Integrand size = 21, antiderivative size = 117 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4-\frac {b d \arctan (c x)}{4 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {i b d \log \left (1+c^2 x^2\right )}{10 c^4} \]
1/4*b*d*x/c^3+1/10*I*b*d*x^2/c^2-1/12*b*d*x^3/c-1/20*I*b*d*x^4-1/4*b*d*arc tan(c*x)/c^4+1/4*d*x^4*(a+b*arctan(c*x))+1/5*I*c*d*x^5*(a+b*arctan(c*x))-1 /10*I*b*d*ln(c^2*x^2+1)/c^4
Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {d \left (3 a c^4 x^4 (5+4 i c x)+b c x \left (15+6 i c x-5 c^2 x^2-3 i c^3 x^3\right )+3 b \left (-5+5 c^4 x^4+4 i c^5 x^5\right ) \arctan (c x)-6 i b \log \left (1+c^2 x^2\right )\right )}{60 c^4} \]
(d*(3*a*c^4*x^4*(5 + (4*I)*c*x) + b*c*x*(15 + (6*I)*c*x - 5*c^2*x^2 - (3*I )*c^3*x^3) + 3*b*(-5 + 5*c^4*x^4 + (4*I)*c^5*x^5)*ArcTan[c*x] - (6*I)*b*Lo g[1 + c^2*x^2]))/(60*c^4)
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5407, 27, 523, 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.
\(\displaystyle \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx\) |
\(\Big \downarrow \) 5407 |
\(\displaystyle -b c \int \frac {d x^4 (4 i c x+5)}{20 \left (c^2 x^2+1\right )}dx+\frac {1}{5} i c d x^5 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{20} b c d \int \frac {x^4 (4 i c x+5)}{c^2 x^2+1}dx+\frac {1}{5} i c d x^5 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))\) |
\(\Big \downarrow \) 523 |
\(\displaystyle -\frac {1}{20} b c d \int \left (\frac {4 i x^3}{c}+\frac {5 x^2}{c^2}-\frac {4 i x}{c^3}+\frac {4 i c x+5}{c^4 \left (c^2 x^2+1\right )}-\frac {5}{c^4}\right )dx+\frac {1}{5} i c d x^5 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} i c d x^5 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))-\frac {1}{20} b c d \left (\frac {5 \arctan (c x)}{c^5}-\frac {5 x}{c^4}-\frac {2 i x^2}{c^3}+\frac {5 x^3}{3 c^2}+\frac {2 i \log \left (c^2 x^2+1\right )}{c^5}+\frac {i x^4}{c}\right )\) |
(d*x^4*(a + b*ArcTan[c*x]))/4 + (I/5)*c*d*x^5*(a + b*ArcTan[c*x]) - (b*c*d *((-5*x)/c^4 - ((2*I)*x^2)/c^3 + (5*x^3)/(3*c^2) + (I*x^4)/c + (5*ArcTan[c *x])/c^5 + ((2*I)*Log[1 + c^2*x^2])/c^5))/20
3.1.1.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[(a + b*ArcTan[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 + c^2*x^2 ), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && IntegerQ [2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]) )
Time = 1.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85
method | result | size |
parts | \(a d \left (\frac {1}{5} i c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(99\) |
derivativedivides | \(\frac {a d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(105\) |
default | \(\frac {a d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(105\) |
parallelrisch | \(-\frac {-12 i c^{5} b d \arctan \left (c x \right ) x^{5}-12 i x^{5} a \,c^{5} d +3 i x^{4} b \,c^{4} d -15 x^{4} \arctan \left (c x \right ) b \,c^{4} d -15 a \,c^{4} d \,x^{4}+5 c^{3} x^{3} d b -6 i x^{2} b \,c^{2} d +6 i b d \ln \left (c^{2} x^{2}+1\right )-15 b c d x +15 b d \arctan \left (c x \right )}{60 c^{4}}\) | \(118\) |
risch | \(\frac {d b \left (4 x^{5} c -5 i x^{4}\right ) \ln \left (i c x +1\right )}{40}-\frac {d c b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {i d c a \,x^{5}}{5}+\frac {a d \,x^{4}}{4}+\frac {i d \,x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {i b d \,x^{4}}{20}-\frac {b d \,x^{3}}{12 c}+\frac {i b d \,x^{2}}{10 c^{2}}+\frac {b d x}{4 c^{3}}-\frac {b d \arctan \left (c x \right )}{4 c^{4}}-\frac {i d b \ln \left (25 c^{2} x^{2}+25\right )}{10 c^{4}}\) | \(142\) |
a*d*(1/5*I*c*x^5+1/4*x^4)+b*d/c^4*(1/5*I*arctan(c*x)*c^5*x^5+1/4*c^4*x^4*a rctan(c*x)+1/4*c*x-1/20*I*c^4*x^4-1/12*c^3*x^3+1/10*I*c^2*x^2-1/10*I*ln(c^ 2*x^2+1)-1/4*arctan(c*x))
Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {24 i \, a c^{5} d x^{5} + 6 \, {\left (5 \, a - i \, b\right )} c^{4} d x^{4} - 10 \, b c^{3} d x^{3} + 12 i \, b c^{2} d x^{2} + 30 \, b c d x - 27 i \, b d \log \left (\frac {c x + i}{c}\right ) + 3 i \, b d \log \left (\frac {c x - i}{c}\right ) - 3 \, {\left (4 \, b c^{5} d x^{5} - 5 i \, b c^{4} d x^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{120 \, c^{4}} \]
1/120*(24*I*a*c^5*d*x^5 + 6*(5*a - I*b)*c^4*d*x^4 - 10*b*c^3*d*x^3 + 12*I* b*c^2*d*x^2 + 30*b*c*d*x - 27*I*b*d*log((c*x + I)/c) + 3*I*b*d*log((c*x - I)/c) - 3*(4*b*c^5*d*x^5 - 5*I*b*c^4*d*x^4)*log(-(c*x + I)/(c*x - I)))/c^4
Time = 1.88 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.57 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {i a c d x^{5}}{5} - \frac {b d x^{3}}{12 c} + \frac {i b d x^{2}}{10 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b d \left (\frac {i \log {\left (25 b c d x - 25 i b d \right )}}{40} - \frac {11 i \log {\left (25 b c d x + 25 i b d \right )}}{60}\right )}{c^{4}} + x^{4} \left (\frac {a d}{4} - \frac {i b d}{20}\right ) + \left (\frac {b c d x^{5}}{10} - \frac {i b d x^{4}}{8}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 12 b c^{5} d x^{5} + 15 i b c^{4} d x^{4} - 5 i b d\right ) \log {\left (- i c x + 1 \right )}}{120 c^{4}} \]
I*a*c*d*x**5/5 - b*d*x**3/(12*c) + I*b*d*x**2/(10*c**2) + b*d*x/(4*c**3) + b*d*(I*log(25*b*c*d*x - 25*I*b*d)/40 - 11*I*log(25*b*c*d*x + 25*I*b*d)/60 )/c**4 + x**4*(a*d/4 - I*b*d/20) + (b*c*d*x**5/10 - I*b*d*x**4/8)*log(I*c* x + 1) + (-12*b*c**5*d*x**5 + 15*I*b*c**4*d*x**4 - 5*I*b*d)*log(-I*c*x + 1 )/(120*c**4)
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{5} i \, a c d x^{5} + \frac {1}{4} \, a d x^{4} + \frac {1}{20} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c d + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d \]
1/5*I*a*c*d*x^5 + 1/4*a*d*x^4 + 1/20*I*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*c*d + 1/12*(3*x^4*arctan(c*x) - c* ((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d
\[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )} x^{3} \,d x } \]
Time = 0.78 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=-\frac {\frac {d\,\left (15\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,6{}\mathrm {i}\right )}{60}-\frac {b\,c\,d\,x}{4}+\frac {b\,c^3\,d\,x^3}{12}-\frac {b\,c^2\,d\,x^2\,1{}\mathrm {i}}{10}}{c^4}+\frac {d\,\left (15\,a\,x^4+15\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,3{}\mathrm {i}\right )}{60}+\frac {c\,d\,\left (a\,x^5\,12{}\mathrm {i}+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,12{}\mathrm {i}\right )}{60} \]