Integrand size = 21, antiderivative size = 230 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {b e \left (e (3+m)-2 c^2 d (5+m)\right ) x^{2+m}}{c^3 (2+m) (3+m) (5+m)}-\frac {b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-\frac {b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c^3 (1+m) (2+m) (3+m) (5+m)} \]
b*e*(e*(3+m)-2*c^2*d*(5+m))*x^(2+m)/c^3/(5+m)/(m^2+5*m+6)-b*e^2*x^(4+m)/c/ (4+m)/(5+m)+d^2*x^(1+m)*(a+b*arctan(c*x))/(1+m)+2*d*e*x^(3+m)*(a+b*arctan( c*x))/(3+m)+e^2*x^(5+m)*(a+b*arctan(c*x))/(5+m)-b*(e^2*(m^2+4*m+3)-2*c^2*d *e*(m^2+6*m+5)+c^4*d^2*(m^2+8*m+15))*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2 *m],-c^2*x^2)/c^3/(m^2+3*m+2)/(m^2+8*m+15)
Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.84 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=x^{1+m} \left (\frac {d^2 (a+b \arctan (c x))}{1+m}+\frac {2 d e x^2 (a+b \arctan (c x))}{3+m}+\frac {e^2 x^4 (a+b \arctan (c x))}{5+m}-\frac {b c d^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+3 m+m^2}-\frac {2 b c d e x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-c^2 x^2\right )}{12+7 m+m^2}-\frac {b c e^2 x^5 \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-c^2 x^2\right )}{(5+m) (6+m)}\right ) \]
x^(1 + m)*((d^2*(a + b*ArcTan[c*x]))/(1 + m) + (2*d*e*x^2*(a + b*ArcTan[c* x]))/(3 + m) + (e^2*x^4*(a + b*ArcTan[c*x]))/(5 + m) - (b*c*d^2*x*Hypergeo metric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(2 + 3*m + m^2) - (2*b*c*d *e*x^3*Hypergeometric2F1[1, (4 + m)/2, (6 + m)/2, -(c^2*x^2)])/(12 + 7*m + m^2) - (b*c*e^2*x^5*Hypergeometric2F1[1, (6 + m)/2, (8 + m)/2, -(c^2*x^2) ])/((5 + m)*(6 + m)))
Time = 0.52 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5511, 1584, 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.
| \(\displaystyle \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx\) |
| \(\Big \downarrow \) 5511 |
| \(\displaystyle -b c \int \frac {x^{m+1} \left (\frac {e^2 x^4}{m+5}+\frac {2 d e x^2}{m+3}+\frac {d^2}{m+1}\right )}{c^2 x^2+1}dx+\frac {d^2 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {2 d e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {e^2 x^{m+5} (a+b \arctan (c x))}{m+5}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -b c \int \left (\frac {e \left (\frac {2 c^2 d}{m+3}-\frac {e}{m+5}\right ) x^{m+1}}{c^4}+\frac {\left (15 d^2 c^4+d^2 m^2 c^4+8 d^2 m c^4-2 d e m^2 c^2-10 d e c^2-12 d e m c^2+3 e^2+e^2 m^2+4 e^2 m\right ) x^{m+1}}{c^4 (m+1) (m+3) (m+5) \left (c^2 x^2+1\right )}+\frac {e^2 x^{m+3}}{c^2 (m+5)}\right )dx+\frac {d^2 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {2 d e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {e^2 x^{m+5} (a+b \arctan (c x))}{m+5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {2 d e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {e^2 x^{m+5} (a+b \arctan (c x))}{m+5}-b c \left (\frac {e^2 x^{m+4}}{c^2 (m+4) (m+5)}+\frac {x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{c^4 (m+1) (m+2) (m+3) (m+5)}+\frac {e x^{m+2} \left (\frac {2 c^2 d}{m+3}-\frac {e}{m+5}\right )}{c^4 (m+2)}\right )\) |
(d^2*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (2*d*e*x^(3 + m)*(a + b*ArcT an[c*x]))/(3 + m) + (e^2*x^(5 + m)*(a + b*ArcTan[c*x]))/(5 + m) - b*c*((e* ((2*c^2*d)/(3 + m) - e/(5 + m))*x^(2 + m))/(c^4*(2 + m)) + (e^2*x^(4 + m)) /(c^2*(4 + m)*(5 + m)) + ((e^2*(3 + 4*m + m^2) - 2*c^2*d*e*(5 + 6*m + m^2) + c^4*d^2*(15 + 8*m + m^2))*x^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(c^4*(1 + m)*(2 + m)*(3 + m)*(5 + m)))
3.13.29.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -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]))
\[\int x^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \arctan \left (c x \right )\right )d x\]
\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d ^2)*arctan(c*x))*x^m, x)
\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]
\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
a*e^2*x^(m + 5)/(m + 5) + 2*a*d*e*x^(m + 3)/(m + 3) + a*d^2*x^(m + 1)/(m + 1) + (((b*e^2*m^2 + 4*b*e^2*m + 3*b*e^2)*x^5 + 2*(b*d*e*m^2 + 6*b*d*e*m + 5*b*d*e)*x^3 + (b*d^2*m^2 + 8*b*d^2*m + 15*b*d^2)*x)*x^m*arctan(c*x) - (m ^3 + 9*m^2 + 23*m + 15)*integrate(((b*c*e^2*m^2 + 4*b*c*e^2*m + 3*b*c*e^2) *x^5 + 2*(b*c*d*e*m^2 + 6*b*c*d*e*m + 5*b*c*d*e)*x^3 + (b*c*d^2*m^2 + 8*b* c*d^2*m + 15*b*c*d^2)*x)*x^m/(m^3 + (c^2*m^3 + 9*c^2*m^2 + 23*c^2*m + 15*c ^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9*m^2 + 23*m + 15)
\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
Timed out. \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2 \,d x \]