Integrand size = 21, antiderivative size = 177 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d^3}{20 x^4}+\frac {b c d^2 \left (c^2 d-5 e\right )}{10 x^2}-\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))+\frac {1}{5} b c d \left (c^4 d^2-5 c^2 d e+15 e^2\right ) \log (x)-\frac {b \left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (1+c^2 x^2\right )}{10 c} \] Output:
-1/20*b*c*d^3/x^4+1/10*b*c*d^2*(c^2*d-5*e)/x^2-1/5*d^3*(a+b*arctan(c*x))/x ^5-d^2*e*(a+b*arctan(c*x))/x^3-3*d*e^2*(a+b*arctan(c*x))/x+e^3*x*(a+b*arct an(c*x))+1/5*b*c*d*(c^4*d^2-5*c^2*d*e+15*e^2)*ln(x)-1/10*b*(c^6*d^3-5*c^4* d^2*e+15*c^2*d*e^2+5*e^3)*ln(c^2*x^2+1)/c
Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=\frac {1}{20} \left (-\frac {4 a d^3}{x^5}-\frac {b c d^3}{x^4}-\frac {20 a d^2 e}{x^3}+\frac {2 b c d^2 \left (c^2 d-5 e\right )}{x^2}-\frac {60 a d e^2}{x}+20 a e^3 x-\frac {4 b \left (d^3+5 d^2 e x^2+15 d e^2 x^4-5 e^3 x^6\right ) \arctan (c x)}{x^5}+4 b c d \left (c^4 d^2-5 c^2 d e+15 e^2\right ) \log (x)-\frac {2 b \left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (1+c^2 x^2\right )}{c}\right ) \] Input:
Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]
Output:
((-4*a*d^3)/x^5 - (b*c*d^3)/x^4 - (20*a*d^2*e)/x^3 + (2*b*c*d^2*(c^2*d - 5 *e))/x^2 - (60*a*d*e^2)/x + 20*a*e^3*x - (4*b*(d^3 + 5*d^2*e*x^2 + 15*d*e^ 2*x^4 - 5*e^3*x^6)*ArcTan[c*x])/x^5 + 4*b*c*d*(c^4*d^2 - 5*c^2*d*e + 15*e^ 2)*Log[x] - (2*b*(c^6*d^3 - 5*c^4*d^2*e + 15*c^2*d*e^2 + 5*e^3)*Log[1 + c^ 2*x^2])/c)/20
Time = 0.66 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5511, 27, 2331, 2123, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int -\frac {-5 e^3 x^6+15 d e^2 x^4+5 d^2 e x^2+d^3}{5 x^5 \left (c^2 x^2+1\right )}dx-\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} b c \int \frac {-5 e^3 x^6+15 d e^2 x^4+5 d^2 e x^2+d^3}{x^5 \left (c^2 x^2+1\right )}dx-\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2331 |
| \(\displaystyle \frac {1}{10} b c \int \frac {-5 e^3 x^6+15 d e^2 x^4+5 d^2 e x^2+d^3}{x^6 \left (c^2 x^2+1\right )}dx^2-\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {1}{10} b c \int \left (\frac {d^3}{x^6}-\frac {\left (c^2 d-5 e\right ) d^2}{x^4}+\frac {\left (d^2 c^4-5 d e c^2+15 e^2\right ) d}{x^2}+\frac {-d^3 c^6+5 d^2 e c^4-15 d e^2 c^2-5 e^3}{c^2 x^2+1}\right )dx^2-\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^3 (a+b \arctan (c x))}{5 x^5}-\frac {d^2 e (a+b \arctan (c x))}{x^3}-\frac {3 d e^2 (a+b \arctan (c x))}{x}+e^3 x (a+b \arctan (c x))+\frac {1}{10} b c \left (\frac {d^2 \left (c^2 d-5 e\right )}{x^2}+d \log \left (x^2\right ) \left (c^4 d^2-5 c^2 d e+15 e^2\right )-\frac {\left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{c^2}-\frac {d^3}{2 x^4}\right )\) |
Input:
Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]
Output:
-1/5*(d^3*(a + b*ArcTan[c*x]))/x^5 - (d^2*e*(a + b*ArcTan[c*x]))/x^3 - (3* d*e^2*(a + b*ArcTan[c*x]))/x + e^3*x*(a + b*ArcTan[c*x]) + (b*c*(-1/2*d^3/ x^4 + (d^2*(c^2*d - 5*e))/x^2 + d*(c^4*d^2 - 5*c^2*d*e + 15*e^2)*Log[x^2] - ((c^6*d^3 - 5*c^4*d^2*e + 15*c^2*d*e^2 + 5*e^3)*Log[1 + c^2*x^2])/c^2))/ 10
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && 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 = 0.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18
| method | result | size |
| derivativedivides | \(c^{5} \left (\frac {a \left (c x \,e^{3}-\frac {c \,d^{3}}{5 x^{5}}-\frac {3 c d \,e^{2}}{x}-\frac {c \,d^{2} e}{x^{3}}\right )}{c^{6}}+\frac {b \left (\arctan \left (c x \right ) c x \,e^{3}-\frac {\arctan \left (c x \right ) c \,d^{3}}{5 x^{5}}-\frac {3 \arctan \left (c x \right ) c d \,e^{2}}{x}-\frac {\arctan \left (c x \right ) c \,d^{2} e}{x^{3}}+\frac {c^{2} d^{2} \left (c^{2} d -5 e \right )}{10 x^{2}}-\frac {c^{2} d^{3}}{20 x^{4}}+\frac {d \,c^{2} \left (c^{4} d^{2}-5 c^{2} d e +15 e^{2}\right ) \ln \left (c x \right )}{5}-\frac {\left (c^{6} d^{3}-5 c^{4} d^{2} e +15 e^{2} d \,c^{2}+5 e^{3}\right ) \ln \left (c^{2} x^{2}+1\right )}{10}\right )}{c^{6}}\right )\) | \(208\) |
| default | \(c^{5} \left (\frac {a \left (c x \,e^{3}-\frac {c \,d^{3}}{5 x^{5}}-\frac {3 c d \,e^{2}}{x}-\frac {c \,d^{2} e}{x^{3}}\right )}{c^{6}}+\frac {b \left (\arctan \left (c x \right ) c x \,e^{3}-\frac {\arctan \left (c x \right ) c \,d^{3}}{5 x^{5}}-\frac {3 \arctan \left (c x \right ) c d \,e^{2}}{x}-\frac {\arctan \left (c x \right ) c \,d^{2} e}{x^{3}}+\frac {c^{2} d^{2} \left (c^{2} d -5 e \right )}{10 x^{2}}-\frac {c^{2} d^{3}}{20 x^{4}}+\frac {d \,c^{2} \left (c^{4} d^{2}-5 c^{2} d e +15 e^{2}\right ) \ln \left (c x \right )}{5}-\frac {\left (c^{6} d^{3}-5 c^{4} d^{2} e +15 e^{2} d \,c^{2}+5 e^{3}\right ) \ln \left (c^{2} x^{2}+1\right )}{10}\right )}{c^{6}}\right )\) | \(208\) |
| parts | \(a \left (e^{3} x -\frac {e \,d^{2}}{x^{3}}-\frac {3 e^{2} d}{x}-\frac {d^{3}}{5 x^{5}}\right )+b \,c^{5} \left (\frac {\arctan \left (c x \right ) x \,e^{3}}{c^{5}}-\frac {\arctan \left (c x \right ) d^{2} e}{c^{5} x^{3}}-\frac {3 \arctan \left (c x \right ) e^{2} d}{c^{5} x}-\frac {\arctan \left (c x \right ) d^{3}}{5 c^{5} x^{5}}-\frac {\frac {\left (c^{6} d^{3}-5 c^{4} d^{2} e +15 e^{2} d \,c^{2}+5 e^{3}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {c^{2} d^{2} \left (c^{2} d -5 e \right )}{2 x^{2}}+\frac {c^{2} d^{3}}{4 x^{4}}-d \,c^{2} \left (c^{4} d^{2}-5 c^{2} d e +15 e^{2}\right ) \ln \left (c x \right )}{5 c^{6}}\right )\) | \(211\) |
| parallelrisch | \(\frac {4 \ln \left (x \right ) b \,c^{6} d^{3} x^{5}-2 \ln \left (c^{2} x^{2}+1\right ) b \,c^{6} d^{3} x^{5}-2 b \,c^{6} d^{3} x^{5}-20 \ln \left (x \right ) b \,c^{4} d^{2} e \,x^{5}+10 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2} e \,x^{5}+10 b \,c^{4} d^{2} e \,x^{5}+60 \ln \left (x \right ) b \,c^{2} d \,e^{2} x^{5}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d \,e^{2} x^{5}+20 x^{6} \arctan \left (c x \right ) b c \,e^{3}+20 x^{6} e^{3} a c +2 b \,c^{4} d^{3} x^{3}-10 \ln \left (c^{2} x^{2}+1\right ) b \,e^{3} x^{5}-60 x^{4} \arctan \left (c x \right ) b c d \,e^{2}-60 a c d \,e^{2} x^{4}-10 b \,c^{2} d^{2} e \,x^{3}-20 x^{2} \arctan \left (c x \right ) b c \,d^{2} e -20 a c \,d^{2} e \,x^{2}-b \,c^{2} d^{3} x -4 \arctan \left (c x \right ) b c \,d^{3}-4 a c \,d^{3}}{20 c \,x^{5}}\) | \(295\) |
| risch | \(\frac {i b \left (-5 x^{6} e^{3}+15 x^{4} e^{2} d +5 x^{2} e \,d^{2}+d^{3}\right ) \ln \left (i c x +1\right )}{10 x^{5}}+\frac {4 \ln \left (x \right ) b \,c^{6} d^{3} x^{5}-2 \ln \left (c^{2} x^{2}+1\right ) b \,c^{6} d^{3} x^{5}-20 \ln \left (x \right ) b \,c^{4} d^{2} e \,x^{5}+10 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2} e \,x^{5}+10 i b c \,e^{3} x^{6} \ln \left (-i c x +1\right )+60 \ln \left (x \right ) b \,c^{2} d \,e^{2} x^{5}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d \,e^{2} x^{5}-10 i b c \,d^{2} e \,x^{2} \ln \left (-i c x +1\right )+20 x^{6} e^{3} a c +2 b \,c^{4} d^{3} x^{3}-10 \ln \left (c^{2} x^{2}+1\right ) b \,e^{3} x^{5}-30 i b c d \,e^{2} x^{4} \ln \left (-i c x +1\right )-60 a c d \,e^{2} x^{4}-10 b \,c^{2} d^{2} e \,x^{3}-2 i b c \,d^{3} \ln \left (-i c x +1\right )-20 a c \,d^{2} e \,x^{2}-b \,c^{2} d^{3} x -4 a c \,d^{3}}{20 c \,x^{5}}\) | \(336\) |
Input:
int((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x,method=_RETURNVERBOSE)
Output:
c^5*(a/c^6*(c*x*e^3-1/5*c*d^3/x^5-3*c*d*e^2/x-c*d^2*e/x^3)+b/c^6*(arctan(c *x)*c*x*e^3-1/5*arctan(c*x)*c*d^3/x^5-3*arctan(c*x)*c*d*e^2/x-arctan(c*x)* c*d^2*e/x^3+1/10*c^2*d^2*(c^2*d-5*e)/x^2-1/20*c^2*d^3/x^4+1/5*d*c^2*(c^4*d ^2-5*c^2*d*e+15*e^2)*ln(c*x)-1/10*(c^6*d^3-5*c^4*d^2*e+15*c^2*d*e^2+5*e^3) *ln(c^2*x^2+1)))
Time = 0.12 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=\frac {20 \, a c e^{3} x^{6} - 60 \, a c d e^{2} x^{4} - b c^{2} d^{3} x - 20 \, a c d^{2} e x^{2} - 2 \, {\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2} + 5 \, b e^{3}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) + 4 \, {\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2}\right )} x^{5} \log \left (x\right ) - 4 \, a c d^{3} + 2 \, {\left (b c^{4} d^{3} - 5 \, b c^{2} d^{2} e\right )} x^{3} + 4 \, {\left (5 \, b c e^{3} x^{6} - 15 \, b c d e^{2} x^{4} - 5 \, b c d^{2} e x^{2} - b c d^{3}\right )} \arctan \left (c x\right )}{20 \, c x^{5}} \] Input:
integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")
Output:
1/20*(20*a*c*e^3*x^6 - 60*a*c*d*e^2*x^4 - b*c^2*d^3*x - 20*a*c*d^2*e*x^2 - 2*(b*c^6*d^3 - 5*b*c^4*d^2*e + 15*b*c^2*d*e^2 + 5*b*e^3)*x^5*log(c^2*x^2 + 1) + 4*(b*c^6*d^3 - 5*b*c^4*d^2*e + 15*b*c^2*d*e^2)*x^5*log(x) - 4*a*c*d ^3 + 2*(b*c^4*d^3 - 5*b*c^2*d^2*e)*x^3 + 4*(5*b*c*e^3*x^6 - 15*b*c*d*e^2*x ^4 - 5*b*c*d^2*e*x^2 - b*c*d^3)*arctan(c*x))/(c*x^5)
Time = 0.58 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.63 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=\begin {cases} - \frac {a d^{3}}{5 x^{5}} - \frac {a d^{2} e}{x^{3}} - \frac {3 a d e^{2}}{x} + a e^{3} x + \frac {b c^{5} d^{3} \log {\left (x \right )}}{5} - \frac {b c^{5} d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d^{3}}{10 x^{2}} - b c^{3} d^{2} e \log {\left (x \right )} + \frac {b c^{3} d^{2} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b c d^{3}}{20 x^{4}} - \frac {b c d^{2} e}{2 x^{2}} + 3 b c d e^{2} \log {\left (x \right )} - \frac {3 b c d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b d^{2} e \operatorname {atan}{\left (c x \right )}}{x^{3}} - \frac {3 b d e^{2} \operatorname {atan}{\left (c x \right )}}{x} + b e^{3} x \operatorname {atan}{\left (c x \right )} - \frac {b e^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{3}}{5 x^{5}} - \frac {d^{2} e}{x^{3}} - \frac {3 d e^{2}}{x} + e^{3} x\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**6,x)
Output:
Piecewise((-a*d**3/(5*x**5) - a*d**2*e/x**3 - 3*a*d*e**2/x + a*e**3*x + b* c**5*d**3*log(x)/5 - b*c**5*d**3*log(x**2 + c**(-2))/10 + b*c**3*d**3/(10* x**2) - b*c**3*d**2*e*log(x) + b*c**3*d**2*e*log(x**2 + c**(-2))/2 - b*c*d **3/(20*x**4) - b*c*d**2*e/(2*x**2) + 3*b*c*d*e**2*log(x) - 3*b*c*d*e**2*l og(x**2 + c**(-2))/2 - b*d**3*atan(c*x)/(5*x**5) - b*d**2*e*atan(c*x)/x**3 - 3*b*d*e**2*atan(c*x)/x + b*e**3*x*atan(c*x) - b*e**3*log(x**2 + c**(-2) )/(2*c), Ne(c, 0)), (a*(-d**3/(5*x**5) - d**2*e/x**3 - 3*d*e**2/x + e**3*x ), True))
Time = 0.05 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{3} + \frac {1}{2} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{2} e - \frac {3}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d e^{2} + a e^{3} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e^{3}}{2 \, c} - \frac {3 \, a d e^{2}}{x} - \frac {a d^{2} e}{x^{3}} - \frac {a d^{3}}{5 \, x^{5}} \] Input:
integrate((e*x^2+d)^3*(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^3 + 1/2*((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^2*e - 3/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d*e^2 + a*e^3*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x ^2 + 1))*b*e^3/c - 3*a*d*e^2/x - a*d^2*e/x^3 - 1/5*a*d^3/x^5
Time = 0.13 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.51 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=-\frac {2 \, b c^{6} d^{3} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, b c^{6} d^{3} x^{5} \log \left (x\right ) - 10 \, b c^{4} d^{2} e x^{5} \log \left (c^{2} x^{2} + 1\right ) + 20 \, b c^{4} d^{2} e x^{5} \log \left (x\right ) - 20 \, b c e^{3} x^{6} \arctan \left (c x\right ) + 30 \, b c^{2} d e^{2} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c^{2} d e^{2} x^{5} \log \left (x\right ) - 2 \, b c^{4} d^{3} x^{3} - 20 \, a c e^{3} x^{6} + 60 \, b c d e^{2} x^{4} \arctan \left (c x\right ) + 10 \, b e^{3} x^{5} \log \left (c^{2} x^{2} + 1\right ) + 10 \, b c^{2} d^{2} e x^{3} + 60 \, a c d e^{2} x^{4} + 20 \, b c d^{2} e x^{2} \arctan \left (c x\right ) + b c^{2} d^{3} x + 20 \, a c d^{2} e x^{2} + 4 \, b c d^{3} \arctan \left (c x\right ) + 4 \, a c d^{3}}{20 \, c x^{5}} \] Input:
integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="giac")
Output:
-1/20*(2*b*c^6*d^3*x^5*log(c^2*x^2 + 1) - 4*b*c^6*d^3*x^5*log(x) - 10*b*c^ 4*d^2*e*x^5*log(c^2*x^2 + 1) + 20*b*c^4*d^2*e*x^5*log(x) - 20*b*c*e^3*x^6* arctan(c*x) + 30*b*c^2*d*e^2*x^5*log(c^2*x^2 + 1) - 60*b*c^2*d*e^2*x^5*log (x) - 2*b*c^4*d^3*x^3 - 20*a*c*e^3*x^6 + 60*b*c*d*e^2*x^4*arctan(c*x) + 10 *b*e^3*x^5*log(c^2*x^2 + 1) + 10*b*c^2*d^2*e*x^3 + 60*a*c*d*e^2*x^4 + 20*b *c*d^2*e*x^2*arctan(c*x) + b*c^2*d^3*x + 20*a*c*d^2*e*x^2 + 4*b*c*d^3*arct an(c*x) + 4*a*c*d^3)/(c*x^5)
Time = 0.96 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=\ln \left (x\right )\,\left (\frac {b\,c^5\,d^3}{5}-b\,c^3\,d^2\,e+3\,b\,c\,d\,e^2\right )-\frac {a\,d^3-x^3\,\left (\frac {b\,c^3\,d^3}{2}-\frac {5\,b\,c\,d^2\,e}{2}\right )+\frac {b\,c\,d^3\,x}{4}+5\,a\,d^2\,e\,x^2+15\,a\,d\,e^2\,x^4}{5\,x^5}-\frac {\ln \left (c^2\,x^2+1\right )\,\left (b\,c^6\,d^3-5\,b\,c^4\,d^2\,e+15\,b\,c^2\,d\,e^2+5\,b\,e^3\right )}{10\,c}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,d^3}{5}+b\,d^2\,e\,x^2+3\,b\,d\,e^2\,x^4-b\,e^3\,x^6\right )}{x^5}+a\,e^3\,x \] Input:
int(((a + b*atan(c*x))*(d + e*x^2)^3)/x^6,x)
Output:
log(x)*((b*c^5*d^3)/5 + 3*b*c*d*e^2 - b*c^3*d^2*e) - (a*d^3 - x^3*((b*c^3* d^3)/2 - (5*b*c*d^2*e)/2) + (b*c*d^3*x)/4 + 5*a*d^2*e*x^2 + 15*a*d*e^2*x^4 )/(5*x^5) - (log(c^2*x^2 + 1)*(5*b*e^3 + b*c^6*d^3 + 15*b*c^2*d*e^2 - 5*b* c^4*d^2*e))/(10*c) - (atan(c*x)*((b*d^3)/5 - b*e^3*x^6 + b*d^2*e*x^2 + 3*b *d*e^2*x^4))/x^5 + a*e^3*x
Time = 0.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.52 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arctan (c x))}{x^6} \, dx=\frac {-4 \mathit {atan} \left (c x \right ) b c \,d^{3}-20 \mathit {atan} \left (c x \right ) b c \,d^{2} e \,x^{2}-60 \mathit {atan} \left (c x \right ) b c d \,e^{2} x^{4}+20 \mathit {atan} \left (c x \right ) b c \,e^{3} x^{6}-2 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{6} d^{3} x^{5}+10 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2} e \,x^{5}-30 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,c^{2} d \,e^{2} x^{5}-10 \,\mathrm {log}\left (c^{2} x^{2}+1\right ) b \,e^{3} x^{5}+4 \,\mathrm {log}\left (x \right ) b \,c^{6} d^{3} x^{5}-20 \,\mathrm {log}\left (x \right ) b \,c^{4} d^{2} e \,x^{5}+60 \,\mathrm {log}\left (x \right ) b \,c^{2} d \,e^{2} x^{5}-4 a c \,d^{3}-20 a c \,d^{2} e \,x^{2}-60 a c d \,e^{2} x^{4}+20 a c \,e^{3} x^{6}+2 b \,c^{4} d^{3} x^{3}-b \,c^{2} d^{3} x -10 b \,c^{2} d^{2} e \,x^{3}}{20 c \,x^{5}} \] Input:
int((e*x^2+d)^3*(a+b*atan(c*x))/x^6,x)
Output:
( - 4*atan(c*x)*b*c*d**3 - 20*atan(c*x)*b*c*d**2*e*x**2 - 60*atan(c*x)*b*c *d*e**2*x**4 + 20*atan(c*x)*b*c*e**3*x**6 - 2*log(c**2*x**2 + 1)*b*c**6*d* *3*x**5 + 10*log(c**2*x**2 + 1)*b*c**4*d**2*e*x**5 - 30*log(c**2*x**2 + 1) *b*c**2*d*e**2*x**5 - 10*log(c**2*x**2 + 1)*b*e**3*x**5 + 4*log(x)*b*c**6* d**3*x**5 - 20*log(x)*b*c**4*d**2*e*x**5 + 60*log(x)*b*c**2*d*e**2*x**5 - 4*a*c*d**3 - 20*a*c*d**2*e*x**2 - 60*a*c*d*e**2*x**4 + 20*a*c*e**3*x**6 + 2*b*c**4*d**3*x**3 - b*c**2*d**3*x - 10*b*c**2*d**2*e*x**3)/(20*c*x**5)