Integrand size = 21, antiderivative size = 150 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d^2}{20 x^4}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}+\frac {1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right ) \] Output:
-1/20*b*c*d^2/x^4+1/30*b*c*d*(3*c^2*d-10*e)/x^2-1/5*d^2*(a+b*arctan(c*x))/ x^5-2/3*d*e*(a+b*arctan(c*x))/x^3-e^2*(a+b*arctan(c*x))/x+1/15*b*c*(3*c^4* d^2-10*c^2*d*e+15*e^2)*ln(x)-1/30*b*c*(3*c^4*d^2-10*c^2*d*e+15*e^2)*ln(c^2 *x^2+1)
Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {1}{60} \left (-\frac {12 d^2 (a+b \arctan (c x))}{x^5}-\frac {40 d e (a+b \arctan (c x))}{x^3}-\frac {60 e^2 (a+b \arctan (c x))}{x}+30 b c e^2 \left (2 \log (x)-\log \left (1+c^2 x^2\right )\right )-20 b c d e \left (\frac {1}{x^2}+2 c^2 \log (x)-c^2 \log \left (1+c^2 x^2\right )\right )-3 b c d^2 \left (\frac {1}{x^4}-\frac {2 c^2}{x^2}-4 c^4 \log (x)+2 c^4 \log \left (1+c^2 x^2\right )\right )\right ) \] Input:
Integrate[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^6,x]
Output:
((-12*d^2*(a + b*ArcTan[c*x]))/x^5 - (40*d*e*(a + b*ArcTan[c*x]))/x^3 - (6 0*e^2*(a + b*ArcTan[c*x]))/x + 30*b*c*e^2*(2*Log[x] - Log[1 + c^2*x^2]) - 20*b*c*d*e*(x^(-2) + 2*c^2*Log[x] - c^2*Log[1 + c^2*x^2]) - 3*b*c*d^2*(x^( -4) - (2*c^2)/x^2 - 4*c^4*Log[x] + 2*c^4*Log[1 + c^2*x^2]))/60
Time = 0.46 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5511, 27, 1578, 1195, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {15 e^2 x^4+10 d e x^2+3 d^2}{15 x^5 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} b c \int \frac {15 e^2 x^4+10 d e x^2+3 d^2}{x^5 \left (c^2 x^2+1\right )}dx-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{30} b c \int \frac {15 e^2 x^4+10 d e x^2+3 d^2}{x^6 \left (c^2 x^2+1\right )}dx^2-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \frac {1}{30} b c \int \left (\frac {3 d^2}{x^6}-\frac {\left (3 c^2 d-10 e\right ) d}{x^4}+\frac {-3 d^2 c^6+10 d e c^4-15 e^2 c^2}{c^2 x^2+1}+\frac {3 d^2 c^4-10 d e c^2+15 e^2}{x^2}\right )dx^2-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}+\frac {1}{30} b c \left (\frac {d \left (3 c^2 d-10 e\right )}{x^2}+\log \left (x^2\right ) \left (3 c^4 d^2-10 c^2 d e+15 e^2\right )-\left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )-\frac {3 d^2}{2 x^4}\right )\) |
Input:
Int[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^6,x]
Output:
-1/5*(d^2*(a + b*ArcTan[c*x]))/x^5 - (2*d*e*(a + b*ArcTan[c*x]))/(3*x^3) - (e^2*(a + b*ArcTan[c*x]))/x + (b*c*((-3*d^2)/(2*x^4) + (d*(3*c^2*d - 10*e ))/x^2 + (3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[x^2] - (3*c^4*d^2 - 10*c^2* d*e + 15*e^2)*Log[1 + c^2*x^2]))/30
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(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 = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11
| method | result | size |
| parts | \(a \left (-\frac {2 d e}{3 x^{3}}-\frac {e^{2}}{x}-\frac {d^{2}}{5 x^{5}}\right )+b \,c^{5} \left (-\frac {2 \arctan \left (c x \right ) d e}{3 c^{5} x^{3}}-\frac {\arctan \left (c x \right ) e^{2}}{c^{5} x}-\frac {\arctan \left (c x \right ) d^{2}}{5 c^{5} x^{5}}-\frac {\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )-\frac {d \left (3 c^{2} d -10 e \right )}{2 x^{2}}+\frac {3 d^{2}}{4 x^{4}}}{15 c^{4}}\right )\) | \(167\) |
| derivativedivides | \(c^{5} \left (\frac {a \left (-\frac {2 d e}{3 c \,x^{3}}-\frac {e^{2}}{c x}-\frac {d^{2}}{5 c \,x^{5}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{3 c \,x^{3}}-\frac {\arctan \left (c x \right ) e^{2}}{c x}-\frac {\arctan \left (c x \right ) d^{2}}{5 c \,x^{5}}-\frac {\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )}{15}+\frac {d \left (3 c^{2} d -10 e \right )}{30 x^{2}}-\frac {d^{2}}{20 x^{4}}-\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}\right )}{c^{4}}\right )\) | \(178\) |
| default | \(c^{5} \left (\frac {a \left (-\frac {2 d e}{3 c \,x^{3}}-\frac {e^{2}}{c x}-\frac {d^{2}}{5 c \,x^{5}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{3 c \,x^{3}}-\frac {\arctan \left (c x \right ) e^{2}}{c x}-\frac {\arctan \left (c x \right ) d^{2}}{5 c \,x^{5}}-\frac {\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )}{15}+\frac {d \left (3 c^{2} d -10 e \right )}{30 x^{2}}-\frac {d^{2}}{20 x^{4}}-\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}\right )}{c^{4}}\right )\) | \(178\) |
| parallelrisch | \(\frac {12 \ln \left (x \right ) b \,c^{5} d^{2} x^{5}-6 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{5} d^{2}-6 b \,c^{5} d^{2} x^{5}-40 \ln \left (x \right ) b \,c^{3} d e \,x^{5}+20 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{3} d e +20 b \,c^{3} d e \,x^{5}+60 \ln \left (x \right ) b c \,e^{2} x^{5}-30 \ln \left (c^{2} x^{2}+1\right ) x^{5} b c \,e^{2}+6 x^{3} b \,c^{3} d^{2}-60 x^{4} \arctan \left (c x \right ) b \,e^{2}-60 a \,e^{2} x^{4}-20 b c d e \,x^{3}-40 x^{2} \arctan \left (c x \right ) b d e -40 a d e \,x^{2}-3 b c \,d^{2} x -12 b \,d^{2} \arctan \left (c x \right )-12 d^{2} a}{60 x^{5}}\) | \(219\) |
| risch | \(\frac {i b \left (15 e^{2} x^{4}+10 d e \,x^{2}+3 d^{2}\right ) \ln \left (i c x +1\right )}{30 x^{5}}-\frac {-12 \ln \left (x \right ) b \,c^{5} d^{2} x^{5}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{5} d^{2} x^{5}+40 \ln \left (x \right ) b \,c^{3} d e \,x^{5}-20 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{3} d e \,x^{5}-60 \ln \left (x \right ) b c \,e^{2} x^{5}+30 \ln \left (-c^{2} x^{2}-1\right ) b c \,e^{2} x^{5}+30 i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}-6 x^{3} b \,c^{3} d^{2}+20 i b d e \,x^{2} \ln \left (-i c x +1\right )+60 a \,e^{2} x^{4}+20 b c d e \,x^{3}+6 i b \,d^{2} \ln \left (-i c x +1\right )+40 a d e \,x^{2}+3 b c \,d^{2} x +12 d^{2} a}{60 x^{5}}\) | \(251\) |
Input:
int((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x,method=_RETURNVERBOSE)
Output:
a*(-2/3*d*e/x^3-e^2/x-1/5*d^2/x^5)+b*c^5*(-2/3*arctan(c*x)/c^5*d*e/x^3-arc tan(c*x)/c^5*e^2/x-1/5*arctan(c*x)*d^2/c^5/x^5-1/15/c^4*(1/2*(3*c^4*d^2-10 *c^2*d*e+15*e^2)*ln(c^2*x^2+1)+(-3*c^4*d^2+10*c^2*d*e-15*e^2)*ln(c*x)-1/2* d*(3*c^2*d-10*e)/x^2+3/4*d^2/x^4))
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {60 \, a e^{2} x^{4} + 2 \, {\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (x\right ) + 3 \, b c d^{2} x + 40 \, a d e x^{2} - 2 \, {\left (3 \, b c^{3} d^{2} - 10 \, b c d e\right )} x^{3} + 12 \, a d^{2} + 4 \, {\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \arctan \left (c x\right )}{60 \, x^{5}} \] Input:
integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")
Output:
-1/60*(60*a*e^2*x^4 + 2*(3*b*c^5*d^2 - 10*b*c^3*d*e + 15*b*c*e^2)*x^5*log( c^2*x^2 + 1) - 4*(3*b*c^5*d^2 - 10*b*c^3*d*e + 15*b*c*e^2)*x^5*log(x) + 3* b*c*d^2*x + 40*a*d*e*x^2 - 2*(3*b*c^3*d^2 - 10*b*c*d*e)*x^3 + 12*a*d^2 + 4 *(15*b*e^2*x^4 + 10*b*d*e*x^2 + 3*b*d^2)*arctan(c*x))/x^5
Time = 0.52 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.57 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\begin {cases} - \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} + \frac {b c^{5} d^{2} \log {\left (x \right )}}{5} - \frac {b c^{5} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d^{2}}{10 x^{2}} - \frac {2 b c^{3} d e \log {\left (x \right )}}{3} + \frac {b c^{3} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} - \frac {b c d^{2}}{20 x^{4}} - \frac {b c d e}{3 x^{2}} + b c e^{2} \log {\left (x \right )} - \frac {b c e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{x} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{5 x^{5}} - \frac {2 d e}{3 x^{3}} - \frac {e^{2}}{x}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x**2+d)**2*(a+b*atan(c*x))/x**6,x)
Output:
Piecewise((-a*d**2/(5*x**5) - 2*a*d*e/(3*x**3) - a*e**2/x + b*c**5*d**2*lo g(x)/5 - b*c**5*d**2*log(x**2 + c**(-2))/10 + b*c**3*d**2/(10*x**2) - 2*b* c**3*d*e*log(x)/3 + b*c**3*d*e*log(x**2 + c**(-2))/3 - b*c*d**2/(20*x**4) - b*c*d*e/(3*x**2) + b*c*e**2*log(x) - b*c*e**2*log(x**2 + c**(-2))/2 - b* d**2*atan(c*x)/(5*x**5) - 2*b*d*e*atan(c*x)/(3*x**3) - b*e**2*atan(c*x)/x, Ne(c, 0)), (a*(-d**2/(5*x**5) - 2*d*e/(3*x**3) - e**2/x), True))
Time = 0.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d e - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b e^{2} - \frac {a e^{2}}{x} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {a d^{2}}{5 \, x^{5}} \] Input:
integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")
Output:
-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*((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*e - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*e^2 - a*e^2/x - 2/3*a*d*e/x^3 - 1/5*a*d^2/x^5
Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.30 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {6 \, b c^{5} d^{2} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{5} d^{2} x^{5} \log \left (x\right ) - 20 \, b c^{3} d e x^{5} \log \left (c^{2} x^{2} + 1\right ) + 40 \, b c^{3} d e x^{5} \log \left (x\right ) + 30 \, b c e^{2} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c e^{2} x^{5} \log \left (x\right ) - 6 \, b c^{3} d^{2} x^{3} + 60 \, b e^{2} x^{4} \arctan \left (c x\right ) + 20 \, b c d e x^{3} + 60 \, a e^{2} x^{4} + 40 \, b d e x^{2} \arctan \left (c x\right ) + 3 \, b c d^{2} x + 40 \, a d e x^{2} + 12 \, b d^{2} \arctan \left (c x\right ) + 12 \, a d^{2}}{60 \, x^{5}} \] Input:
integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x, algorithm="giac")
Output:
-1/60*(6*b*c^5*d^2*x^5*log(c^2*x^2 + 1) - 12*b*c^5*d^2*x^5*log(x) - 20*b*c ^3*d*e*x^5*log(c^2*x^2 + 1) + 40*b*c^3*d*e*x^5*log(x) + 30*b*c*e^2*x^5*log (c^2*x^2 + 1) - 60*b*c*e^2*x^5*log(x) - 6*b*c^3*d^2*x^3 + 60*b*e^2*x^4*arc tan(c*x) + 20*b*c*d*e*x^3 + 60*a*e^2*x^4 + 40*b*d*e*x^2*arctan(c*x) + 3*b* c*d^2*x + 40*a*d*e*x^2 + 12*b*d^2*arctan(c*x) + 12*a*d^2)/x^5
Time = 0.53 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {b\,c^3\,d^2}{10\,x^2}-\frac {a\,e^2}{x}-\frac {b\,c^5\,d^2\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {a\,d^2}{5\,x^5}+\frac {b\,c^5\,d^2\,\ln \left (x\right )}{5}-\frac {2\,a\,d\,e}{3\,x^3}-\frac {b\,c\,e^2\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c\,d^2}{20\,x^4}+b\,c\,e^2\,\ln \left (x\right )-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{5\,x^5}-\frac {b\,e^2\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,c^3\,d\,e\,\ln \left (c^2\,x^2+1\right )}{3}-\frac {2\,b\,c^3\,d\,e\,\ln \left (x\right )}{3}-\frac {b\,c\,d\,e}{3\,x^2}-\frac {2\,b\,d\,e\,\mathrm {atan}\left (c\,x\right )}{3\,x^3} \] Input:
int(((a + b*atan(c*x))*(d + e*x^2)^2)/x^6,x)
Output:
(b*c^3*d^2)/(10*x^2) - (a*e^2)/x - (b*c^5*d^2*log(c^2*x^2 + 1))/10 - (a*d^ 2)/(5*x^5) + (b*c^5*d^2*log(x))/5 - (2*a*d*e)/(3*x^3) - (b*c*e^2*log(c^2*x ^2 + 1))/2 - (b*c*d^2)/(20*x^4) + b*c*e^2*log(x) - (b*d^2*atan(c*x))/(5*x^ 5) - (b*e^2*atan(c*x))/x + (b*c^3*d*e*log(c^2*x^2 + 1))/3 - (2*b*c^3*d*e*l og(x))/3 - (b*c*d*e)/(3*x^2) - (2*b*d*e*atan(c*x))/(3*x^3)
Time = 0.20 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.30 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {-12 \mathit {atan} \left (c x \right ) b \,d^{2}-40 \mathit {atan} \left (c x \right ) b d e \,x^{2}-60 \mathit {atan} \left (c x \right ) b \,e^{2} x^{4}-6 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{5} d^{2} x^{5}+20 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{3} d e \,x^{5}-30 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b c \,e^{2} x^{5}+12 \,\mathrm {log}\left (x \right ) b \,c^{5} d^{2} x^{5}-40 \,\mathrm {log}\left (x \right ) b \,c^{3} d e \,x^{5}+60 \,\mathrm {log}\left (x \right ) b c \,e^{2} x^{5}-12 a \,d^{2}-40 a d e \,x^{2}-60 a \,e^{2} x^{4}+6 b \,c^{3} d^{2} x^{3}-3 b c \,d^{2} x -20 b c d e \,x^{3}}{60 x^{5}} \] Input:
int((e*x^2+d)^2*(a+b*atan(c*x))/x^6,x)
Output:
( - 12*atan(c*x)*b*d**2 - 40*atan(c*x)*b*d*e*x**2 - 60*atan(c*x)*b*e**2*x* *4 - 6*log(c**2*x**2 + 1)*b*c**5*d**2*x**5 + 20*log(c**2*x**2 + 1)*b*c**3* d*e*x**5 - 30*log(c**2*x**2 + 1)*b*c*e**2*x**5 + 12*log(x)*b*c**5*d**2*x** 5 - 40*log(x)*b*c**3*d*e*x**5 + 60*log(x)*b*c*e**2*x**5 - 12*a*d**2 - 40*a *d*e*x**2 - 60*a*e**2*x**4 + 6*b*c**3*d**2*x**3 - 3*b*c*d**2*x - 20*b*c*d* e*x**3)/(60*x**5)