Integrand size = 21, antiderivative size = 161 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=-\frac {b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) x^2}{210 c^5}-\frac {b \left (14 c^2 d-5 e\right ) e x^4}{140 c^3}-\frac {b e^2 x^6}{42 c}+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))+\frac {b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right )}{210 c^7} \]
-1/210*b*(35*c^4*d^2-42*c^2*d*e+15*e^2)*x^2/c^5-1/140*b*(14*c^2*d-5*e)*e*x ^4/c^3-1/42*b*e^2*x^6/c+1/3*d^2*x^3*(a+b*arctan(c*x))+2/5*d*e*x^5*(a+b*arc tan(c*x))+1/7*e^2*x^7*(a+b*arctan(c*x))+1/210*b*(35*c^4*d^2-42*c^2*d*e+15* e^2)*ln(c^2*x^2+1)/c^7
Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {c^2 x^2 \left (-30 b e^2+3 b c^2 e \left (28 d+5 e x^2\right )-2 b c^4 \left (35 d^2+21 d e x^2+5 e^2 x^4\right )+4 a c^5 x \left (35 d^2+42 d e x^2+15 e^2 x^4\right )\right )+4 b c^7 x^3 \left (35 d^2+42 d e x^2+15 e^2 x^4\right ) \arctan (c x)+2 b \left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right )}{420 c^7} \]
(c^2*x^2*(-30*b*e^2 + 3*b*c^2*e*(28*d + 5*e*x^2) - 2*b*c^4*(35*d^2 + 21*d* e*x^2 + 5*e^2*x^4) + 4*a*c^5*x*(35*d^2 + 42*d*e*x^2 + 15*e^2*x^4)) + 4*b*c ^7*x^3*(35*d^2 + 42*d*e*x^2 + 15*e^2*x^4)*ArcTan[c*x] + 2*b*(35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/(420*c^7)
Time = 0.43 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98, 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.
| \(\displaystyle \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int \frac {x^3 \left (15 e^2 x^4+42 d e x^2+35 d^2\right )}{105 \left (c^2 x^2+1\right )}dx+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{105} b c \int \frac {x^3 \left (15 e^2 x^4+42 d e x^2+35 d^2\right )}{c^2 x^2+1}dx+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {1}{210} b c \int \frac {x^2 \left (15 e^2 x^4+42 d e x^2+35 d^2\right )}{c^2 x^2+1}dx^2+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle -\frac {1}{210} b c \int \left (\frac {15 e^2 x^4}{c^2}+\frac {3 \left (14 c^2 d-5 e\right ) e x^2}{c^4}+\frac {35 d^2 c^4-42 d e c^2+15 e^2}{c^6}+\frac {-35 d^2 c^4+42 d e c^2-15 e^2}{c^6 \left (c^2 x^2+1\right )}\right )dx^2+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {2}{5} d e x^5 (a+b \arctan (c x))+\frac {1}{7} e^2 x^7 (a+b \arctan (c x))-\frac {1}{210} b c \left (\frac {5 e^2 x^6}{c^2}+\frac {3 e x^4 \left (14 c^2 d-5 e\right )}{2 c^4}-\frac {\left (35 c^4 d^2-42 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )}{c^8}+\frac {x^2 \left (35 c^4 d^2-42 c^2 d e+15 e^2\right )}{c^6}\right )\) |
(d^2*x^3*(a + b*ArcTan[c*x]))/3 + (2*d*e*x^5*(a + b*ArcTan[c*x]))/5 + (e^2 *x^7*(a + b*ArcTan[c*x]))/7 - (b*c*(((35*c^4*d^2 - 42*c^2*d*e + 15*e^2)*x^ 2)/c^6 + (3*(14*c^2*d - 5*e)*e*x^4)/(2*c^4) + (5*e^2*x^6)/c^2 - ((35*c^4*d ^2 - 42*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/c^8))/210
3.12.26.3.1 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[((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 = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(a \left (\frac {1}{7} e^{2} x^{7}+\frac {2}{5} e d \,x^{5}+\frac {1}{3} d^{2} x^{3}\right )+\frac {b \left (\frac {\arctan \left (c x \right ) c^{3} e^{2} x^{7}}{7}+\frac {2 \arctan \left (c x \right ) c^{3} d e \,x^{5}}{5}+\frac {\arctan \left (c x \right ) d^{2} c^{3} x^{3}}{3}-\frac {\frac {35 d^{2} c^{6} x^{2}}{2}+\frac {21 d \,c^{6} e \,x^{4}}{2}+\frac {5 e^{2} c^{6} x^{6}}{2}-21 d \,c^{4} e \,x^{2}-\frac {15 e^{2} c^{4} x^{4}}{4}+\frac {15 e^{2} c^{2} x^{2}}{2}+\frac {\left (-35 c^{4} d^{2}+42 c^{2} d e -15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{105 c^{4}}\right )}{c^{3}}\) | \(181\) |
| derivativedivides | \(\frac {\frac {a \left (\frac {1}{3} d^{2} c^{7} x^{3}+\frac {2}{5} d \,c^{7} e \,x^{5}+\frac {1}{7} e^{2} c^{7} x^{7}\right )}{c^{4}}+\frac {b \left (\frac {\arctan \left (c x \right ) d^{2} c^{7} x^{3}}{3}+\frac {2 \arctan \left (c x \right ) d \,c^{7} e \,x^{5}}{5}+\frac {\arctan \left (c x \right ) e^{2} c^{7} x^{7}}{7}-\frac {d^{2} c^{6} x^{2}}{6}-\frac {d \,c^{6} e \,x^{4}}{10}+\frac {d \,c^{4} e \,x^{2}}{5}-\frac {e^{2} c^{6} x^{6}}{42}+\frac {e^{2} c^{4} x^{4}}{28}-\frac {e^{2} c^{2} x^{2}}{14}-\frac {\left (-35 c^{4} d^{2}+42 c^{2} d e -15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}\right )}{c^{4}}}{c^{3}}\) | \(191\) |
| default | \(\frac {\frac {a \left (\frac {1}{3} d^{2} c^{7} x^{3}+\frac {2}{5} d \,c^{7} e \,x^{5}+\frac {1}{7} e^{2} c^{7} x^{7}\right )}{c^{4}}+\frac {b \left (\frac {\arctan \left (c x \right ) d^{2} c^{7} x^{3}}{3}+\frac {2 \arctan \left (c x \right ) d \,c^{7} e \,x^{5}}{5}+\frac {\arctan \left (c x \right ) e^{2} c^{7} x^{7}}{7}-\frac {d^{2} c^{6} x^{2}}{6}-\frac {d \,c^{6} e \,x^{4}}{10}+\frac {d \,c^{4} e \,x^{2}}{5}-\frac {e^{2} c^{6} x^{6}}{42}+\frac {e^{2} c^{4} x^{4}}{28}-\frac {e^{2} c^{2} x^{2}}{14}-\frac {\left (-35 c^{4} d^{2}+42 c^{2} d e -15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}\right )}{c^{4}}}{c^{3}}\) | \(191\) |
| parallelrisch | \(\frac {60 x^{7} \arctan \left (c x \right ) b \,c^{7} e^{2}+60 a \,c^{7} e^{2} x^{7}+168 x^{5} \arctan \left (c x \right ) b \,c^{7} d e -10 b \,c^{6} e^{2} x^{6}+168 a \,c^{7} d e \,x^{5}+140 x^{3} \arctan \left (c x \right ) b \,c^{7} d^{2}-42 b \,c^{6} d e \,x^{4}+140 a \,c^{7} d^{2} x^{3}+15 b \,c^{4} e^{2} x^{4}-70 b \,c^{6} d^{2} x^{2}+84 b \,c^{4} d e \,x^{2}+70 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2}-30 b \,c^{2} e^{2} x^{2}-84 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d e +30 \ln \left (c^{2} x^{2}+1\right ) b \,e^{2}}{420 c^{7}}\) | \(212\) |
| risch | \(\frac {i b d e \,x^{5} \ln \left (-i c x +1\right )}{5}-\frac {i b \left (15 e^{2} x^{7}+42 e d \,x^{5}+35 d^{2} x^{3}\right ) \ln \left (i c x +1\right )}{210}+\frac {i b \,d^{2} x^{3} \ln \left (-i c x +1\right )}{6}+\frac {x^{7} e^{2} a}{7}+\frac {i b \,e^{2} x^{7} \ln \left (-i c x +1\right )}{14}+\frac {2 x^{5} e d a}{5}-\frac {b \,e^{2} x^{6}}{42 c}+\frac {x^{3} d^{2} a}{3}-\frac {b d e \,x^{4}}{10 c}-\frac {b \,d^{2} x^{2}}{6 c}+\frac {b \,e^{2} x^{4}}{28 c^{3}}+\frac {b d e \,x^{2}}{5 c^{3}}+\frac {\ln \left (-c^{2} x^{2}-1\right ) b \,d^{2}}{6 c^{3}}-\frac {b \,e^{2} x^{2}}{14 c^{5}}-\frac {\ln \left (-c^{2} x^{2}-1\right ) b d e}{5 c^{5}}+\frac {\ln \left (-c^{2} x^{2}-1\right ) b \,e^{2}}{14 c^{7}}\) | \(246\) |
a*(1/7*e^2*x^7+2/5*e*d*x^5+1/3*d^2*x^3)+b/c^3*(1/7*arctan(c*x)*c^3*e^2*x^7 +2/5*arctan(c*x)*c^3*d*e*x^5+1/3*arctan(c*x)*d^2*c^3*x^3-1/105/c^4*(35/2*d ^2*c^6*x^2+21/2*d*c^6*e*x^4+5/2*e^2*c^6*x^6-21*d*c^4*e*x^2-15/4*e^2*c^4*x^ 4+15/2*e^2*c^2*x^2+1/2*(-35*c^4*d^2+42*c^2*d*e-15*e^2)*ln(c^2*x^2+1)))
Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.16 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {60 \, a c^{7} e^{2} x^{7} + 168 \, a c^{7} d e x^{5} - 10 \, b c^{6} e^{2} x^{6} + 140 \, a c^{7} d^{2} x^{3} - 3 \, {\left (14 \, b c^{6} d e - 5 \, b c^{4} e^{2}\right )} x^{4} - 2 \, {\left (35 \, b c^{6} d^{2} - 42 \, b c^{4} d e + 15 \, b c^{2} e^{2}\right )} x^{2} + 4 \, {\left (15 \, b c^{7} e^{2} x^{7} + 42 \, b c^{7} d e x^{5} + 35 \, b c^{7} d^{2} x^{3}\right )} \arctan \left (c x\right ) + 2 \, {\left (35 \, b c^{4} d^{2} - 42 \, b c^{2} d e + 15 \, b e^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{420 \, c^{7}} \]
1/420*(60*a*c^7*e^2*x^7 + 168*a*c^7*d*e*x^5 - 10*b*c^6*e^2*x^6 + 140*a*c^7 *d^2*x^3 - 3*(14*b*c^6*d*e - 5*b*c^4*e^2)*x^4 - 2*(35*b*c^6*d^2 - 42*b*c^4 *d*e + 15*b*c^2*e^2)*x^2 + 4*(15*b*c^7*e^2*x^7 + 42*b*c^7*d*e*x^5 + 35*b*c ^7*d^2*x^3)*arctan(c*x) + 2*(35*b*c^4*d^2 - 42*b*c^2*d*e + 15*b*e^2)*log(c ^2*x^2 + 1))/c^7
Time = 0.47 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.52 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\begin {cases} \frac {a d^{2} x^{3}}{3} + \frac {2 a d e x^{5}}{5} + \frac {a e^{2} x^{7}}{7} + \frac {b d^{2} x^{3} \operatorname {atan}{\left (c x \right )}}{3} + \frac {2 b d e x^{5} \operatorname {atan}{\left (c x \right )}}{5} + \frac {b e^{2} x^{7} \operatorname {atan}{\left (c x \right )}}{7} - \frac {b d^{2} x^{2}}{6 c} - \frac {b d e x^{4}}{10 c} - \frac {b e^{2} x^{6}}{42 c} + \frac {b d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6 c^{3}} + \frac {b d e x^{2}}{5 c^{3}} + \frac {b e^{2} x^{4}}{28 c^{3}} - \frac {b d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{5 c^{5}} - \frac {b e^{2} x^{2}}{14 c^{5}} + \frac {b e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{14 c^{7}} & \text {for}\: c \neq 0 \\a \left (\frac {d^{2} x^{3}}{3} + \frac {2 d e x^{5}}{5} + \frac {e^{2} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a*d**2*x**3/3 + 2*a*d*e*x**5/5 + a*e**2*x**7/7 + b*d**2*x**3*at an(c*x)/3 + 2*b*d*e*x**5*atan(c*x)/5 + b*e**2*x**7*atan(c*x)/7 - b*d**2*x* *2/(6*c) - b*d*e*x**4/(10*c) - b*e**2*x**6/(42*c) + b*d**2*log(x**2 + c**( -2))/(6*c**3) + b*d*e*x**2/(5*c**3) + b*e**2*x**4/(28*c**3) - b*d*e*log(x* *2 + c**(-2))/(5*c**5) - b*e**2*x**2/(14*c**5) + b*e**2*log(x**2 + c**(-2) )/(14*c**7), Ne(c, 0)), (a*(d**2*x**3/3 + 2*d*e*x**5/5 + e**2*x**7/7), Tru e))
Time = 0.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.12 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {1}{7} \, a e^{2} x^{7} + \frac {2}{5} \, a d e x^{5} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} + \frac {1}{10} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b d e + \frac {1}{84} \, {\left (12 \, x^{7} \arctan \left (c x\right ) - c {\left (\frac {2 \, c^{4} x^{6} - 3 \, c^{2} x^{4} + 6 \, x^{2}}{c^{6}} - \frac {6 \, \log \left (c^{2} x^{2} + 1\right )}{c^{8}}\right )}\right )} b e^{2} \]
1/7*a*e^2*x^7 + 2/5*a*d*e*x^5 + 1/3*a*d^2*x^3 + 1/6*(2*x^3*arctan(c*x) - c *(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d^2 + 1/10*(4*x^5*arctan(c*x) - c*((c ^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*d*e + 1/84*(12*x^7*arctan (c*x) - c*((2*c^4*x^6 - 3*c^2*x^4 + 6*x^2)/c^6 - 6*log(c^2*x^2 + 1)/c^8))* b*e^2
\[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
Time = 0.91 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.19 \[ \int x^2 \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {a\,d^2\,x^3}{3}+\frac {a\,e^2\,x^7}{7}+\frac {b\,d^2\,\ln \left (c^2\,x^2+1\right )}{6\,c^3}+\frac {b\,e^2\,\ln \left (c^2\,x^2+1\right )}{14\,c^7}-\frac {b\,d^2\,x^2}{6\,c}-\frac {b\,e^2\,x^6}{42\,c}+\frac {b\,e^2\,x^4}{28\,c^3}-\frac {b\,e^2\,x^2}{14\,c^5}+\frac {2\,a\,d\,e\,x^5}{5}+\frac {b\,d^2\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,e^2\,x^7\,\mathrm {atan}\left (c\,x\right )}{7}-\frac {b\,d\,e\,\ln \left (c^2\,x^2+1\right )}{5\,c^5}-\frac {b\,d\,e\,x^4}{10\,c}+\frac {b\,d\,e\,x^2}{5\,c^3}+\frac {2\,b\,d\,e\,x^5\,\mathrm {atan}\left (c\,x\right )}{5} \]
(a*d^2*x^3)/3 + (a*e^2*x^7)/7 + (b*d^2*log(c^2*x^2 + 1))/(6*c^3) + (b*e^2* log(c^2*x^2 + 1))/(14*c^7) - (b*d^2*x^2)/(6*c) - (b*e^2*x^6)/(42*c) + (b*e ^2*x^4)/(28*c^3) - (b*e^2*x^2)/(14*c^5) + (2*a*d*e*x^5)/5 + (b*d^2*x^3*ata n(c*x))/3 + (b*e^2*x^7*atan(c*x))/7 - (b*d*e*log(c^2*x^2 + 1))/(5*c^5) - ( b*d*e*x^4)/(10*c) + (b*d*e*x^2)/(5*c^3) + (2*b*d*e*x^5*atan(c*x))/5