Integrand size = 19, antiderivative size = 110 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d}{20 x^4}+\frac {b c \left (3 c^2 d-5 e\right )}{30 x^2}-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}+\frac {1}{15} b c^3 \left (3 c^2 d-5 e\right ) \log (x)-\frac {1}{30} b c^3 \left (3 c^2 d-5 e\right ) \log \left (1+c^2 x^2\right ) \]
-1/20*b*c*d/x^4+1/30*b*c*(3*c^2*d-5*e)/x^2-1/5*d*(a+b*arctan(c*x))/x^5-1/3 *e*(a+b*arctan(c*x))/x^3+1/15*b*c^3*(3*c^2*d-5*e)*ln(x)-1/30*b*c^3*(3*c^2* d-5*e)*ln(c^2*x^2+1)
Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {a d}{5 x^5}-\frac {b c d}{20 x^4}-\frac {a e}{3 x^3}+\frac {b c^3 d}{10 x^2}-\frac {b d \arctan (c x)}{5 x^5}-\frac {b e \arctan (c x)}{3 x^3}+\frac {1}{5} b c^5 d \log (x)-\frac {1}{10} b c^5 d \log \left (1+c^2 x^2\right )+\frac {1}{6} b c e \left (-\frac {1}{x^2}-2 c^2 \log (x)+c^2 \log \left (1+c^2 x^2\right )\right ) \]
-1/5*(a*d)/x^5 - (b*c*d)/(20*x^4) - (a*e)/(3*x^3) + (b*c^3*d)/(10*x^2) - ( b*d*ArcTan[c*x])/(5*x^5) - (b*e*ArcTan[c*x])/(3*x^3) + (b*c^5*d*Log[x])/5 - (b*c^5*d*Log[1 + c^2*x^2])/10 + (b*c*e*(-x^(-2) - 2*c^2*Log[x] + c^2*Log [1 + c^2*x^2]))/6
Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5511, 27, 354, 86, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {5 e x^2+3 d}{15 x^5 \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} b c \int \frac {5 e x^2+3 d}{x^5 \left (c^2 x^2+1\right )}dx-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{30} b c \int \frac {5 e x^2+3 d}{x^6 \left (c^2 x^2+1\right )}dx^2-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {1}{30} b c \int \left (\frac {3 d}{x^6}+\frac {5 c^4 e-3 c^6 d}{c^2 x^2+1}+\frac {3 c^4 d-5 c^2 e}{x^2}+\frac {5 e-3 c^2 d}{x^4}\right )dx^2-\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d (a+b \arctan (c x))}{5 x^5}-\frac {e (a+b \arctan (c x))}{3 x^3}+\frac {1}{30} b c \left (\frac {3 c^2 d-5 e}{x^2}+c^2 \log \left (x^2\right ) \left (3 c^2 d-5 e\right )-c^2 \left (3 c^2 d-5 e\right ) \log \left (c^2 x^2+1\right )-\frac {3 d}{2 x^4}\right )\) |
-1/5*(d*(a + b*ArcTan[c*x]))/x^5 - (e*(a + b*ArcTan[c*x]))/(3*x^3) + (b*c* ((-3*d)/(2*x^4) + (3*c^2*d - 5*e)/x^2 + c^2*(3*c^2*d - 5*e)*Log[x^2] - c^2 *(3*c^2*d - 5*e)*Log[1 + c^2*x^2]))/30
3.12.23.3.1 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[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Sim p[(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, m, q}, x] && ((IGtQ[q, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] && !ILt Q[(m - 1)/2, 0]))
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08
| method | result | size |
| parts | \(a \left (-\frac {e}{3 x^{3}}-\frac {d}{5 x^{5}}\right )+b \,c^{5} \left (-\frac {\arctan \left (c x \right ) e}{3 c^{5} x^{3}}-\frac {\arctan \left (c x \right ) d}{5 c^{5} x^{5}}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )-\frac {3 c^{2} d -5 e}{2 c^{2} x^{2}}+\frac {3 d}{4 c^{2} x^{4}}+\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{15 c^{2}}\right )\) | \(119\) |
| derivativedivides | \(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )}{15}+\frac {3 c^{2} d -5 e}{30 c^{2} x^{2}}-\frac {d}{20 c^{2} x^{4}}\right )}{c^{2}}\right )\) | \(127\) |
| default | \(c^{5} \left (\frac {a \left (-\frac {d}{5 c^{3} x^{5}}-\frac {e}{3 c^{3} x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right ) d}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\left (3 c^{2} d -5 e \right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{2} d +5 e \right ) \ln \left (c x \right )}{15}+\frac {3 c^{2} d -5 e}{30 c^{2} x^{2}}-\frac {d}{20 c^{2} x^{4}}\right )}{c^{2}}\right )\) | \(127\) |
| parallelrisch | \(\frac {12 \ln \left (x \right ) b \,c^{5} d \,x^{5}-6 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{5} d -6 b \,c^{5} d \,x^{5}-20 \ln \left (x \right ) b \,c^{3} e \,x^{5}+10 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{3} e +10 b \,c^{3} e \,x^{5}+6 b \,c^{3} d \,x^{3}-10 b c e \,x^{3}-20 \arctan \left (c x \right ) b e \,x^{2}-20 a e \,x^{2}-3 b c d x -12 \arctan \left (c x \right ) b d -12 a d}{60 x^{5}}\) | \(145\) |
| risch | \(\frac {i b \left (5 e \,x^{2}+3 d \right ) \ln \left (i c x +1\right )}{30 x^{5}}-\frac {-12 \ln \left (x \right ) b \,c^{5} d \,x^{5}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{5} d \,x^{5}+20 \ln \left (x \right ) b \,c^{3} e \,x^{5}-10 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{3} e \,x^{5}-6 b \,c^{3} d \,x^{3}+10 i b e \ln \left (-i c x +1\right ) x^{2}+10 b c e \,x^{3}+6 i b d \ln \left (-i c x +1\right )+20 a e \,x^{2}+3 b c d x +12 a d}{60 x^{5}}\) | \(163\) |
a*(-1/3*e/x^3-1/5*d/x^5)+b*c^5*(-1/3*arctan(c*x)/c^5*e/x^3-1/5*arctan(c*x) *d/c^5/x^5-1/15/c^2*((-3*c^2*d+5*e)*ln(c*x)-1/2*(3*c^2*d-5*e)/c^2/x^2+3/4* d/c^2/x^4+1/2*(3*c^2*d-5*e)*ln(c^2*x^2+1)))
Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {2 \, {\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, b c^{5} d - 5 \, b c^{3} e\right )} x^{5} \log \left (x\right ) + 3 \, b c d x + 20 \, a e x^{2} - 2 \, {\left (3 \, b c^{3} d - 5 \, b c e\right )} x^{3} + 12 \, a d + 4 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \arctan \left (c x\right )}{60 \, x^{5}} \]
-1/60*(2*(3*b*c^5*d - 5*b*c^3*e)*x^5*log(c^2*x^2 + 1) - 4*(3*b*c^5*d - 5*b *c^3*e)*x^5*log(x) + 3*b*c*d*x + 20*a*e*x^2 - 2*(3*b*c^3*d - 5*b*c*e)*x^3 + 12*a*d + 4*(5*b*e*x^2 + 3*b*d)*arctan(c*x))/x^5
Time = 0.50 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.39 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\begin {cases} - \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} + \frac {b c^{5} d \log {\left (x \right )}}{5} - \frac {b c^{5} d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d}{10 x^{2}} - \frac {b c^{3} e \log {\left (x \right )}}{3} + \frac {b c^{3} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d}{20 x^{4}} - \frac {b c e}{6 x^{2}} - \frac {b d \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b e \operatorname {atan}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{5 x^{5}} - \frac {e}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*d/(5*x**5) - a*e/(3*x**3) + b*c**5*d*log(x)/5 - b*c**5*d*log (x**2 + c**(-2))/10 + b*c**3*d/(10*x**2) - b*c**3*e*log(x)/3 + b*c**3*e*lo g(x**2 + c**(-2))/6 - b*c*d/(20*x**4) - b*c*e/(6*x**2) - b*d*atan(c*x)/(5* x**5) - b*e*atan(c*x)/(3*x**3), Ne(c, 0)), (a*(-d/(5*x**5) - e/(3*x**3)), True))
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e - \frac {a e}{3 \, x^{3}} - \frac {a d}{5 \, x^{5}} \]
-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 + 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*e - 1/3*a*e/x^3 - 1/5*a*d/x^5
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{6}} \,d x } \]
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^6} \, dx=\frac {b\,c^3\,e\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,c^5\,d\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {x^3\,\left (\frac {b\,c\,e}{6}-\frac {b\,c^3\,d}{10}\right )+\frac {a\,d}{5}+x^2\,\left (\frac {a\,e}{3}+\frac {b\,e\,\mathrm {atan}\left (c\,x\right )}{3}\right )+\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{5}+\frac {b\,c\,d\,x}{20}}{x^5}+\frac {b\,c^5\,d\,\ln \left (x\right )}{5}-\frac {b\,c^3\,e\,\ln \left (x\right )}{3} \]