Integrand size = 23, antiderivative size = 191 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {11 i b d^3 x}{12 c^2}-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} i b c^2 d^3 x^5-\frac {11 i b d^3 \arctan (c x)}{12 c^3}+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))+\frac {7 b d^3 \log \left (1+c^2 x^2\right )}{15 c^3} \]
11/12*I*b*d^3*x/c^2-7/15*b*d^3*x^2/c-11/36*I*b*d^3*x^3+3/20*b*c*d^3*x^4+1/ 30*I*b*c^2*d^3*x^5-11/12*I*b*d^3*arctan(c*x)/c^3+1/3*d^3*x^3*(a+b*arctan(c *x))+3/4*I*c*d^3*x^4*(a+b*arctan(c*x))-3/5*c^2*d^3*x^5*(a+b*arctan(c*x))-1 /6*I*c^3*d^3*x^6*(a+b*arctan(c*x))+7/15*b*d^3*ln(c^2*x^2+1)/c^3
Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.76 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {d^3 \left (3 a c^3 x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )+b c x \left (165 i-84 c x-55 i c^2 x^2+27 c^3 x^3+6 i c^4 x^4\right )+3 b \left (-55 i+20 c^3 x^3+45 i c^4 x^4-36 c^5 x^5-10 i c^6 x^6\right ) \arctan (c x)+84 b \log \left (1+c^2 x^2\right )\right )}{180 c^3} \]
(d^3*(3*a*c^3*x^3*(20 + (45*I)*c*x - 36*c^2*x^2 - (10*I)*c^3*x^3) + b*c*x* (165*I - 84*c*x - (55*I)*c^2*x^2 + 27*c^3*x^3 + (6*I)*c^4*x^4) + 3*b*(-55* I + 20*c^3*x^3 + (45*I)*c^4*x^4 - 36*c^5*x^5 - (10*I)*c^6*x^6)*ArcTan[c*x] + 84*b*Log[1 + c^2*x^2]))/(180*c^3)
Time = 0.42 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5407, 27, 2333, 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.
| \(\displaystyle \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx\) |
| \(\Big \downarrow \) 5407 |
| \(\displaystyle -b c \int \frac {d^3 x^3 \left (-10 i c^3 x^3-36 c^2 x^2+45 i c x+20\right )}{60 \left (c^2 x^2+1\right )}dx-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{60} b c d^3 \int \frac {x^3 \left (-10 i c^3 x^3-36 c^2 x^2+45 i c x+20\right )}{c^2 x^2+1}dx-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle -\frac {1}{60} b c d^3 \int \left (-10 i c x^4-36 x^3+\frac {55 i x^2}{c}+\frac {56 x}{c^2}+\frac {55 i-56 c x}{c^3 \left (c^2 x^2+1\right )}-\frac {55 i}{c^3}\right )dx-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))-\frac {1}{60} b c d^3 \left (\frac {55 i \arctan (c x)}{c^4}-\frac {55 i x}{c^3}+\frac {28 x^2}{c^2}-\frac {28 \log \left (c^2 x^2+1\right )}{c^4}-2 i c x^5+\frac {55 i x^3}{3 c}-9 x^4\right )\) |
(d^3*x^3*(a + b*ArcTan[c*x]))/3 + ((3*I)/4)*c*d^3*x^4*(a + b*ArcTan[c*x]) - (3*c^2*d^3*x^5*(a + b*ArcTan[c*x]))/5 - (I/6)*c^3*d^3*x^6*(a + b*ArcTan[ c*x]) - (b*c*d^3*(((-55*I)*x)/c^3 + (28*x^2)/c^2 + (((55*I)/3)*x^3)/c - 9* x^4 - (2*I)*c*x^5 + ((55*I)*ArcTan[c*x])/c^4 - (28*Log[1 + c^2*x^2])/c^4)) /60
3.1.21.3.1 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[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 1.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.81
| method | result | size |
| parts | \(a \,d^{3} \left (-\frac {1}{6} i c^{3} x^{6}-\frac {3}{5} c^{2} x^{5}+\frac {3}{4} i c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(154\) |
| derivativedivides | \(\frac {a \,d^{3} \left (-\frac {1}{6} i c^{6} x^{6}-\frac {3}{5} c^{5} x^{5}+\frac {3}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(160\) |
| default | \(\frac {a \,d^{3} \left (-\frac {1}{6} i c^{6} x^{6}-\frac {3}{5} c^{5} x^{5}+\frac {3}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(160\) |
| parallelrisch | \(-\frac {-165 i b \,d^{3} x c +55 i x^{3} b \,c^{3} d^{3}+165 i b \,d^{3} \arctan \left (c x \right )+108 c^{5} b \,d^{3} \arctan \left (c x \right ) x^{5}-6 i x^{5} b \,c^{5} d^{3}+108 a \,c^{5} d^{3} x^{5}-135 i x^{4} a \,c^{4} d^{3}-27 b \,c^{4} d^{3} x^{4}+30 i x^{6} a \,c^{6} d^{3}-60 x^{3} \arctan \left (c x \right ) b \,d^{3} c^{3}-60 a \,c^{3} d^{3} x^{3}+84 b \,c^{2} d^{3} x^{2}-135 i x^{4} \arctan \left (c x \right ) b \,c^{4} d^{3}+30 i x^{6} \arctan \left (c x \right ) b \,c^{6} d^{3}-84 b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{180 c^{3}}\) | \(209\) |
| risch | \(-\frac {d^{3} b \left (10 c^{3} x^{6}-36 i c^{2} x^{5}-45 c \,x^{4}+20 i x^{3}\right ) \ln \left (i c x +1\right )}{120}+\frac {11 i b \,d^{3} x}{12 c^{2}}+\frac {d^{3} c^{3} x^{6} b \ln \left (-i c x +1\right )}{12}-\frac {11 i b \,d^{3} \arctan \left (c x \right )}{12 c^{3}}-\frac {11 i b \,d^{3} x^{3}}{36}-\frac {3 a \,c^{2} d^{3} x^{5}}{5}+\frac {i b \,c^{2} d^{3} x^{5}}{30}-\frac {3 d^{3} c \,x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {i a \,c^{3} d^{3} x^{6}}{6}+\frac {3 b c \,d^{3} x^{4}}{20}-\frac {3 i d^{3} c^{2} b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {a \,d^{3} x^{3}}{3}-\frac {7 b \,d^{3} x^{2}}{15 c}+\frac {3 i a c \,d^{3} x^{4}}{4}+\frac {i d^{3} b \,x^{3} \ln \left (-i c x +1\right )}{6}+\frac {7 b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{15 c^{3}}\) | \(257\) |
a*d^3*(-1/6*I*c^3*x^6-3/5*c^2*x^5+3/4*I*c*x^4+1/3*x^3)+b*d^3/c^3*(-1/6*I*a rctan(c*x)*c^6*x^6-3/5*c^5*x^5*arctan(c*x)+3/4*I*arctan(c*x)*c^4*x^4+1/3*c ^3*x^3*arctan(c*x)+11/12*I*c*x+1/30*I*c^5*x^5+3/20*c^4*x^4-11/36*I*c^3*x^3 -7/15*c^2*x^2+7/15*ln(c^2*x^2+1)-11/12*I*arctan(c*x))
Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.99 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {-60 i \, a c^{6} d^{3} x^{6} - 12 \, {\left (18 \, a - i \, b\right )} c^{5} d^{3} x^{5} - 54 \, {\left (-5 i \, a - b\right )} c^{4} d^{3} x^{4} + 10 \, {\left (12 \, a - 11 i \, b\right )} c^{3} d^{3} x^{3} - 168 \, b c^{2} d^{3} x^{2} + 330 i \, b c d^{3} x + 333 \, b d^{3} \log \left (\frac {c x + i}{c}\right ) + 3 \, b d^{3} \log \left (\frac {c x - i}{c}\right ) + 3 \, {\left (10 \, b c^{6} d^{3} x^{6} - 36 i \, b c^{5} d^{3} x^{5} - 45 \, b c^{4} d^{3} x^{4} + 20 i \, b c^{3} d^{3} x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{360 \, c^{3}} \]
1/360*(-60*I*a*c^6*d^3*x^6 - 12*(18*a - I*b)*c^5*d^3*x^5 - 54*(-5*I*a - b) *c^4*d^3*x^4 + 10*(12*a - 11*I*b)*c^3*d^3*x^3 - 168*b*c^2*d^3*x^2 + 330*I* b*c*d^3*x + 333*b*d^3*log((c*x + I)/c) + 3*b*d^3*log((c*x - I)/c) + 3*(10* b*c^6*d^3*x^6 - 36*I*b*c^5*d^3*x^5 - 45*b*c^4*d^3*x^4 + 20*I*b*c^3*d^3*x^3 )*log(-(c*x + I)/(c*x - I)))/c^3
Time = 2.48 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.65 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=- \frac {i a c^{3} d^{3} x^{6}}{6} - \frac {7 b d^{3} x^{2}}{15 c} + \frac {11 i b d^{3} x}{12 c^{2}} - \frac {b d^{3} \left (- \frac {\log {\left (310 b c d^{3} x - 310 i b d^{3} \right )}}{120} - \frac {209 \log {\left (310 b c d^{3} x + 310 i b d^{3} \right )}}{280}\right )}{c^{3}} - x^{5} \cdot \left (\frac {3 a c^{2} d^{3}}{5} - \frac {i b c^{2} d^{3}}{30}\right ) - x^{4} \left (- \frac {3 i a c d^{3}}{4} - \frac {3 b c d^{3}}{20}\right ) - x^{3} \left (- \frac {a d^{3}}{3} + \frac {11 i b d^{3}}{36}\right ) + \left (- \frac {b c^{3} d^{3} x^{6}}{12} + \frac {3 i b c^{2} d^{3} x^{5}}{10} + \frac {3 b c d^{3} x^{4}}{8} - \frac {i b d^{3} x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (70 b c^{6} d^{3} x^{6} - 252 i b c^{5} d^{3} x^{5} - 315 b c^{4} d^{3} x^{4} + 140 i b c^{3} d^{3} x^{3} + 150 b d^{3}\right ) \log {\left (- i c x + 1 \right )}}{840 c^{3}} \]
-I*a*c**3*d**3*x**6/6 - 7*b*d**3*x**2/(15*c) + 11*I*b*d**3*x/(12*c**2) - b *d**3*(-log(310*b*c*d**3*x - 310*I*b*d**3)/120 - 209*log(310*b*c*d**3*x + 310*I*b*d**3)/280)/c**3 - x**5*(3*a*c**2*d**3/5 - I*b*c**2*d**3/30) - x**4 *(-3*I*a*c*d**3/4 - 3*b*c*d**3/20) - x**3*(-a*d**3/3 + 11*I*b*d**3/36) + ( -b*c**3*d**3*x**6/12 + 3*I*b*c**2*d**3*x**5/10 + 3*b*c*d**3*x**4/8 - I*b*d **3*x**3/6)*log(I*c*x + 1) + (70*b*c**6*d**3*x**6 - 252*I*b*c**5*d**3*x**5 - 315*b*c**4*d**3*x**4 + 140*I*b*c**3*d**3*x**3 + 150*b*d**3)*log(-I*c*x + 1)/(840*c**3)
Time = 0.31 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.27 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {1}{6} i \, a c^{3} d^{3} x^{6} - \frac {3}{5} \, a c^{2} d^{3} x^{5} + \frac {3}{4} i \, a c d^{3} x^{4} - \frac {1}{90} i \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{3} d^{3} - \frac {3}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{3} + \frac {1}{3} \, a d^{3} x^{3} + \frac {1}{4} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{3} \]
-1/6*I*a*c^3*d^3*x^6 - 3/5*a*c^2*d^3*x^5 + 3/4*I*a*c*d^3*x^4 - 1/90*I*(15* x^6*arctan(c*x) - c*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c ^7))*b*c^3*d^3 - 3/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*lo g(c^2*x^2 + 1)/c^6))*b*c^2*d^3 + 1/3*a*d^3*x^3 + 1/4*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*c*d^3 + 1/6*(2*x^3*arctan (c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d^3
\[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
Time = 0.92 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {\frac {d^3\,\left (-84\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,165{}\mathrm {i}\right )}{180}+\frac {7\,b\,c^2\,d^3\,x^2}{15}-\frac {b\,c\,d^3\,x\,11{}\mathrm {i}}{12}}{c^3}+\frac {d^3\,\left (60\,a\,x^3+60\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,55{}\mathrm {i}\right )}{180}-\frac {c^3\,d^3\,\left (a\,x^6\,30{}\mathrm {i}+b\,x^6\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{180}+\frac {c\,d^3\,\left (a\,x^4\,135{}\mathrm {i}+27\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,135{}\mathrm {i}\right )}{180}-\frac {c^2\,d^3\,\left (108\,a\,x^5+108\,b\,x^5\,\mathrm {atan}\left (c\,x\right )-b\,x^5\,6{}\mathrm {i}\right )}{180} \]
(d^3*(60*a*x^3 - b*x^3*55i + 60*b*x^3*atan(c*x)))/180 - ((d^3*(b*atan(c*x) *165i - 84*b*log(c^2*x^2 + 1)))/180 + (7*b*c^2*d^3*x^2)/15 - (b*c*d^3*x*11 i)/12)/c^3 - (c^3*d^3*(a*x^6*30i + b*x^6*atan(c*x)*30i))/180 + (c*d^3*(a*x ^4*135i + 27*b*x^4 + b*x^4*atan(c*x)*135i))/180 - (c^2*d^3*(108*a*x^5 - b* x^5*6i + 108*b*x^5*atan(c*x)))/180