{4}}\)	\(85\)
risch	\(-\frac {x^{4}}{4}-\frac {2 i x^{3}}{3 b}+\frac {x^{2}}{b^{2}}+\frac {i a \,x^{2}}{b^{2}}-\frac {4 a x}{b^{3}}-\frac {2 i a^{2} x}{b^{3}}+\frac {2 i x}{b^{3}}+\frac {3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{4}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{4}}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{4}}-\frac {3 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{4}}+\frac {6 i \arctan \left (b x +a \right ) a^{2}}{b^{4}}-\frac {2 \arctan \left (b x +a \right ) a^{3}}{b^{4}}-\frac {2 i \arctan \left (b x +a \right )}{b^{4}}+\frac {6 \arctan \left (b x +a \right ) a}{b^{4}}\)	\(211\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 73, normalized size = 0.95 \begin {gather*} -\frac {i \, {\left (-3 i \, b^{3} x^{4} + 8 \, b^{2} x^{3} - 12 \, {\left (a - i\right )} b x^{2} + 24 \, {\left (a^{2} - 2 i \, a - 1\right )} x\right )}}{12 \, b^{3}} - \frac {2 \, {\left (-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{4}} \end {gather*}

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

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Fricas [A]
time = 2.90, size = 77, normalized size = 1.00 \begin {gather*} -\frac {3 \, b^{4} x^{4} + 8 i \, b^{3} x^{3} + 12 \, {\left (-i \, a - 1\right )} b^{2} x^{2} + 24 \, {\left (i \, a^{2} + 2 \, a - i\right )} b x + 24 \, {\left (-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1\right )} \log \left (\frac {b x + a - i}{b}\right )}{12 \, b^{4}} \end {gather*}

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

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Sympy [A]
time = 0.20, size = 76, normalized size = 0.99 \begin {gather*} - \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 (a + b x - i \right )}}{b^{4}} \end {gather*}

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(a + b*x - I)/b**4

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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. 158 vs. \(2 (59) = 118\).
time = 0.43, size = 158, normalized size = 2.05 \begin {gather*} -\frac {{\left (i \, b x + i \, a + 1\right )}^{4} {\left (-\frac {4 i \, {\left (3 \, a b - 5 i \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} - \frac {18 \, {\left (a^{2} b^{2} - 4 i \, a b^{2} - 3 \, b^{2}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{2} b^{2}} + \frac {12 i \, {\left (a^{3} b^{3} - 9 i \, a^{2} b^{3} - 15 \, a b^{3} + 7 i \, b^{3}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{3} b^{3}} + 3\right )}}{12 \, b^{4}} - \frac {2 \, {\left (i \, a^{3} + 3 \, a^{2} - 3 i \, a - 1\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{4}} \end {gather*}

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]

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

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Mupad [B]
time = 0.54, size = 129, normalized size = 1.68 \begin {gather*} 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} \end {gather*}

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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