Integrand size = 23, antiderivative size = 161 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=-\frac {b c d^2}{12 x^3}-\frac {i b c^2 d^2}{3 x^2}+\frac {3 b c^3 d^2}{4 x}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {2 i c d^2 (a+b \arctan (c x))}{3 x^3}+\frac {c^2 d^2 (a+b \arctan (c x))}{2 x^2}-\frac {2}{3} i b c^4 d^2 \log (x)-\frac {1}{24} i b c^4 d^2 \log (i-c x)+\frac {17}{24} i b c^4 d^2 \log (i+c x) \] Output:
-1/12*b*c*d^2/x^3-1/3*I*b*c^2*d^2/x^2+3/4*b*c^3*d^2/x-1/4*d^2*(a+b*arctan( c*x))/x^4-2/3*I*c*d^2*(a+b*arctan(c*x))/x^3+1/2*c^2*d^2*(a+b*arctan(c*x))/ x^2-2/3*I*b*c^4*d^2*ln(x)-1/24*I*b*c^4*d^2*ln(I-c*x)+17/24*I*b*c^4*d^2*ln( I+c*x)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=\frac {d^2 \left (-3 a-8 i a c x+6 a c^2 x^2-4 i b c^2 x^2-3 b \arctan (c x)-8 i b c x \arctan (c x)+6 b c^2 x^2 \arctan (c x)-b c x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )+6 b c^3 x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )-8 i b c^4 x^4 \log (x)+4 i b c^4 x^4 \log \left (1+c^2 x^2\right )\right )}{12 x^4} \] Input:
Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^5,x]
Output:
(d^2*(-3*a - (8*I)*a*c*x + 6*a*c^2*x^2 - (4*I)*b*c^2*x^2 - 3*b*ArcTan[c*x] - (8*I)*b*c*x*ArcTan[c*x] + 6*b*c^2*x^2*ArcTan[c*x] - b*c*x*Hypergeometri c2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 6*b*c^3*x^3*Hypergeometric2F1[-1/2, 1, 1 /2, -(c^2*x^2)] - (8*I)*b*c^4*x^4*Log[x] + (4*I)*b*c^4*x^4*Log[1 + c^2*x^2 ]))/(12*x^4)
Time = 0.42 (sec) , antiderivative size = 136, 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.
| \(\displaystyle \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx\) |
| \(\Big \downarrow \) 5407 |
| \(\displaystyle -b c \int -\frac {d^2 \left (-6 c^2 x^2+8 i c x+3\right )}{12 x^4 \left (c^2 x^2+1\right )}dx+\frac {c^2 d^2 (a+b \arctan (c x))}{2 x^2}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {2 i c d^2 (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{12} b c d^2 \int \frac {-6 c^2 x^2+8 i c x+3}{x^4 \left (c^2 x^2+1\right )}dx+\frac {c^2 d^2 (a+b \arctan (c x))}{2 x^2}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {2 i c d^2 (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {1}{12} b c d^2 \int \left (-\frac {i c^4}{2 (c x-i)}+\frac {17 i c^4}{2 (c x+i)}-\frac {8 i c^3}{x}-\frac {9 c^2}{x^2}+\frac {8 i c}{x^3}+\frac {3}{x^4}\right )dx+\frac {c^2 d^2 (a+b \arctan (c x))}{2 x^2}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {2 i c d^2 (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c^2 d^2 (a+b \arctan (c x))}{2 x^2}-\frac {d^2 (a+b \arctan (c x))}{4 x^4}-\frac {2 i c d^2 (a+b \arctan (c x))}{3 x^3}+\frac {1}{12} b c d^2 \left (-8 i c^3 \log (x)-\frac {1}{2} i c^3 \log (-c x+i)+\frac {17}{2} i c^3 \log (c x+i)+\frac {9 c^2}{x}-\frac {4 i c}{x^2}-\frac {1}{x^3}\right )\) |
Input:
Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^5,x]
Output:
-1/4*(d^2*(a + b*ArcTan[c*x]))/x^4 - (((2*I)/3)*c*d^2*(a + b*ArcTan[c*x])) /x^3 + (c^2*d^2*(a + b*ArcTan[c*x]))/(2*x^2) + (b*c*d^2*(-x^(-3) - ((4*I)* c)/x^2 + (9*c^2)/x - (8*I)*c^3*Log[x] - (I/2)*c^3*Log[I - c*x] + ((17*I)/2 )*c^3*Log[I + c*x]))/12
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 = 0.47 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78
| method | result | size |
| parts | \(d^{2} a \left (\frac {c^{2}}{2 x^{2}}-\frac {2 i c}{3 x^{3}}-\frac {1}{4 x^{4}}\right )+d^{2} b \,c^{4} \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {2 i \arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{3 c^{2} x^{2}}-\frac {2 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {3}{4 c x}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {3 \arctan \left (c x \right )}{4}\right )\) | \(125\) |
| derivativedivides | \(c^{4} \left (d^{2} a \left (-\frac {1}{4 c^{4} x^{4}}+\frac {1}{2 c^{2} x^{2}}-\frac {2 i}{3 c^{3} x^{3}}\right )+d^{2} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {2 i \arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{3 c^{2} x^{2}}-\frac {2 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {3}{4 c x}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {3 \arctan \left (c x \right )}{4}\right )\right )\) | \(131\) |
| default | \(c^{4} \left (d^{2} a \left (-\frac {1}{4 c^{4} x^{4}}+\frac {1}{2 c^{2} x^{2}}-\frac {2 i}{3 c^{3} x^{3}}\right )+d^{2} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}+\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {2 i \arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i}{3 c^{2} x^{2}}-\frac {2 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {3}{4 c x}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {3 \arctan \left (c x \right )}{4}\right )\right )\) | \(131\) |
| risch | \(-\frac {i d^{2} b \left (6 c^{2} x^{2}-8 i c x -3\right ) \ln \left (i c x +1\right )}{24 x^{4}}+\frac {i d^{2} \left (17 b \,c^{4} \ln \left (-99 c x -99 i\right ) x^{4}-16 b \,c^{4} \ln \left (-165 c x \right ) x^{4}-b \,c^{4} \ln \left (45 c x -45 i\right ) x^{4}+6 x^{2} b \ln \left (-i c x +1\right ) c^{2}-18 i b \,c^{3} x^{3}-8 b \,c^{2} x^{2}-12 i a \,c^{2} x^{2}-16 a c x -8 i b c x \ln \left (-i c x +1\right )-3 b \ln \left (-i c x +1\right )+2 i b c x +6 i a \right )}{24 x^{4}}\) | \(179\) |
| parallelrisch | \(-\frac {8 i c^{4} b \,d^{2} \ln \left (x \right ) x^{4}-4 i c^{4} b \,d^{2} \ln \left (c^{2} x^{2}+1\right ) x^{4}-4 i x^{4} b \,c^{4} d^{2}-9 d^{2} b \arctan \left (c x \right ) x^{4} c^{4}+6 a \,c^{4} d^{2} x^{4}-9 b \,c^{3} d^{2} x^{3}+4 i x^{2} b \,c^{2} d^{2}-6 d^{2} b \arctan \left (c x \right ) x^{2} c^{2}+8 i x \arctan \left (c x \right ) b c \,d^{2}-6 a \,c^{2} d^{2} x^{2}+8 i a c \,d^{2} x +b c \,d^{2} x +3 b \,d^{2} \arctan \left (c x \right )+3 d^{2} a}{12 x^{4}}\) | \(184\) |
Input:
int((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^5,x,method=_RETURNVERBOSE)
Output:
d^2*a*(1/2*c^2/x^2-2/3*I*c/x^3-1/4/x^4)+d^2*b*c^4*(-1/4/c^4/x^4*arctan(c*x )+1/2/c^2/x^2*arctan(c*x)-2/3*I*arctan(c*x)/c^3/x^3-1/3*I/c^2/x^2-2/3*I*ln (c*x)-1/12/c^3/x^3+3/4/c/x+1/3*I*ln(c^2*x^2+1)+3/4*arctan(c*x))
Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=\frac {-16 i \, b c^{4} d^{2} x^{4} \log \left (x\right ) + 17 i \, b c^{4} d^{2} x^{4} \log \left (\frac {c x + i}{c}\right ) - i \, b c^{4} d^{2} x^{4} \log \left (\frac {c x - i}{c}\right ) + 18 \, b c^{3} d^{2} x^{3} + 4 \, {\left (3 \, a - 2 i \, b\right )} c^{2} d^{2} x^{2} - 2 \, {\left (8 i \, a + b\right )} c d^{2} x - 6 \, a d^{2} + {\left (6 i \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x - 3 i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{24 \, x^{4}} \] Input:
integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^5,x, algorithm="fricas")
Output:
1/24*(-16*I*b*c^4*d^2*x^4*log(x) + 17*I*b*c^4*d^2*x^4*log((c*x + I)/c) - I *b*c^4*d^2*x^4*log((c*x - I)/c) + 18*b*c^3*d^2*x^3 + 4*(3*a - 2*I*b)*c^2*d ^2*x^2 - 2*(8*I*a + b)*c*d^2*x - 6*a*d^2 + (6*I*b*c^2*d^2*x^2 + 8*b*c*d^2* x - 3*I*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^4
Time = 9.38 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.71 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=- \frac {2 i b c^{4} d^{2} \log {\left (1485 b^{2} c^{9} d^{4} x \right )}}{3} - \frac {i b c^{4} d^{2} \log {\left (1485 b^{2} c^{9} d^{4} x - 1485 i b^{2} c^{8} d^{4} \right )}}{24} + \frac {17 i b c^{4} d^{2} \log {\left (1485 b^{2} c^{9} d^{4} x + 1485 i b^{2} c^{8} d^{4} \right )}}{24} + \frac {\left (- 6 i b c^{2} d^{2} x^{2} - 8 b c d^{2} x + 3 i b d^{2}\right ) \log {\left (i c x + 1 \right )}}{24 x^{4}} + \frac {\left (6 i b c^{2} d^{2} x^{2} + 8 b c d^{2} x - 3 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{24 x^{4}} - \frac {3 a d^{2} - 9 b c^{3} d^{2} x^{3} + x^{2} \left (- 6 a c^{2} d^{2} + 4 i b c^{2} d^{2}\right ) + x \left (8 i a c d^{2} + b c d^{2}\right )}{12 x^{4}} \] Input:
integrate((d+I*c*d*x)**2*(a+b*atan(c*x))/x**5,x)
Output:
-2*I*b*c**4*d**2*log(1485*b**2*c**9*d**4*x)/3 - I*b*c**4*d**2*log(1485*b** 2*c**9*d**4*x - 1485*I*b**2*c**8*d**4)/24 + 17*I*b*c**4*d**2*log(1485*b**2 *c**9*d**4*x + 1485*I*b**2*c**8*d**4)/24 + (-6*I*b*c**2*d**2*x**2 - 8*b*c* d**2*x + 3*I*b*d**2)*log(I*c*x + 1)/(24*x**4) + (6*I*b*c**2*d**2*x**2 + 8* b*c*d**2*x - 3*I*b*d**2)*log(-I*c*x + 1)/(24*x**4) - (3*a*d**2 - 9*b*c**3* d**2*x**3 + x**2*(-6*a*c**2*d**2 + 4*I*b*c**2*d**2) + x*(8*I*a*c*d**2 + b* c*d**2))/(12*x**4)
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=\frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c^{2} d^{2} + \frac {1}{3} i \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c d^{2} + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d^{2} + \frac {a c^{2} d^{2}}{2 \, x^{2}} - \frac {2 i \, a c d^{2}}{3 \, x^{3}} - \frac {a d^{2}}{4 \, x^{4}} \] Input:
integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^5,x, algorithm="maxima")
Output:
1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c^2*d^2 + 1/3*I*((c^2*lo g(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*c*d^2 + 1/ 12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^2 + 1/2*a*c^2*d^2/x^2 - 2/3*I*a*c*d^2/x^3 - 1/4*a*d^2/x^4
Time = 0.14 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=-\frac {-17 i \, b c^{4} d^{2} x^{4} \log \left (i \, c x - 1\right ) + i \, b c^{4} d^{2} x^{4} \log \left (-i \, c x - 1\right ) + 16 i \, b c^{4} d^{2} x^{4} \log \left (x\right ) - 18 \, b c^{3} d^{2} x^{3} - 12 \, b c^{2} d^{2} x^{2} \arctan \left (c x\right ) - 12 \, a c^{2} d^{2} x^{2} + 8 i \, b c^{2} d^{2} x^{2} + 16 i \, b c d^{2} x \arctan \left (c x\right ) + 16 i \, a c d^{2} x + 2 \, b c d^{2} x + 6 \, b d^{2} \arctan \left (c x\right ) + 6 \, a d^{2}}{24 \, x^{4}} \] Input:
integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^5,x, algorithm="giac")
Output:
-1/24*(-17*I*b*c^4*d^2*x^4*log(I*c*x - 1) + I*b*c^4*d^2*x^4*log(-I*c*x - 1 ) + 16*I*b*c^4*d^2*x^4*log(x) - 18*b*c^3*d^2*x^3 - 12*b*c^2*d^2*x^2*arctan (c*x) - 12*a*c^2*d^2*x^2 + 8*I*b*c^2*d^2*x^2 + 16*I*b*c*d^2*x*arctan(c*x) + 16*I*a*c*d^2*x + 2*b*c*d^2*x + 6*b*d^2*arctan(c*x) + 6*a*d^2)/x^4
Time = 1.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.88 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=\frac {d^2\,\left (9\,b\,c^3\,\mathrm {atan}\left (x\,\sqrt {c^2}\right )\,\sqrt {c^2}+b\,c^4\,\ln \left (c^2\,x^2+1\right )\,4{}\mathrm {i}-b\,c^4\,\ln \left (x\right )\,8{}\mathrm {i}\right )}{12}-\frac {\frac {d^2\,\left (3\,a+3\,b\,\mathrm {atan}\left (c\,x\right )\right )}{12}+\frac {d^2\,x\,\left (a\,c\,8{}\mathrm {i}+b\,c+b\,c\,\mathrm {atan}\left (c\,x\right )\,8{}\mathrm {i}\right )}{12}-\frac {d^2\,x^2\,\left (6\,a\,c^2+6\,b\,c^2\,\mathrm {atan}\left (c\,x\right )-b\,c^2\,4{}\mathrm {i}\right )}{12}-\frac {3\,b\,c^3\,d^2\,x^3}{4}}{x^4} \] Input:
int(((a + b*atan(c*x))*(d + c*d*x*1i)^2)/x^5,x)
Output:
(d^2*(b*c^4*log(c^2*x^2 + 1)*4i - b*c^4*log(x)*8i + 9*b*c^3*atan(x*(c^2)^( 1/2))*(c^2)^(1/2)))/12 - ((d^2*(3*a + 3*b*atan(c*x)))/12 + (d^2*x*(a*c*8i + b*c + b*c*atan(c*x)*8i))/12 - (d^2*x^2*(6*a*c^2 - b*c^2*4i + 6*b*c^2*ata n(c*x)))/12 - (3*b*c^3*d^2*x^3)/4)/x^4
Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))}{x^5} \, dx=\frac {d^{2} \left (9 \mathit {atan} \left (c x \right ) b \,c^{4} x^{4}+6 \mathit {atan} \left (c x \right ) b \,c^{2} x^{2}-8 \mathit {atan} \left (c x \right ) b c i x -3 \mathit {atan} \left (c x \right ) b +4 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{4} i \,x^{4}-8 \,\mathrm {log}\left (x \right ) b \,c^{4} i \,x^{4}+6 a \,c^{2} x^{2}-8 a c i x -3 a +9 b \,c^{3} x^{3}-4 b \,c^{2} i \,x^{2}-b c x \right )}{12 x^{4}} \] Input:
int((d+I*c*d*x)^2*(a+b*atan(c*x))/x^5,x)
Output:
(d**2*(9*atan(c*x)*b*c**4*x**4 + 6*atan(c*x)*b*c**2*x**2 - 8*atan(c*x)*b*c *i*x - 3*atan(c*x)*b + 4*log(c**2*x**2 + 1)*b*c**4*i*x**4 - 8*log(x)*b*c** 4*i*x**4 + 6*a*c**2*x**2 - 8*a*c*i*x - 3*a + 9*b*c**3*x**3 - 4*b*c**2*i*x* *2 - b*c*x))/(12*x**4)