\(\int x^m (d+e x^2)^2 (a+b \arctan (c x)) \, dx\) [1228]

Optimal result

Integrand size = 21, antiderivative size = 202 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=-\frac {b e \left (\frac {2 c^2 d}{3+m}-\frac {e}{5+m}\right ) x^{2+m}}{c^3 (2+m)}-\frac {b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-\frac {b \left (\frac {c^4 d^2}{1+m}-\frac {2 c^2 d e}{3+m}+\frac {e^2}{5+m}\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c^3 (2+m)} \] Output:

-b*e*(2*c^2*d/(3+m)-e/(5+m))*x^(2+m)/c^3/(2+m)-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*(c^4*d^2/(1+m)-2*c^2*d*e/(3+m)+e^2/ 
(5+m))*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-c^2*x^2)/c^3/(2+m)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=x^{1+m} \left (\frac {d^2 (a+b \arctan (c x))}{1+m}+\frac {2 d e x^2 (a+b \arctan (c x))}{3+m}+\frac {e^2 x^4 (a+b \arctan (c x))}{5+m}-\frac {b c d^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+3 m+m^2}-\frac {2 b c d e x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-c^2 x^2\right )}{12+7 m+m^2}-\frac {b c e^2 x^5 \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-c^2 x^2\right )}{(5+m) (6+m)}\right ) \] Input:

Integrate[x^m*(d + e*x^2)^2*(a + b*ArcTan[c*x]),x]
 

Output:

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)))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.11, 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

\(\displaystyle \frac {d^2 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {2 d e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {e^2 x^{m+5} (a+b \arctan (c x))}{m+5}-b c \left (\frac {e^2 x^{m+4}}{c^2 (m+4) (m+5)}+\frac {x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{c^4 (m+1) (m+2) (m+3) (m+5)}+\frac {e x^{m+2} \left (\frac {2 c^2 d}{m+3}-\frac {e}{m+5}\right )}{c^4 (m+2)}\right )\)

Input:

Int[x^m*(d + e*x^2)^2*(a + b*ArcTan[c*x]),x]
 

Output:

(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)))
 

Defintions of rubi rules used

rule 1584

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5511

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]))
 

Maple [F]

\[\int x^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \arctan \left (c x \right )\right )d x\]

Input:

int(x^m*(e*x^2+d)^2*(a+b*arctan(c*x)),x)
 

Output:

int(x^m*(e*x^2+d)^2*(a+b*arctan(c*x)),x)
 

Fricas [F]

\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \] Input:

integrate(x**m*(e*x**2+d)**2*(a+b*atan(c*x)),x)
 

Output:

Integral(x**m*(a + b*atan(c*x))*(d + e*x**2)**2, x)
 

Maxima [F]

\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="maxima")
 

Output:

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)
 

Giac [F]

\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^2*(a+b*arctan(c*x)),x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)^2*(b*arctan(c*x) + a)*x^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2 \,d x \] Input:

int(x^m*(a + b*atan(c*x))*(d + e*x^2)^2,x)
 

Output:

int(x^m*(a + b*atan(c*x))*(d + e*x^2)^2, x)
 

Reduce [F]

\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\text {too large to display} \] Input:

int(x^m*(e*x^2+d)^2*(a+b*atan(c*x)),x)
 

Output:

(x**m*atan(c*x)*b*c**5*d**2*m**5*x + 14*x**m*atan(c*x)*b*c**5*d**2*m**4*x 
+ 71*x**m*atan(c*x)*b*c**5*d**2*m**3*x + 154*x**m*atan(c*x)*b*c**5*d**2*m* 
*2*x + 120*x**m*atan(c*x)*b*c**5*d**2*m*x + 2*x**m*atan(c*x)*b*c**5*d*e*m* 
*5*x**3 + 24*x**m*atan(c*x)*b*c**5*d*e*m**4*x**3 + 98*x**m*atan(c*x)*b*c** 
5*d*e*m**3*x**3 + 156*x**m*atan(c*x)*b*c**5*d*e*m**2*x**3 + 80*x**m*atan(c 
*x)*b*c**5*d*e*m*x**3 + x**m*atan(c*x)*b*c**5*e**2*m**5*x**5 + 10*x**m*ata 
n(c*x)*b*c**5*e**2*m**4*x**5 + 35*x**m*atan(c*x)*b*c**5*e**2*m**3*x**5 + 5 
0*x**m*atan(c*x)*b*c**5*e**2*m**2*x**5 + 24*x**m*atan(c*x)*b*c**5*e**2*m*x 
**5 + x**m*a*c**5*d**2*m**5*x + 14*x**m*a*c**5*d**2*m**4*x + 71*x**m*a*c** 
5*d**2*m**3*x + 154*x**m*a*c**5*d**2*m**2*x + 120*x**m*a*c**5*d**2*m*x + 2 
*x**m*a*c**5*d*e*m**5*x**3 + 24*x**m*a*c**5*d*e*m**4*x**3 + 98*x**m*a*c**5 
*d*e*m**3*x**3 + 156*x**m*a*c**5*d*e*m**2*x**3 + 80*x**m*a*c**5*d*e*m*x**3 
 + x**m*a*c**5*e**2*m**5*x**5 + 10*x**m*a*c**5*e**2*m**4*x**5 + 35*x**m*a* 
c**5*e**2*m**3*x**5 + 50*x**m*a*c**5*e**2*m**2*x**5 + 24*x**m*a*c**5*e**2* 
m*x**5 - x**m*b*c**4*d**2*m**4 - 14*x**m*b*c**4*d**2*m**3 - 71*x**m*b*c**4 
*d**2*m**2 - 154*x**m*b*c**4*d**2*m - 120*x**m*b*c**4*d**2 - 2*x**m*b*c**4 
*d*e*m**4*x**2 - 20*x**m*b*c**4*d*e*m**3*x**2 - 58*x**m*b*c**4*d*e*m**2*x* 
*2 - 40*x**m*b*c**4*d*e*m*x**2 - x**m*b*c**4*e**2*m**4*x**4 - 6*x**m*b*c** 
4*e**2*m**3*x**4 - 11*x**m*b*c**4*e**2*m**2*x**4 - 6*x**m*b*c**4*e**2*m*x* 
*4 + 2*x**m*b*c**2*d*e*m**4 + 24*x**m*b*c**2*d*e*m**3 + 98*x**m*b*c**2*...