Integrand size = 16, antiderivative size = 92 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=-\frac {2 (1-i a)^3 x}{b^4}+\frac {i (i+a)^2 x^2}{b^3}+\frac {2 (1-i a) x^3}{3 b^2}+\frac {i x^4}{2 b}-\frac {x^5}{5}+\frac {2 i (i+a)^4 \log (i+a+b x)}{b^5} \] Output:
-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 = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=-\frac {2 (1-i a)^3 x}{b^4}+\frac {i (i+a)^2 x^2}{b^3}+\frac {2 (1-i a) x^3}{3 b^2}+\frac {i x^4}{2 b}-\frac {x^5}{5}+\frac {2 i (i+a)^4 \log (i+a+b x)}{b^5} \] Input:
Integrate[E^((2*I)*ArcTan[a + b*x])*x^4,x]
Output:
(-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 = 0.46 (sec) , antiderivative size = 92, 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 |
\(\displaystyle \frac {2 i (a+i)^4 \log (a+b x+i)}{b^5}-\frac {2 (1-i a)^3 x}{b^4}+\frac {i (a+i)^2 x^2}{b^3}+\frac {2 (1-i a) x^3}{3 b^2}+\frac {i x^4}{2 b}-\frac {x^5}{5}\) |
Input:
Int[E^((2*I)*ArcTan[a + b*x])*x^4,x]
Output:
(-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
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (78 ) = 156\).
Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(\frac {-6 x^{5} b^{5}-30 i b^{2} x^{2}+60 i \ln \left (b x +a +i\right ) a^{4}+180 i a b x -240 \ln \left (b x +a +i\right ) a^{3}+30 i x^{2} a^{2} b^{2}-60 i x \,a^{3} b +20 b^{3} x^{3}-20 i x^{3} a \,b^{3}-60 a \,b^{2} x^{2}+240 \ln \left (b x +a +i\right ) a +60 i \ln \left (b x +a +i\right )+15 i x^{4} b^{4}+180 a^{2} b x -360 i \ln \left (b x +a +i\right ) a^{2}-60 b x}{30 b^{5}}\) | \(160\) |
default | \(-\frac {i \left (-\frac {1}{5} i b^{4} x^{5}-\frac {1}{2} b^{3} x^{4}+\frac {2}{3} i b^{2} x^{3}+\frac {2}{3} a \,b^{2} x^{3}-2 i a b \,x^{2}-a^{2} b \,x^{2}+6 i x \,a^{2}+2 a^{3} x +b \,x^{2}-2 i x -6 a x \right )}{b^{4}}+\frac {\frac {\left (2 i a^{4} b -8 a^{3} b -12 i a^{2} b +8 a b +2 i b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (2 i a^{5}-4 i a^{3}-6 a^{4}-6 i a -4 a^{2}+2-\frac {\left (2 i a^{4} b -8 a^{3} b -12 i a^{2} b +8 a b +2 i b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{4}}\) | \(228\) |
risch | \(-\frac {x^{5}}{5}-\frac {6 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{5}}+\frac {2 x^{3}}{3 b^{2}}+\frac {8 i \arctan \left (b x +a \right ) a^{3}}{b^{5}}-\frac {2 a \,x^{2}}{b^{3}}-\frac {8 i \arctan \left (b x +a \right ) a}{b^{5}}+\frac {6 x \,a^{2}}{b^{4}}+\frac {i a^{2} x^{2}}{b^{3}}-\frac {2 i a \,x^{3}}{3 b^{2}}-\frac {2 x}{b^{4}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{b^{5}}-\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{5}}-\frac {2 i a^{3} x}{b^{4}}+\frac {4 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{5}}-\frac {i x^{2}}{b^{3}}+\frac {6 i a x}{b^{4}}+\frac {i x^{4}}{2 b}+\frac {2 \arctan \left (b x +a \right ) a^{4}}{b^{5}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{5}}-\frac {12 \arctan \left (b x +a \right ) a^{2}}{b^{5}}+\frac {2 \arctan \left (b x +a \right )}{b^{5}}\) | \(292\) |
Input:
int((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,x,method=_RETURNVERBOSE)
Output:
1/30*(-6*x^5*b^5-30*I*x^2*b^2+60*I*ln(I+a+b*x)*a^4+180*I*x*a*b-240*ln(I+a+ b*x)*a^3+30*I*x^2*a^2*b^2-60*I*x*a^3*b+20*b^3*x^3-20*I*x^3*a*b^3-60*a*b^2* x^2+240*ln(I+a+b*x)*a+60*I*ln(I+a+b*x)+15*I*x^4*b^4+180*a^2*b*x-360*I*ln(I +a+b*x)*a^2-60*b*x)/b^5
Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=-\frac {6 \, b^{5} x^{5} - 15 i \, b^{4} x^{4} + 20 \, {\left (i \, a - 1\right )} b^{3} x^{3} + 30 \, {\left (-i \, a^{2} + 2 \, a + i\right )} b^{2} x^{2} + 60 \, {\left (i \, a^{3} - 3 \, a^{2} - 3 i \, a + 1\right )} b x + 60 \, {\left (-i \, a^{4} + 4 \, a^{3} + 6 i \, a^{2} - 4 \, a - i\right )} \log \left (\frac {b x + a + i}{b}\right )}{30 \, b^{5}} \] Input:
integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,x, algorithm="fricas")
Output:
-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 = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=- \frac {x^{5}}{5} - x^{3} \cdot \left (\frac {2 i a}{3 b^{2}} - \frac {2}{3 b^{2}}\right ) - x^{2} \left (- \frac {i a^{2}}{b^{3}} + \frac {2 a}{b^{3}} + \frac {i}{b^{3}}\right ) - x \left (\frac {2 i a^{3}}{b^{4}} - \frac {6 a^{2}}{b^{4}} - \frac {6 i a}{b^{4}} + \frac {2}{b^{4}}\right ) + \frac {i x^{4}}{2 b} + \frac {2 i \left (a + i\right )^{4} \log {\left (a + b x + i \right )}}{b^{5}} \] Input:
integrate((1+I*(b*x+a))**2/(1+(b*x+a)**2)*x**4,x)
Output:
-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
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (70) = 140\).
Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.62 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=-\frac {6 \, b^{4} x^{5} - 15 i \, b^{3} x^{4} + 20 \, {\left (i \, a - 1\right )} b^{2} x^{3} + 30 \, {\left (-i \, a^{2} + 2 \, a + i\right )} b x^{2} + 60 \, {\left (i \, a^{3} - 3 \, a^{2} - 3 i \, a + 1\right )} x}{30 \, b^{4}} + \frac {2 \, {\left (a^{4} + 4 i \, a^{3} - 6 \, a^{2} - 4 i \, a + 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{5}} + \frac {{\left (i \, a^{4} - 4 \, a^{3} - 6 i \, a^{2} + 4 \, a + i\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}} \] Input:
integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,x, algorithm="maxima")
Output:
-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*(a^4 + 4*I*a^3 - 6* a^2 - 4*I*a + 1)*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
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=-\frac {2 \, {\left (-i \, a^{4} + 4 \, a^{3} + 6 i \, a^{2} - 4 \, a - i\right )} \log \left (b x + a + i\right )}{b^{5}} - \frac {6 \, b^{5} x^{5} - 15 i \, b^{4} x^{4} + 20 i \, a b^{3} x^{3} - 30 i \, a^{2} b^{2} x^{2} - 20 \, b^{3} x^{3} + 60 i \, a^{3} b x + 60 \, a b^{2} x^{2} - 180 \, a^{2} b x + 30 i \, b^{2} x^{2} - 180 i \, a b x + 60 \, b x}{30 \, b^{5}} \] Input:
integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,x, algorithm="giac")
Output:
-2*(-I*a^4 + 4*a^3 + 6*I*a^2 - 4*a - I)*log(b*x + a + I)/b^5 - 1/30*(6*b^5 *x^5 - 15*I*b^4*x^4 + 20*I*a*b^3*x^3 - 30*I*a^2*b^2*x^2 - 20*b^3*x^3 + 60* I*a^3*b*x + 60*a*b^2*x^2 - 180*a^2*b*x + 30*I*b^2*x^2 - 180*I*a*b*x + 60*b *x)/b^5
Time = 0.21 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.18 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (\frac {8\,a-8\,a^3}{b^5}+\frac {\left (2\,a^4-12\,a^2+2\right )\,1{}\mathrm {i}}{b^5}\right )-x^4\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,b}\right )-\frac {x^5}{5}+\frac {x^2\,{\left (-1+a\,1{}\mathrm {i}\right )}^2\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )}{2\,b^2}-\frac {x^3\,\left (-1+a\,1{}\mathrm {i}\right )\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{3\,b}+\frac {x\,{\left (-1+a\,1{}\mathrm {i}\right )}^3\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{b^3} \] Input:
int((x^4*(a*1i + b*x*1i + 1)^2)/((a + b*x)^2 + 1),x)
Output:
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)*1i)/b^3
Time = 0.15 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.84 \[ \int e^{2 i \arctan (a+b x)} x^4 \, dx=\frac {60 \mathit {atan} \left (b x +a \right ) a^{4}+240 \mathit {atan} \left (b x +a \right ) a^{3} i -360 \mathit {atan} \left (b x +a \right ) a^{2}-240 \mathit {atan} \left (b x +a \right ) a i +60 \mathit {atan} \left (b x +a \right )+30 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4} i -120 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}-180 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2} i +120 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a +30 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) i -60 a^{3} b i x +30 a^{2} b^{2} i \,x^{2}+180 a^{2} b x -20 a \,b^{3} i \,x^{3}-60 a \,b^{2} x^{2}+180 a b i x -6 b^{5} x^{5}+15 b^{4} i \,x^{4}+20 b^{3} x^{3}-30 b^{2} i \,x^{2}-60 b x}{30 b^{5}} \] Input:
int((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x^4,x)
Output:
(60*atan(a + b*x)*a**4 + 240*atan(a + b*x)*a**3*i - 360*atan(a + b*x)*a**2 - 240*atan(a + b*x)*a*i + 60*atan(a + b*x) + 30*log(a**2 + 2*a*b*x + b**2 *x**2 + 1)*a**4*i - 120*log(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**3 - 180*log (a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2*i + 120*log(a**2 + 2*a*b*x + b**2*x* *2 + 1)*a + 30*log(a**2 + 2*a*b*x + b**2*x**2 + 1)*i - 60*a**3*b*i*x + 30* a**2*b**2*i*x**2 + 180*a**2*b*x - 20*a*b**3*i*x**3 - 60*a*b**2*x**2 + 180* a*b*i*x - 6*b**5*x**5 + 15*b**4*i*x**4 + 20*b**3*x**3 - 30*b**2*i*x**2 - 6 0*b*x)/(30*b**5)