3.2.98 \(\int e^{-2 i \text {ArcTan}(a+b x)} x^4 \, dx\) [198]

Optimal. Leaf size=99 \[ -\frac {2 (1+i a)^3 x}{b^4}-\frac {i (i-a)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}-\frac {2 i (i-a)^4 \log (i-a-b x)}{b^5} \]

[Out]

-2*(1+I*a)^3*x/b^4-I*(I-a)^2*x^2/b^3+2/3*(1+I*a)*x^3/b^2-1/2*I*x^4/b-1/5*x^5-2*I*(I-a)^4*ln(I-a-b*x)/b^5

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Rubi [A]
time = 0.07, antiderivative size = 99, 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)^4 \log (-a-b x+i)}{b^5}-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (-a+i)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

(-2*(1 + I*a)^3*x)/b^4 - (I*(I - a)^2*x^2)/b^3 + (2*(1 + I*a)*x^3)/(3*b^2) - ((I/2)*x^4)/b - x^5/5 - ((2*I)*(I
- a)^4*Log[I - a - b*x])/b^5

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^4 \, dx &=\int \frac {x^4 (1-i a-i b x)}{1+i a+i b x} \, dx\\ &=\int \left (\frac {2 (-1-i a)^3}{b^4}-\frac {2 i (-i+a)^2 x}{b^3}+\frac {2 (1+i a) x^2}{b^2}-\frac {2 i x^3}{b}-x^4-\frac {2 i (-i+a)^4}{b^4 (-i+a+b x)}\right ) \, dx\\ &=-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (i-a)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}-\frac {2 i (i-a)^4 \log (i-a-b x)}{b^5}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 95, normalized size = 0.96 \begin {gather*} -\frac {2 (1+i a)^3 x}{b^4}-\frac {i (-i+a)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}-\frac {2 i (-i+a)^4 \log (i-a-b x)}{b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

(-2*(1 + I*a)^3*x)/b^4 - (I*(-I + a)^2*x^2)/b^3 + (2*(1 + I*a)*x^3)/(3*b^2) - ((I/2)*x^4)/b - x^5/5 - ((2*I)*(
-I + a)^4*Log[I - a - b*x])/b^5

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Maple [A]
time = 0.14, size = 125, normalized size = 1.26

method	result	size
default	\(-\frac {i \left (-\frac {1}{5} i b^{4} x^{5}+\frac {1}{2} b^{3} x^{4}+\frac {2}{3} i b^{2} x^{3}-\frac {2}{3} a \,b^{2} x^{3}-2 i a b \,x^{2}+a^{2} b \,x^{2}+6 i a^{2} x -2 a^{3} x -x^{2} b -2 i x +6 a x \right )}{b^{4}}+\frac {\left (-2 i a^{4}-8 a^{3}+12 i a^{2}+8 a -2 i\right ) \ln \left (-b x -a +i\right )}{b^{5}}\)	\(125\)
risch	\(-\frac {x^{5}}{5}+\frac {2 i a \,x^{3}}{3 b^{2}}+\frac {2 x^{3}}{3 b^{2}}-\frac {6 i a x}{b^{4}}-\frac {2 a \,x^{2}}{b^{3}}-\frac {i a^{2} x^{2}}{b^{3}}+\frac {6 a^{2} x}{b^{4}}+\frac {2 i a^{3} x}{b^{4}}+\frac {i x^{2}}{b^{3}}-\frac {2 x}{b^{4}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{5}}-\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{5}}+\frac {6 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{5}}+\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{5}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{b^{5}}+\frac {8 i \arctan \left (b x +a \right ) a}{b^{5}}-\frac {8 i \arctan \left (b x +a \right ) a^{3}}{b^{5}}+\frac {2 \arctan \left (b x +a \right ) a^{4}}{b^{5}}-\frac {i x^{4}}{2 b}-\frac {12 \arctan \left (b x +a \right ) a^{2}}{b^{5}}+\frac {2 \arctan \left (b x +a \right )}{b^{5}}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x,method=_RETURNVERBOSE)

[Out]

-I/b^4*(-1/5*I*b^4*x^5+1/2*b^3*x^4+2/3*I*b^2*x^3-2/3*a*b^2*x^3-2*I*a*b*x^2+a^2*b*x^2+6*I*a^2*x-2*a^3*x-x^2*b-2
*I*x+6*a*x)+(-2*I*a^4+12*I*a^2-8*a^3-2*I+8*a)/b^5*ln(I-a-b*x)

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Maxima [A]
time = 0.26, size = 105, normalized size = 1.06 \begin {gather*} -\frac {6 \, b^{4} x^{5} + 15 i \, b^{3} x^{4} - 20 \, {\left (i \, a + 1\right )} b^{2} x^{3} - 30 \, {\left (-i \, a^{2} - 2 \, a + i\right )} b x^{2} - 60 \, {\left (i \, a^{3} + 3 \, a^{2} - 3 i \, a - 1\right )} x}{30 \, b^{4}} - \frac {2 \, {\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="maxima")

[Out]

-1/30*(6*b^4*x^5 + 15*I*b^3*x^4 - 20*(I*a + 1)*b^2*x^3 - 30*(-I*a^2 - 2*a + I)*b*x^2 - 60*(I*a^3 + 3*a^2 - 3*I
*a - 1)*x)/b^4 - 2*(I*a^4 + 4*a^3 - 6*I*a^2 - 4*a + I)*log(I*b*x + I*a + 1)/b^5

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Fricas [A]
time = 3.21, size = 105, normalized size = 1.06 \begin {gather*} -\frac {6 \, b^{5} x^{5} + 15 i \, b^{4} x^{4} + 20 \, {\left (-i \, a - 1\right )} b^{3} x^{3} + 30 \, {\left (i \, a^{2} + 2 \, a - i\right )} b^{2} x^{2} + 60 \, {\left (-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1\right )} b x + 60 \, {\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} \log \left (\frac {b x + a - i}{b}\right )}{30 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="fricas")

[Out]

-1/30*(6*b^5*x^5 + 15*I*b^4*x^4 + 20*(-I*a - 1)*b^3*x^3 + 30*(I*a^2 + 2*a - I)*b^2*x^2 + 60*(-I*a^3 - 3*a^2 +
3*I*a + 1)*b*x + 60*(I*a^4 + 4*a^3 - 6*I*a^2 - 4*a + I)*log((b*x + a - I)/b))/b^5

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Sympy [A]
time = 0.25, size = 114, normalized size = 1.15 \begin {gather*} - \frac {x^{5}}{5} - x^{3} \left (- \frac {2 i a}{3 b^{2}} - \frac {2}{3 b^{2}}\right ) - x^{2} \left (\frac {i a^{2}}{b^{3}} + \frac {2 a}{b^{3}} - \frac {i}{b^{3}}\right ) - x \left (- \frac {2 i a^{3}}{b^{4}} - \frac {6 a^{2}}{b^{4}} + \frac {6 i a}{b^{4}} + \frac {2}{b^{4}}\right ) - \frac {i x^{4}}{2 b} - \frac {2 i \left (a - i\right )^{4} \log {\left (a + b x - i \right )}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(1+I*(b*x+a))**2*(1+(b*x+a)**2),x)

[Out]

-x**5/5 - x**3*(-2*I*a/(3*b**2) - 2/(3*b**2)) - x**2*(I*a**2/b**3 + 2*a/b**3 - I/b**3) - x*(-2*I*a**3/b**4 - 6
*a**2/b**4 + 6*I*a/b**4 + 2/b**4) - I*x**4/(2*b) - 2*I*(a - I)**4*log(a + b*x - I)/b**5

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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. 215 vs. \(2 (73) = 146\).
time = 0.45, size = 215, normalized size = 2.17 \begin {gather*} \frac {i \, {\left (i \, b x + i \, a + 1\right )}^{5} {\left (-\frac {15 i \, {\left (2 \, a b - 3 i \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} - \frac {20 \, {\left (3 \, a^{2} b^{2} - 10 i \, a b^{2} - 7 \, b^{2}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{2} b^{2}} + \frac {60 i \, {\left (a^{3} b^{3} - 6 i \, a^{2} b^{3} - 9 \, a b^{3} + 4 i \, b^{3}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{3} b^{3}} + \frac {30 \, {\left (a^{4} b^{4} - 12 i \, a^{3} b^{4} - 30 \, a^{2} b^{4} + 28 i \, a b^{4} + 9 \, b^{4}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{4} b^{4}} + 6\right )}}{30 \, b^{5}} - \frac {2 \, {\left (-i \, a^{4} - 4 \, a^{3} + 6 i \, a^{2} + 4 \, a - i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1+I*(b*x+a))^2*(1+(b*x+a)^2),x, algorithm="giac")

[Out]

1/30*I*(I*b*x + I*a + 1)^5*(-15*I*(2*a*b - 3*I*b)/((I*b*x + I*a + 1)*b) - 20*(3*a^2*b^2 - 10*I*a*b^2 - 7*b^2)/
((I*b*x + I*a + 1)^2*b^2) + 60*I*(a^3*b^3 - 6*I*a^2*b^3 - 9*a*b^3 + 4*I*b^3)/((I*b*x + I*a + 1)^3*b^3) + 30*(a
^4*b^4 - 12*I*a^3*b^4 - 30*a^2*b^4 + 28*I*a*b^4 + 9*b^4)/((I*b*x + I*a + 1)^4*b^4) + 6)/b^5 - 2*(-I*a^4 - 4*a^
3 + 6*I*a^2 + 4*a - I)*log(1/(sqrt((b*x + a)^2 + 1)*abs(b)))/b^5

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Mupad [B]
time = 0.17, size = 165, normalized size = 1.67 \begin {gather*} \ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {8\,a-8\,a^3}{b^5}-\frac {\left (2\,a^4-12\,a^2+2\right )\,1{}\mathrm {i}}{b^5}\right )+x^4\,\left (\frac {a-\mathrm {i}}{4\,b}-\frac {a+1{}\mathrm {i}}{4\,b}\right )-\frac {x^5}{5}+\frac {x^2\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,{\left (a-\mathrm {i}\right )}^2}{2\,b^2}-\frac {x^3\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{3\,b}-\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,{\left (a-\mathrm {i}\right )}^3}{b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*((a + b*x)^2 + 1))/(a*1i + b*x*1i + 1)^2,x)

[Out]

log(x + (a - 1i)/b)*((8*a - 8*a^3)/b^5 - ((2*a^4 - 12*a^2 + 2)*1i)/b^5) + x^4*((a - 1i)/(4*b) - (a + 1i)/(4*b)
) - x^5/5 + (x^2*((a - 1i)/b - (a + 1i)/b)*(a - 1i)^2)/(2*b^2) - (x^3*((a - 1i)/b - (a + 1i)/b)*(a - 1i))/(3*b
) - (x*((a - 1i)/b - (a + 1i)/b)*(a - 1i)^3)/b^3

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