Integrand size = 23, antiderivative size = 171 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d^2}{20 x^4}-\frac {i b c^2 d^2}{6 x^3}+\frac {4 b c^3 d^2}{15 x^2}+\frac {i b c^4 d^2}{2 x}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {i c d^2 (a+b \arctan (c x))}{2 x^4}+\frac {c^2 d^2 (a+b \arctan (c x))}{3 x^3}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {1}{60} b c^5 d^2 \log (i-c x)-\frac {31}{60} b c^5 d^2 \log (i+c x) \]
-1/20*b*c*d^2/x^4-1/6*I*b*c^2*d^2/x^3+4/15*b*c^3*d^2/x^2+1/2*I*b*c^4*d^2/x -1/5*d^2*(a+b*arctan(c*x))/x^5-1/2*I*c*d^2*(a+b*arctan(c*x))/x^4+1/3*c^2*d ^2*(a+b*arctan(c*x))/x^3+8/15*b*c^5*d^2*ln(x)-1/60*b*c^5*d^2*ln(I-c*x)-31/ 60*b*c^5*d^2*ln(c*x+I)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {d^2 \left (-12 a-30 i a c x-3 b c x+20 a c^2 x^2+16 b c^3 x^3+2 b \left (-6-15 i c x+10 c^2 x^2\right ) \arctan (c x)-10 i b c^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )+32 b c^5 x^5 \log (x)-16 b c^5 x^5 \log \left (1+c^2 x^2\right )\right )}{60 x^5} \]
(d^2*(-12*a - (30*I)*a*c*x - 3*b*c*x + 20*a*c^2*x^2 + 16*b*c^3*x^3 + 2*b*( -6 - (15*I)*c*x + 10*c^2*x^2)*ArcTan[c*x] - (10*I)*b*c^2*x^2*Hypergeometri c2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 32*b*c^5*x^5*Log[x] - 16*b*c^5*x^5*Log[1 + c^2*x^2]))/(60*x^5)
Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.83, 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.
| \(\displaystyle \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 5407 |
| \(\displaystyle -b c \int -\frac {d^2 \left (-10 c^2 x^2+15 i c x+6\right )}{30 x^5 \left (c^2 x^2+1\right )}dx+\frac {c^2 d^2 (a+b \arctan (c x))}{3 x^3}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {i c d^2 (a+b \arctan (c x))}{2 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{30} b c d^2 \int \frac {-10 c^2 x^2+15 i c x+6}{x^5 \left (c^2 x^2+1\right )}dx+\frac {c^2 d^2 (a+b \arctan (c x))}{3 x^3}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {i c d^2 (a+b \arctan (c x))}{2 x^4}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {1}{30} b c d^2 \int \left (-\frac {c^5}{2 (c x-i)}-\frac {31 c^5}{2 (c x+i)}+\frac {16 c^4}{x}-\frac {15 i c^3}{x^2}-\frac {16 c^2}{x^3}+\frac {15 i c}{x^4}+\frac {6}{x^5}\right )dx+\frac {c^2 d^2 (a+b \arctan (c x))}{3 x^3}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {i c d^2 (a+b \arctan (c x))}{2 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c^2 d^2 (a+b \arctan (c x))}{3 x^3}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {i c d^2 (a+b \arctan (c x))}{2 x^4}+\frac {1}{30} b c d^2 \left (16 c^4 \log (x)-\frac {1}{2} c^4 \log (-c x+i)-\frac {31}{2} c^4 \log (c x+i)+\frac {15 i c^3}{x}+\frac {8 c^2}{x^2}-\frac {5 i c}{x^3}-\frac {3}{2 x^4}\right )\) |
-1/5*(d^2*(a + b*ArcTan[c*x]))/x^5 - ((I/2)*c*d^2*(a + b*ArcTan[c*x]))/x^4 + (c^2*d^2*(a + b*ArcTan[c*x]))/(3*x^3) + (b*c*d^2*(-3/(2*x^4) - ((5*I)*c )/x^3 + (8*c^2)/x^2 + ((15*I)*c^3)/x + 16*c^4*Log[x] - (c^4*Log[I - c*x])/ 2 - (31*c^4*Log[I + c*x])/2))/30
3.1.19.3.1 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 = 0.96 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.78
| method | result | size |
| parts | \(a \,d^{2} \left (-\frac {i c}{2 x^{4}}-\frac {1}{5 x^{5}}+\frac {c^{2}}{3 x^{3}}\right )+b \,d^{2} c^{5} \left (-\frac {\arctan \left (c x \right )}{5 c^{5} x^{5}}-\frac {i \arctan \left (c x \right )}{2 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{6 c^{3} x^{3}}+\frac {i}{2 c x}-\frac {1}{20 c^{4} x^{4}}+\frac {4}{15 c^{2} x^{2}}+\frac {8 \ln \left (c x \right )}{15}-\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}+\frac {i \arctan \left (c x \right )}{2}\right )\) | \(133\) |
| derivativedivides | \(c^{5} \left (a \,d^{2} \left (-\frac {1}{5 c^{5} x^{5}}-\frac {i}{2 c^{4} x^{4}}+\frac {1}{3 c^{3} x^{3}}\right )+b \,d^{2} \left (-\frac {\arctan \left (c x \right )}{5 c^{5} x^{5}}-\frac {i \arctan \left (c x \right )}{2 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{6 c^{3} x^{3}}+\frac {i}{2 c x}-\frac {1}{20 c^{4} x^{4}}+\frac {4}{15 c^{2} x^{2}}+\frac {8 \ln \left (c x \right )}{15}-\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}+\frac {i \arctan \left (c x \right )}{2}\right )\right )\) | \(139\) |
| default | \(c^{5} \left (a \,d^{2} \left (-\frac {1}{5 c^{5} x^{5}}-\frac {i}{2 c^{4} x^{4}}+\frac {1}{3 c^{3} x^{3}}\right )+b \,d^{2} \left (-\frac {\arctan \left (c x \right )}{5 c^{5} x^{5}}-\frac {i \arctan \left (c x \right )}{2 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{6 c^{3} x^{3}}+\frac {i}{2 c x}-\frac {1}{20 c^{4} x^{4}}+\frac {4}{15 c^{2} x^{2}}+\frac {8 \ln \left (c x \right )}{15}-\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}+\frac {i \arctan \left (c x \right )}{2}\right )\right )\) | \(139\) |
| risch | \(-\frac {i b \,d^{2} \left (10 c^{2} x^{2}-15 i c x -6\right ) \ln \left (i c x +1\right )}{60 x^{5}}-\frac {d^{2} \left (31 b \,c^{5} \ln \left (-c x -i\right ) x^{5}+b \,c^{5} \ln \left (c x -i\right ) x^{5}-32 b \,c^{5} \ln \left (-x \right ) x^{5}-30 i b \,c^{4} x^{4}-10 i b \,x^{2} \ln \left (-i c x +1\right ) c^{2}-16 b \,c^{3} x^{3}+10 i b \,c^{2} x^{2}-20 c^{2} x^{2} a +30 i x a c -15 b c x \ln \left (-i c x +1\right )+6 i b \ln \left (-i c x +1\right )+3 x b c +12 a \right )}{60 x^{5}}\) | \(184\) |
| parallelrisch | \(\frac {30 i c^{5} b \,d^{2} \arctan \left (c x \right ) x^{5}-16 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{5} d^{2}+32 \ln \left (x \right ) x^{5} b \,c^{5} d^{2}-16 b \,c^{5} d^{2} x^{5}+30 i x^{4} b \,c^{4} d^{2}+16 b \,c^{3} d^{2} x^{3}-10 i x^{2} b \,c^{2} d^{2}+20 x^{2} \arctan \left (c x \right ) b \,c^{2} d^{2}-30 i x \arctan \left (c x \right ) b c \,d^{2}+20 x^{2} d^{2} c^{2} a -30 i a c \,d^{2} x -3 b c \,d^{2} x -12 b \arctan \left (c x \right ) d^{2}-12 a \,d^{2}}{60 x^{5}}\) | \(184\) |
a*d^2*(-1/2*I*c/x^4-1/5/x^5+1/3*c^2/x^3)+b*d^2*c^5*(-1/5/c^5/x^5*arctan(c* x)-1/2*I*arctan(c*x)/c^4/x^4+1/3*arctan(c*x)/c^3/x^3-1/6*I/c^3/x^3+1/2*I/c /x-1/20/c^4/x^4+4/15/c^2/x^2+8/15*ln(c*x)-4/15*ln(c^2*x^2+1)+1/2*I*arctan( c*x))
Time = 0.26 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.98 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {32 \, b c^{5} d^{2} x^{5} \log \left (x\right ) - 31 \, b c^{5} d^{2} x^{5} \log \left (\frac {c x + i}{c}\right ) - b c^{5} d^{2} x^{5} \log \left (\frac {c x - i}{c}\right ) + 30 i \, b c^{4} d^{2} x^{4} + 16 \, b c^{3} d^{2} x^{3} + 10 \, {\left (2 \, a - i \, b\right )} c^{2} d^{2} x^{2} - 3 \, {\left (10 i \, a + b\right )} c d^{2} x - 12 \, a d^{2} + {\left (10 i \, b c^{2} d^{2} x^{2} + 15 \, b c d^{2} x - 6 i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{60 \, x^{5}} \]
1/60*(32*b*c^5*d^2*x^5*log(x) - 31*b*c^5*d^2*x^5*log((c*x + I)/c) - b*c^5* d^2*x^5*log((c*x - I)/c) + 30*I*b*c^4*d^2*x^4 + 16*b*c^3*d^2*x^3 + 10*(2*a - I*b)*c^2*d^2*x^2 - 3*(10*I*a + b)*c*d^2*x - 12*a*d^2 + (10*I*b*c^2*d^2* x^2 + 15*b*c*d^2*x - 6*I*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^5
Time = 13.61 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.68 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {8 b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x \right )}}{15} - \frac {b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x - 10395 i b^{2} c^{10} d^{4} \right )}}{60} - \frac {31 b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x + 10395 i b^{2} c^{10} d^{4} \right )}}{60} + \frac {\left (- 10 i b c^{2} d^{2} x^{2} - 15 b c d^{2} x + 6 i b d^{2}\right ) \log {\left (i c x + 1 \right )}}{60 x^{5}} + \frac {\left (10 i b c^{2} d^{2} x^{2} + 15 b c d^{2} x - 6 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{60 x^{5}} - \frac {12 a d^{2} - 30 i b c^{4} d^{2} x^{4} - 16 b c^{3} d^{2} x^{3} + x^{2} \left (- 20 a c^{2} d^{2} + 10 i b c^{2} d^{2}\right ) + x \left (30 i a c d^{2} + 3 b c d^{2}\right )}{60 x^{5}} \]
8*b*c**5*d**2*log(10395*b**2*c**11*d**4*x)/15 - b*c**5*d**2*log(10395*b**2 *c**11*d**4*x - 10395*I*b**2*c**10*d**4)/60 - 31*b*c**5*d**2*log(10395*b** 2*c**11*d**4*x + 10395*I*b**2*c**10*d**4)/60 + (-10*I*b*c**2*d**2*x**2 - 1 5*b*c*d**2*x + 6*I*b*d**2)*log(I*c*x + 1)/(60*x**5) + (10*I*b*c**2*d**2*x* *2 + 15*b*c*d**2*x - 6*I*b*d**2)*log(-I*c*x + 1)/(60*x**5) - (12*a*d**2 - 30*I*b*c**4*d**2*x**4 - 16*b*c**3*d**2*x**3 + x**2*(-20*a*c**2*d**2 + 10*I *b*c**2*d**2) + x*(30*I*a*c*d**2 + 3*b*c*d**2))/(60*x**5)
Time = 0.26 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.07 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c^{2} d^{2} + \frac {1}{6} i \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b c d^{2} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac {a c^{2} d^{2}}{3 \, x^{3}} - \frac {i \, a c d^{2}}{2 \, x^{4}} - \frac {a d^{2}}{5 \, x^{5}} \]
-1/6*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3) *b*c^2*d^2 + 1/6*I*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan (c*x)/x^4)*b*c*d^2 - 1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c ^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d^2 + 1/3*a*c^2*d^2/x^3 - 1/2*I* a*c*d^2/x^4 - 1/5*a*d^2/x^5
\[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{6}} \,d x } \]
Time = 0.98 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.43 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {8\,b\,c^5\,d^2\,\ln \left (x\right )}{15}-\frac {4\,b\,c^5\,d^2\,\ln \left (c^2\,x^2+1\right )}{15}-\frac {\frac {a\,d^2}{5}+\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {4\,b\,c^5\,d^2\,x^5}{15}-\frac {b\,c^6\,d^2\,x^6\,1{}\mathrm {i}}{2}-\frac {c^4\,d^2\,x^4\,\left (a+b\,1{}\mathrm {i}\right )}{3}+\frac {c\,d^2\,x\,\left (b+a\,10{}\mathrm {i}\right )}{20}-\frac {c^2\,d^2\,x^2\,\left (4\,a-b\,5{}\mathrm {i}\right )}{30}+\frac {c^3\,d^2\,x^3\,\left (-13\,b+a\,30{}\mathrm {i}\right )}{60}-\frac {2\,b\,c^2\,d^2\,x^2\,\mathrm {atan}\left (c\,x\right )}{15}+\frac {b\,c^3\,d^2\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}-\frac {b\,c^4\,d^2\,x^4\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,c\,d^2\,x\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}}{c^2\,x^7+x^5}+\frac {b\,c^8\,d^2\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {c^2}}\right )\,1{}\mathrm {i}}{2\,{\left (c^2\right )}^{3/2}} \]
(8*b*c^5*d^2*log(x))/15 - (4*b*c^5*d^2*log(c^2*x^2 + 1))/15 - ((a*d^2)/5 + (b*d^2*atan(c*x))/5 - (4*b*c^5*d^2*x^5)/15 - (b*c^6*d^2*x^6*1i)/2 - (c^4* d^2*x^4*(a + b*1i))/3 + (c*d^2*x*(a*10i + b))/20 - (c^2*d^2*x^2*(4*a - b*5 i))/30 + (c^3*d^2*x^3*(a*30i - 13*b))/60 - (2*b*c^2*d^2*x^2*atan(c*x))/15 + (b*c^3*d^2*x^3*atan(c*x)*1i)/2 - (b*c^4*d^2*x^4*atan(c*x))/3 + (b*c*d^2* x*atan(c*x)*1i)/2)/(x^5 + c^2*x^7) + (b*c^8*d^2*atan((c^2*x)/(c^2)^(1/2))* 1i)/(2*(c^2)^(3/2))