Integrand size = 21, antiderivative size = 106 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d}{6 x^2}-\frac {i b c^2 d}{2 x}-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {i c d (a+b \arctan (c x))}{2 x^2}-\frac {1}{3} b c^3 d \log (x)-\frac {1}{12} b c^3 d \log (i-c x)+\frac {5}{12} b c^3 d \log (i+c x) \]
-1/6*b*c*d/x^2-1/2*I*b*c^2*d/x-1/3*d*(a+b*arctan(c*x))/x^3-1/2*I*c*d*(a+b* arctan(c*x))/x^2-1/3*b*c^3*d*ln(x)-1/12*b*c^3*d*ln(I-c*x)+5/12*b*c^3*d*ln( c*x+I)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=-\frac {d \left (2 a+3 i a c x+b c x+b (2+3 i c x) \arctan (c x)+3 i b c^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )+2 b c^3 x^3 \log (x)-b c^3 x^3 \log \left (1+c^2 x^2\right )\right )}{6 x^3} \]
-1/6*(d*(2*a + (3*I)*a*c*x + b*c*x + b*(2 + (3*I)*c*x)*ArcTan[c*x] + (3*I) *b*c^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 2*b*c^3*x^3*Log[x ] - b*c^3*x^3*Log[1 + c^2*x^2]))/x^3
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 5407 |
\(\displaystyle -b c \int -\frac {d (3 i c x+2)}{6 x^3 \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {i c d (a+b \arctan (c x))}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} b c d \int \frac {3 i c x+2}{x^3 \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {i c d (a+b \arctan (c x))}{2 x^2}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {1}{6} b c d \int \left (-\frac {c^3}{2 (c x-i)}+\frac {5 c^3}{2 (c x+i)}-\frac {2 c^2}{x}+\frac {3 i c}{x^2}+\frac {2}{x^3}\right )dx-\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {i c d (a+b \arctan (c x))}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d (a+b \arctan (c x))}{3 x^3}-\frac {i c d (a+b \arctan (c x))}{2 x^2}+\frac {1}{6} b c d \left (-2 c^2 \log (x)-\frac {1}{2} c^2 \log (-c x+i)+\frac {5}{2} c^2 \log (c x+i)-\frac {3 i c}{x}-\frac {1}{x^2}\right )\) |
-1/3*(d*(a + b*ArcTan[c*x]))/x^3 - ((I/2)*c*d*(a + b*ArcTan[c*x]))/x^2 + ( b*c*d*(-x^(-2) - ((3*I)*c)/x - 2*c^2*Log[x] - (c^2*Log[I - c*x])/2 + (5*c^ 2*Log[I + c*x])/2))/6
3.1.8.3.1 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 = 0.48 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87
method | result | size |
parts | \(a d \left (-\frac {i c}{2 x^{2}}-\frac {1}{3 x^{3}}\right )+b d \,c^{3} \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{2}\right )\) | \(92\) |
derivativedivides | \(c^{3} \left (a d \left (-\frac {1}{3 c^{3} x^{3}}-\frac {i}{2 c^{2} x^{2}}\right )+b d \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{2}\right )\right )\) | \(98\) |
default | \(c^{3} \left (a d \left (-\frac {1}{3 c^{3} x^{3}}-\frac {i}{2 c^{2} x^{2}}\right )+b d \left (-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-\frac {i \arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {i}{2 c x}-\frac {1}{6 c^{2} x^{2}}-\frac {\ln \left (c x \right )}{3}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{2}\right )\right )\) | \(98\) |
parallelrisch | \(-\frac {3 i x^{3} \arctan \left (c x \right ) b \,c^{3} d -3 i a \,c^{3} d \,x^{3}-b \,c^{3} d \ln \left (c^{2} x^{2}+1\right ) x^{3}+2 b \,c^{3} d \ln \left (x \right ) x^{3}-c^{3} x^{3} d b +3 i x^{2} b \,c^{2} d +3 i x \arctan \left (c x \right ) b c d +3 i a c d x +b c d x +2 b d \arctan \left (c x \right )+2 a d}{6 x^{3}}\) | \(121\) |
risch | \(-\frac {\left (3 b c d x -2 i b d \right ) \ln \left (i c x +1\right )}{12 x^{3}}+\frac {d \left (5 b \,c^{3} \ln \left (-c x -i\right ) x^{3}-b \,c^{3} \ln \left (c x -i\right ) x^{3}-4 b \,c^{3} \ln \left (-x \right ) x^{3}-6 i b \,c^{2} x^{2}-6 i x a c +3 b c x \ln \left (-i c x +1\right )-2 i b \ln \left (-i c x +1\right )-2 x b c -4 a \right )}{12 x^{3}}\) | \(129\) |
a*d*(-1/2*I*c/x^2-1/3/x^3)+b*d*c^3*(-1/3*arctan(c*x)/c^3/x^3-1/2*I*arctan( c*x)/c^2/x^2-1/2*I/c/x-1/6/c^2/x^2-1/3*ln(c*x)+1/6*ln(c^2*x^2+1)-1/2*I*arc tan(c*x))
Time = 0.25 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=-\frac {4 \, b c^{3} d x^{3} \log \left (x\right ) - 5 \, b c^{3} d x^{3} \log \left (\frac {c x + i}{c}\right ) + b c^{3} d x^{3} \log \left (\frac {c x - i}{c}\right ) + 6 i \, b c^{2} d x^{2} + 2 \, {\left (3 i \, a + b\right )} c d x + 4 \, a d - {\left (3 \, b c d x - 2 i \, b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{12 \, x^{3}} \]
-1/12*(4*b*c^3*d*x^3*log(x) - 5*b*c^3*d*x^3*log((c*x + I)/c) + b*c^3*d*x^3 *log((c*x - I)/c) + 6*I*b*c^2*d*x^2 + 2*(3*I*a + b)*c*d*x + 4*a*d - (3*b*c *d*x - 2*I*b*d)*log(-(c*x + I)/(c*x - I)))/x^3
Time = 2.60 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.86 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=- \frac {b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x \right )}}{3} - \frac {b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x - 27 i b^{2} c^{6} d^{2} \right )}}{12} + \frac {5 b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x + 27 i b^{2} c^{6} d^{2} \right )}}{12} + \frac {\left (- 3 b c d x + 2 i b d\right ) \log {\left (i c x + 1 \right )}}{12 x^{3}} + \frac {\left (3 b c d x - 2 i b d\right ) \log {\left (- i c x + 1 \right )}}{12 x^{3}} + \frac {- 2 a d - 3 i b c^{2} d x^{2} + x \left (- 3 i a c d - b c d\right )}{6 x^{3}} \]
-b*c**3*d*log(27*b**2*c**7*d**2*x)/3 - b*c**3*d*log(27*b**2*c**7*d**2*x - 27*I*b**2*c**6*d**2)/12 + 5*b*c**3*d*log(27*b**2*c**7*d**2*x + 27*I*b**2*c **6*d**2)/12 + (-3*b*c*d*x + 2*I*b*d)*log(I*c*x + 1)/(12*x**3) + (3*b*c*d* x - 2*I*b*d)*log(-I*c*x + 1)/(12*x**3) + (-2*a*d - 3*I*b*c**2*d*x**2 + x*( -3*I*a*c*d - b*c*d))/(6*x**3)
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=-\frac {1}{2} i \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c d + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d - \frac {i \, a c d}{2 \, x^{2}} - \frac {a d}{3 \, x^{3}} \]
-1/2*I*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c*d + 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*d - 1/2*I*a*c *d/x^2 - 1/3*a*d/x^3
\[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=\int { \frac {{\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{4}} \,d x } \]
Time = 0.98 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.66 \[ \int \frac {(d+i c d x) (a+b \arctan (c x))}{x^4} \, dx=\frac {b\,c^3\,d\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {\frac {a\,d}{3}-x^5\,\left (\frac {b\,c^5\,d}{6}+\frac {a\,c^5\,d\,1{}\mathrm {i}}{2}\right )+\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {c\,d\,x\,\left (b+a\,3{}\mathrm {i}\right )}{6}+\frac {c^2\,d\,x^2\,\left (2\,a+b\,3{}\mathrm {i}\right )}{6}+\frac {b\,c^4\,d\,x^4\,1{}\mathrm {i}}{2}+\frac {b\,c^2\,d\,x^2\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,c^3\,d\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}+\frac {b\,c\,d\,x\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}}{c^2\,x^5+x^3}-\frac {b\,c^3\,d\,\ln \left (x\right )}{3}-\frac {b\,d\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {c^2}}\right )\,{\left (c^2\right )}^{3/2}\,1{}\mathrm {i}}{2} \]
(b*c^3*d*log(c^2*x^2 + 1))/6 - (b*d*atan((c^2*x)/(c^2)^(1/2))*(c^2)^(3/2)* 1i)/2 - ((a*d)/3 - x^5*((a*c^5*d*1i)/2 + (b*c^5*d)/6) + (b*d*atan(c*x))/3 + (c*d*x*(a*3i + b))/6 + (c^2*d*x^2*(2*a + b*3i))/6 + (b*c^4*d*x^4*1i)/2 + (b*c^2*d*x^2*atan(c*x))/3 + (b*c^3*d*x^3*atan(c*x)*1i)/2 + (b*c*d*x*atan( c*x)*1i)/2)/(x^3 + c^2*x^5) - (b*c^3*d*log(x))/3