Integrand size = 16, antiderivative size = 99 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (i-a)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}-\frac {2 i (i-a)^4 \log (i-a-b x)}{b^5} \]
-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
Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (-i+a)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}-\frac {2 i (-i+a)^4 \log (i-a-b x)}{b^5} \]
(-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
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5618, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^4 e^{-2 i \arctan (a+b x)} \, dx\) |
\(\Big \downarrow \) 5618 |
\(\displaystyle \int \frac {x^4 (-i a-i b x+1)}{i a+i b x+1}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {2 i (a-i)^4}{b^4 (a+b x-i)}+\frac {2 (-1-i a)^3}{b^4}-\frac {2 i (a-i)^2 x}{b^3}+\frac {2 (1+i a) x^2}{b^2}-\frac {2 i x^3}{b}-x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 i (-a+i)^4 \log (-a-b x+i)}{b^5}-\frac {2 (1+i a)^3 x}{b^4}-\frac {i (-a+i)^2 x^2}{b^3}+\frac {2 (1+i a) x^3}{3 b^2}-\frac {i x^4}{2 b}-\frac {x^5}{5}\) |
(-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
3.2.98.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((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]))))
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]
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 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 | \(-\frac {x^{5}}{5}+\frac {2 i a \,x^{3}}{3 b^{2}}+\frac {2 x^{3}}{3 b^{2}}-\frac {i x^{4}}{2 b}-\frac {2 a \,x^{2}}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{b^{5}}+\frac {6 a^{2} x}{b^{4}}-\frac {8 i \arctan \left (b x +a \right ) a^{3}}{b^{5}}+\frac {2 i a^{3} x}{b^{4}}-\frac {2 x}{b^{4}}+\frac {8 i \arctan \left (b x +a \right ) a}{b^{5}}-\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{5}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{5}}+\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{5}}+\frac {i x^{2}}{b^{3}}+\frac {6 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{5}}-\frac {6 i a x}{b^{4}}+\frac {2 \arctan \left (b x +a \right ) a^{4}}{b^{5}}-\frac {i a^{2} x^{2}}{b^{3}}-\frac {12 \arctan \left (b x +a \right ) a^{2}}{b^{5}}+\frac {2 \arctan \left (b x +a \right )}{b^{5}}\) | \(292\) |
parallelrisch | \(\frac {60-90 a^{2} b^{2} x^{2}-600 a^{2}+300 a^{4}+300 \ln \left (b x +a -i\right ) a^{4}-600 \ln \left (b x +a -i\right ) a^{2}-600 i \ln \left (b x +a -i\right ) a^{3}+300 i \ln \left (b x +a -i\right ) a +9 i x^{5} b^{5}+60 i \ln \left (b x +a -i\right ) a^{5}+240 \ln \left (b x +a -i\right ) x \,a^{3} b -240 \ln \left (b x +a -i\right ) x a b +60 i \ln \left (b x +a -i\right ) x b +60 i \ln \left (b x +a -i\right ) x \,a^{4} b -360 i \ln \left (b x +a -i\right ) x \,a^{2} b +300 i a +60 \ln \left (b x +a -i\right )+6 b^{6} x^{6}+30 b^{2} x^{2}-5 x^{4} b^{4}+20 a \,b^{3} x^{3}+60 i a^{5}+90 i a \,b^{2} x^{2}-600 i a^{3}-5 i a \,b^{4} x^{4}+10 i a^{2} b^{3} x^{3}+6 a \,b^{5} x^{5}-30 i x^{2} a^{3} b^{2}-10 i b^{3} x^{3}}{30 b^{5} \left (-b x -a +i\right )}\) | \(310\) |
-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)
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=-\frac {6 \, b^{5} x^{5} + 15 i \, b^{4} x^{4} + 20 \, {\left (-i \, a - 1\right )} b^{3} x^{3} + 30 \, {\left (i \, a^{2} + 2 \, a - i\right )} b^{2} x^{2} + 60 \, {\left (-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1\right )} b x + 60 \, {\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} \log \left (\frac {b x + a - i}{b}\right )}{30 \, b^{5}} \]
-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
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.15 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=- \frac {x^{5}}{5} - x^{3} \left (- \frac {2 i a}{3 b^{2}} - \frac {2}{3 b^{2}}\right ) - x^{2} \left (\frac {i a^{2}}{b^{3}} + \frac {2 a}{b^{3}} - \frac {i}{b^{3}}\right ) - x \left (- \frac {2 i a^{3}}{b^{4}} - \frac {6 a^{2}}{b^{4}} + \frac {6 i a}{b^{4}} + \frac {2}{b^{4}}\right ) - \frac {i x^{4}}{2 b} - \frac {2 i \left (a - i\right )^{4} \log {\left (a + b x - i \right )}}{b^{5}} \]
-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
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=-\frac {6 \, b^{4} x^{5} + 15 i \, b^{3} x^{4} - 20 \, {\left (i \, a + 1\right )} b^{2} x^{3} - 30 \, {\left (-i \, a^{2} - 2 \, a + i\right )} b x^{2} - 60 \, {\left (i \, a^{3} + 3 \, a^{2} - 3 i \, a - 1\right )} x}{30 \, b^{4}} - \frac {2 \, {\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{5}} \]
-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
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 = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.17 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=\frac {i \, {\left (i \, b x + i \, a + 1\right )}^{5} {\left (-\frac {15 i \, {\left (2 \, a b - 3 i \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} - \frac {20 \, {\left (3 \, a^{2} b^{2} - 10 i \, a b^{2} - 7 \, b^{2}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{2} b^{2}} + \frac {60 i \, {\left (a^{3} b^{3} - 6 i \, a^{2} b^{3} - 9 \, a b^{3} + 4 i \, b^{3}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{3} b^{3}} + \frac {30 \, {\left (a^{4} b^{4} - 12 i \, a^{3} b^{4} - 30 \, a^{2} b^{4} + 28 i \, a b^{4} + 9 \, b^{4}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{4} b^{4}} + 6\right )}}{30 \, b^{5}} - \frac {2 \, {\left (-i \, a^{4} - 4 \, a^{3} + 6 i \, a^{2} + 4 \, a - i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{5}} \]
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
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.67 \[ \int e^{-2 i \arctan (a+b x)} x^4 \, dx=\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {8\,a-8\,a^3}{b^5}-\frac {\left (2\,a^4-12\,a^2+2\right )\,1{}\mathrm {i}}{b^5}\right )+x^4\,\left (\frac {a-\mathrm {i}}{4\,b}-\frac {a+1{}\mathrm {i}}{4\,b}\right )-\frac {x^5}{5}+\frac {x^2\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,{\left (a-\mathrm {i}\right )}^2}{2\,b^2}-\frac {x^3\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{3\,b}-\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,{\left (a-\mathrm {i}\right )}^3}{b^3} \]