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

Optimal. Leaf size=92 \[ \frac {2 i (a+i)^4 \log (a+b x+i)}{b^5}-\frac {2 (1-i a)^3 x}{b^4}+\frac {i (a+i)^2 x^2}{b^3}+\frac {2 (1-i a) x^3}{3 b^2}+\frac {i x^4}{2 b}-\frac {x^5}{5} \]

[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 = 0.08, antiderivative size = 92, 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[E^((2*I)*ArcTan[a + b*x])*x^4,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 = 0.08, size = 92, normalized size = 1.00 \[ \frac {2 i (a+i)^4 \log (a+b x+i)}{b^5}-\frac {2 (1-i a)^3 x}{b^4}+\frac {i (a+i)^2 x^2}{b^3}+\frac {2 (1-i a) x^3}{3 b^2}+\frac {i x^4}{2 b}-\frac {x^5}{5} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((2*I)*ArcTan[a + b*x])*x^4,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 = 0.48, size = 105, normalized size = 1.14 \[ -\frac {6 \, b^{5} x^{5} - 15 i \, b^{4} x^{4} + 20 \, {\left (i \, a - 1\right )} b^{3} x^{3} - {\left (30 i \, a^{2} - 60 \, a - 30 i\right )} b^{2} x^{2} - {\left (-60 i \, a^{3} + 180 \, a^{2} + 180 i \, a - 60\right )} b x - {\left (60 i \, a^{4} - 240 \, a^{3} - 360 i \, a^{2} + 240 \, a + 60 i\right )} \log \left (\frac {b x + a + i}{b}\right )}{30 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,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 [A]  time = 0.12, size = 130, normalized size = 1.41 \[ \frac {2 \, {\left (a^{4} i - 4 \, a^{3} - 6 \, a^{2} i + 4 \, a + i\right )} \log \left (b x + a + i\right )}{b^{5}} - \frac {6 \, b^{5} x^{5} - 15 \, b^{4} i x^{4} + 20 \, a b^{3} i x^{3} - 30 \, a^{2} b^{2} i x^{2} + 60 \, a^{3} b i x - 20 \, b^{3} x^{3} + 60 \, a b^{2} x^{2} + 30 \, b^{2} i x^{2} - 180 \, a^{2} b x - 180 \, a b i x + 60 \, b x}{30 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a^4*i - 4*a^3 - 6*a^2*i + 4*a + i)*log(b*x + a + i)/b^5 - 1/30*(6*b^5*x^5 - 15*b^4*i*x^4 + 20*a*b^3*i*x^3 -
30*a^2*b^2*i*x^2 + 60*a^3*b*i*x - 20*b^3*x^3 + 60*a*b^2*x^2 + 30*b^2*i*x^2 - 180*a^2*b*x - 180*a*b*i*x + 60*b
*x)/b^5

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maple [B]  time = 0.04, size = 347, normalized size = 3.77 \[ -\frac {x^{5}}{5}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{5}}-\frac {2 i x^{3} a}{3 b^{2}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{b^{5}}+\frac {i x^{4}}{2 b}+\frac {2 x^{3}}{3 b^{2}}-\frac {i x^{2}}{b^{3}}-\frac {2 a \,x^{2}}{b^{3}}-\frac {6 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{5}}+\frac {6 x \,a^{2}}{b^{4}}-\frac {2 x}{b^{4}}-\frac {8 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{b^{5}}+\frac {8 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{3}}{b^{5}}-\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{5}}-\frac {2 i a^{3} x}{b^{4}}+\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{5}}+\frac {i x^{2} a^{2}}{b^{3}}+\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{4}}{b^{5}}+\frac {6 i a x}{b^{4}}-\frac {12 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{b^{5}}+\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 150, normalized size = 1.63 \[ -\frac {6 \, b^{4} x^{5} - 15 i \, b^{3} x^{4} + 20 \, {\left (i \, a - 1\right )} b^{2} x^{3} - {\left (30 i \, a^{2} - 60 \, a - 30 i\right )} b x^{2} - {\left (-60 i \, a^{3} + 180 \, a^{2} + 180 i \, a - 60\right )} x}{30 \, b^{4}} + \frac {{\left (2 \, a^{4} + 8 i \, a^{3} - 12 \, a^{2} - 8 i \, a + 2\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{5}} + \frac {{\left (i \, a^{4} - 4 \, a^{3} - 6 i \, a^{2} + 4 \, a + i\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.58, size = 201, normalized size = 2.18 \[ \ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (\frac {8\,a-8\,a^3}{b^5}+\frac {\left (2\,a^4-12\,a^2+2\right )\,1{}\mathrm {i}}{b^5}\right )-x^4\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,b}\right )-\frac {x^5}{5}+\frac {x^2\,{\left (-1+a\,1{}\mathrm {i}\right )}^2\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )}{2\,b^2}-\frac {x^3\,\left (-1+a\,1{}\mathrm {i}\right )\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{3\,b}+\frac {x\,{\left (-1+a\,1{}\mathrm {i}\right )}^3\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a*1i + b*x*1i + 1)^2)/((a + b*x)^2 + 1),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 - 1)*1i)/(4*b) - ((a*1i
+ 1)*1i)/(4*b)) - x^5/5 + (x^2*(a*1i - 1)^2*(((a*1i - 1)*1i)/b - ((a*1i + 1)*1i)/b))/(2*b^2) - (x^3*(a*1i - 1)
*(((a*1i - 1)*1i)/b - ((a*1i + 1)*1i)/b)*1i)/(3*b) + (x*(a*1i - 1)^3*(((a*1i - 1)*1i)/b - ((a*1i + 1)*1i)/b)*1
i)/b^3

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sympy [A]  time = 0.50, size = 114, normalized size = 1.24 \[ - \frac {x^{5}}{5} - x^{3} \left (\frac {2 i a}{3 b^{2}} - \frac {2}{3 b^{2}}\right ) - x^{2} \left (- \frac {i a^{2}}{b^{3}} + \frac {2 a}{b^{3}} + \frac {i}{b^{3}}\right ) - x \left (\frac {2 i a^{3}}{b^{4}} - \frac {6 a^{2}}{b^{4}} - \frac {6 i a}{b^{4}} + \frac {2}{b^{4}}\right ) + \frac {i x^{4}}{2 b} + \frac {2 i \left (a + i\right )^{4} \log {\left (i a + i b x - 1 \right )}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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