3.198 \(\int e^{-2 i \tan ^{-1}(a+b x)} x^4 \, dx\)

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

[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 = 0.09, 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 = {5095, 77} \[ \frac {2 (1+i a) x^3}{3 b^2}-\frac {i (-a+i)^2 x^2}{b^3}-\frac {2 (1+i a)^3 x}{b^4}-\frac {2 i (-a+i)^4 \log (-a-b x+i)}{b^5}-\frac {i x^4}{2 b}-\frac {x^5}{5} \]

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 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^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 = 0.07, size = 95, normalized size = 0.96 \[ -\frac {2 i (a-i)^4 \log (-a-b x+i)}{b^5}-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (a-i)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5} \]

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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fricas [A]  time = 0.46, size = 105, normalized size = 1.06 \[ -\frac {6 \, b^{5} x^{5} + 15 i \, b^{4} x^{4} + 20 \, {\left (-i \, a - 1\right )} b^{3} x^{3} - {\left (-30 i \, a^{2} - 60 \, a + 30 i\right )} b^{2} x^{2} - {\left (60 i \, a^{3} + 180 \, a^{2} - 180 i \, a - 60\right )} b x - {\left (-60 i \, a^{4} - 240 \, a^{3} + 360 i \, a^{2} + 240 \, a - 60 i\right )} \log \left (\frac {b x + a - i}{b}\right )}{30 \, b^{5}} \]

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 - 60*a + 30*I)*b^2*x^2 - (60*I*a^3 + 180*
a^2 - 180*I*a - 60)*b*x - (-60*I*a^4 - 240*a^3 + 360*I*a^2 + 240*a - 60*I)*log((b*x + a - I)/b))/b^5

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giac [B]  time = 0.12, size = 235, normalized size = 2.37 \[ \frac {2 \, {\left (a^{4} i + 4 \, a^{3} - 6 \, a^{2} i - 4 \, a + i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{5}} + \frac {{\left (b i x + a i + 1\right )}^{5} {\left (\frac {15 \, {\left (2 \, a b - 3 \, b i\right )} i}{{\left (b i x + a i + 1\right )} b} - \frac {20 \, {\left (3 \, a^{2} b^{2} - 10 \, a b^{2} i - 7 \, b^{2}\right )} i^{2}}{{\left (b i x + a i + 1\right )}^{2} b^{2}} + \frac {60 \, {\left (a^{3} b^{3} - 6 \, a^{2} b^{3} i - 9 \, a b^{3} + 4 \, b^{3} i\right )} i^{3}}{{\left (b i x + a i + 1\right )}^{3} b^{3}} - \frac {30 \, {\left (a^{4} b^{4} - 12 \, a^{3} b^{4} i - 30 \, a^{2} b^{4} + 28 \, a b^{4} i + 9 \, b^{4}\right )} i^{4}}{{\left (b i x + a i + 1\right )}^{4} b^{4}} - 6\right )}}{30 \, b^{5} i^{5}} \]

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]

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

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maple [B]  time = 0.06, size = 292, normalized size = 2.95 \[ -\frac {x^{5}}{5}+\frac {8 i \arctan \left (b x +a \right ) a}{b^{5}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{b^{5}}-\frac {i x^{2} a^{2}}{b^{3}}-\frac {6 i a x}{b^{4}}+\frac {2 x^{3}}{3 b^{2}}+\frac {2 i x^{3} a}{3 b^{2}}-\frac {2 a \,x^{2}}{b^{3}}+\frac {6 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{5}}+\frac {6 x \,a^{2}}{b^{4}}-\frac {2 x}{b^{4}}+\frac {2 \arctan \left (b x +a \right ) a^{4}}{b^{5}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{5}}-\frac {12 \arctan \left (b x +a \right ) a^{2}}{b^{5}}+\frac {i x^{2}}{b^{3}}+\frac {2 i a^{3} x}{b^{4}}-\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{5}}+\frac {2 \arctan \left (b x +a \right )}{b^{5}}-\frac {i x^{4}}{2 b}-\frac {8 i \arctan \left (b x +a \right ) a^{3}}{b^{5}}+\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{5}} \]

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)

[Out]

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

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maxima [A]  time = 0.32, size = 102, normalized size = 1.03 \[ -\frac {6 \, b^{4} x^{5} + 15 i \, b^{3} x^{4} - 20 \, {\left (i \, a + 1\right )} b^{2} x^{3} + {\left (30 i \, a^{2} + 60 \, a - 30 i\right )} b x^{2} + {\left (-60 i \, a^{3} - 180 \, a^{2} + 180 i \, a + 60\right )} x}{30 \, b^{4}} + \frac {{\left (-2 i \, a^{4} - 8 \, a^{3} + 12 i \, a^{2} + 8 \, a - 2 i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{5}} \]

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 + 60*a - 30*I)*b*x^2 + (-60*I*a^3 - 180*a^2
+ 180*I*a + 60)*x)/b^4 + (-2*I*a^4 - 8*a^3 + 12*I*a^2 + 8*a - 2*I)*log(I*b*x + I*a + 1)/b^5

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

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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sympy [A]  time = 0.52, size = 117, normalized size = 1.18 \[ - \frac {x^{5}}{5} - x^{3} \left (- \frac {2 i a}{3 b^{2}} - \frac {2}{3 b^{2}}\right ) - x^{2} \left (\frac {i a^{2}}{b^{3}} + \frac {2 a}{b^{3}} - \frac {i}{b^{3}}\right ) - x \left (- \frac {2 i a^{3}}{b^{4}} - \frac {6 a^{2}}{b^{4}} + \frac {6 i a}{b^{4}} + \frac {2}{b^{4}}\right ) - \frac {i x^{4}}{2 b} - \frac {2 i \left (a - i\right )^{4} \log {\left (i a + i b x + 1 \right )}}{b^{5}} \]

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(I*a + I*b*x + 1)/b**5

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