\(\int x^m (d+e x^2)^3 (a+b \arctan (c x)) \, dx\) [1227]

Optimal result

Integrand size = 21, antiderivative size = 351 \[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=-\frac {b e \left (e^2 \left (15+8 m+m^2\right )-3 c^2 d e \left (21+10 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )\right ) x^{2+m}}{c^5 \left (6+5 m+m^2\right ) \left (35+12 m+m^2\right )}+\frac {b e^2 \left (e (5+m)-3 c^2 d (7+m)\right ) x^{4+m}}{c^3 (4+m) \left (35+12 m+m^2\right )}-\frac {b e^3 x^{6+m}}{c (6+m) (7+m)}+\frac {d^3 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {3 d^2 e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {3 d e^2 x^{5+m} (a+b \arctan (c x))}{5+m}+\frac {e^3 x^{7+m} (a+b \arctan (c x))}{7+m}-\frac {b \left (\frac {c^6 d^3}{1+m}-\frac {e \left (e^2 \left (15+8 m+m^2\right )-3 c^2 d e \left (21+10 m+m^2\right )+3 c^4 d^2 \left (35+12 m+m^2\right )\right )}{(3+m) \left (35+12 m+m^2\right )}\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c^5 (2+m)} \] Output:

-b*e*(e^2*(m^2+8*m+15)-3*c^2*d*e*(m^2+10*m+21)+3*c^4*d^2*(m^2+12*m+35))*x^ 
(2+m)/c^5/(m^2+5*m+6)/(m^2+12*m+35)+b*e^2*(e*(5+m)-3*c^2*d*(7+m))*x^(4+m)/ 
c^3/(4+m)/(m^2+12*m+35)-b*e^3*x^(6+m)/c/(6+m)/(7+m)+d^3*x^(1+m)*(a+b*arcta 
n(c*x))/(1+m)+3*d^2*e*x^(3+m)*(a+b*arctan(c*x))/(3+m)+3*d*e^2*x^(5+m)*(a+b 
*arctan(c*x))/(5+m)+e^3*x^(7+m)*(a+b*arctan(c*x))/(7+m)-b*(c^6*d^3/(1+m)-e 
*(e^2*(m^2+8*m+15)-3*c^2*d*e*(m^2+10*m+21)+3*c^4*d^2*(m^2+12*m+35))/(3+m)/ 
(m^2+12*m+35))*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-c^2*x^2)/c^5/(2+m 
)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.75 \[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=x^{1+m} \left (\frac {d^3 (a+b \arctan (c x))}{1+m}+\frac {3 d^2 e x^2 (a+b \arctan (c x))}{3+m}+\frac {3 d e^2 x^4 (a+b \arctan (c x))}{5+m}+\frac {e^3 x^6 (a+b \arctan (c x))}{7+m}-\frac {b c e^3 x^7 \operatorname {Hypergeometric2F1}\left (1,4+\frac {m}{2},5+\frac {m}{2},-c^2 x^2\right )}{(7+m) (8+m)}-\frac {b c d^3 x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+3 m+m^2}-\frac {3 b c d^2 e x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-c^2 x^2\right )}{12+7 m+m^2}-\frac {3 b c d e^2 x^5 \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-c^2 x^2\right )}{(5+m) (6+m)}\right ) \] Input:

Integrate[x^m*(d + e*x^2)^3*(a + b*ArcTan[c*x]),x]
 

Output:

x^(1 + m)*((d^3*(a + b*ArcTan[c*x]))/(1 + m) + (3*d^2*e*x^2*(a + b*ArcTan[ 
c*x]))/(3 + m) + (3*d*e^2*x^4*(a + b*ArcTan[c*x]))/(5 + m) + (e^3*x^6*(a + 
 b*ArcTan[c*x]))/(7 + m) - (b*c*e^3*x^7*Hypergeometric2F1[1, 4 + m/2, 5 + 
m/2, -(c^2*x^2)])/((7 + m)*(8 + m)) - (b*c*d^3*x*Hypergeometric2F1[1, (2 + 
 m)/2, (4 + m)/2, -(c^2*x^2)])/(2 + 3*m + m^2) - (3*b*c*d^2*e*x^3*Hypergeo 
metric2F1[1, (4 + m)/2, (6 + m)/2, -(c^2*x^2)])/(12 + 7*m + m^2) - (3*b*c* 
d*e^2*x^5*Hypergeometric2F1[1, (6 + m)/2, (8 + m)/2, -(c^2*x^2)])/((5 + m) 
*(6 + m)))
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5511, 2333, 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.

\(\displaystyle \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx\)

\(\Big \downarrow \) 5511

\(\displaystyle -b c \int \frac {x^{m+1} \left (\frac {e^3 x^6}{m+7}+\frac {3 d e^2 x^4}{m+5}+\frac {3 d^2 e x^2}{m+3}+\frac {d^3}{m+1}\right )}{c^2 x^2+1}dx+\frac {d^3 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {3 d^2 e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {3 d e^2 x^{m+5} (a+b \arctan (c x))}{m+5}+\frac {e^3 x^{m+7} (a+b \arctan (c x))}{m+7}\)

\(\Big \downarrow \) 2333

\(\displaystyle -b c \int \left (\frac {e \left (3 d^2 \left (m^2+12 m+35\right ) c^4-3 d e \left (m^2+10 m+21\right ) c^2+e^2 \left (m^2+8 m+15\right )\right ) x^{m+1}}{c^6 (m+3) (m+5) (m+7)}+\frac {\left (105 d^3 c^6+d^3 m^3 c^6+15 d^3 m^2 c^6+71 d^3 m c^6-3 d^2 e m^3 c^4-39 d^2 e m^2 c^4-105 d^2 e c^4-141 d^2 e m c^4+3 d e^2 m^3 c^2+63 d e^2 c^2+33 d e^2 m^2 c^2+93 d e^2 m c^2-15 e^3-e^3 m^3-9 e^3 m^2-23 e^3 m\right ) x^{m+1}}{c^6 (m+1) (m+3) (m+5) (m+7) \left (c^2 x^2+1\right )}+\frac {e^2 \left (\frac {3 c^2 d}{m+5}-\frac {e}{m+7}\right ) x^{m+3}}{c^4}+\frac {e^3 x^{m+5}}{c^2 (m+7)}\right )dx+\frac {d^3 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {3 d^2 e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {3 d e^2 x^{m+5} (a+b \arctan (c x))}{m+5}+\frac {e^3 x^{m+7} (a+b \arctan (c x))}{m+7}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^3 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {3 d^2 e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {3 d e^2 x^{m+5} (a+b \arctan (c x))}{m+5}+\frac {e^3 x^{m+7} (a+b \arctan (c x))}{m+7}-b c \left (\frac {e^3 x^{m+6}}{c^2 (m+6) (m+7)}+\frac {e^2 x^{m+4} \left (\frac {3 c^2 d}{m+5}-\frac {e}{m+7}\right )}{c^4 (m+4)}+\frac {e x^{m+2} \left (3 c^4 d^2 \left (m^2+12 m+35\right )-3 c^2 d e \left (m^2+10 m+21\right )+e^2 \left (m^2+8 m+15\right )\right )}{c^6 (m+2) (m+3) (m+5) (m+7)}-\frac {x^{m+2} \left (c^6 \left (-d^3\right ) \left (m^3+15 m^2+71 m+105\right )+3 c^4 d^2 e \left (m^3+13 m^2+47 m+35\right )-3 c^2 d e^2 \left (m^3+11 m^2+31 m+21\right )+e^3 \left (m^3+9 m^2+23 m+15\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{c^6 (m+1) (m+2) (m+3) (m+5) (m+7)}\right )\)

Input:

Int[x^m*(d + e*x^2)^3*(a + b*ArcTan[c*x]),x]
 

Output:

(d^3*x^(1 + m)*(a + b*ArcTan[c*x]))/(1 + m) + (3*d^2*e*x^(3 + m)*(a + b*Ar 
cTan[c*x]))/(3 + m) + (3*d*e^2*x^(5 + m)*(a + b*ArcTan[c*x]))/(5 + m) + (e 
^3*x^(7 + m)*(a + b*ArcTan[c*x]))/(7 + m) - b*c*((e*(e^2*(15 + 8*m + m^2) 
- 3*c^2*d*e*(21 + 10*m + m^2) + 3*c^4*d^2*(35 + 12*m + m^2))*x^(2 + m))/(c 
^6*(2 + m)*(3 + m)*(5 + m)*(7 + m)) + (e^2*((3*c^2*d)/(5 + m) - e/(7 + m)) 
*x^(4 + m))/(c^4*(4 + m)) + (e^3*x^(6 + m))/(c^2*(6 + m)*(7 + m)) - ((e^3* 
(15 + 23*m + 9*m^2 + m^3) - 3*c^2*d*e^2*(21 + 31*m + 11*m^2 + m^3) + 3*c^4 
*d^2*e*(35 + 47*m + 13*m^2 + m^3) - c^6*d^3*(105 + 71*m + 15*m^2 + m^3))*x 
^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(c^6*(1 + 
 m)*(2 + m)*(3 + m)*(5 + m)*(7 + m)))
 

Defintions of rubi rules used

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

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]

\[\int x^{m} \left (e \,x^{2}+d \right )^{3} \left (a +b \arctan \left (c x \right )\right )d x\]

Input:

int(x^m*(e*x^2+d)^3*(a+b*arctan(c*x)),x)
 

Output:

int(x^m*(e*x^2+d)^3*(a+b*arctan(c*x)),x)
 

Fricas [F]

\[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^3*(a+b*arctan(c*x)),x, algorithm="fricas")
 

Output:

integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 
 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d^3)*arctan(c*x))*x^m, x)
 

Sympy [F]

\[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}\, dx \] Input:

integrate(x**m*(e*x**2+d)**3*(a+b*atan(c*x)),x)
 

Output:

Integral(x**m*(a + b*atan(c*x))*(d + e*x**2)**3, x)
 

Maxima [F]

\[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^3*(a+b*arctan(c*x)),x, algorithm="maxima")
 

Output:

a*e^3*x^(m + 7)/(m + 7) + 3*a*d*e^2*x^(m + 5)/(m + 5) + 3*a*d^2*e*x^(m + 3 
)/(m + 3) + a*d^3*x^(m + 1)/(m + 1) + (((b*e^3*m^3 + 9*b*e^3*m^2 + 23*b*e^ 
3*m + 15*b*e^3)*x^7 + 3*(b*d*e^2*m^3 + 11*b*d*e^2*m^2 + 31*b*d*e^2*m + 21* 
b*d*e^2)*x^5 + 3*(b*d^2*e*m^3 + 13*b*d^2*e*m^2 + 47*b*d^2*e*m + 35*b*d^2*e 
)*x^3 + (b*d^3*m^3 + 15*b*d^3*m^2 + 71*b*d^3*m + 105*b*d^3)*x)*x^m*arctan( 
c*x) - (m^4 + 16*m^3 + 86*m^2 + 176*m + 105)*integrate(((b*c*e^3*m^3 + 9*b 
*c*e^3*m^2 + 23*b*c*e^3*m + 15*b*c*e^3)*x^7 + 3*(b*c*d*e^2*m^3 + 11*b*c*d* 
e^2*m^2 + 31*b*c*d*e^2*m + 21*b*c*d*e^2)*x^5 + 3*(b*c*d^2*e*m^3 + 13*b*c*d 
^2*e*m^2 + 47*b*c*d^2*e*m + 35*b*c*d^2*e)*x^3 + (b*c*d^3*m^3 + 15*b*c*d^3* 
m^2 + 71*b*c*d^3*m + 105*b*c*d^3)*x)*x^m/(m^4 + 16*m^3 + (c^2*m^4 + 16*c^2 
*m^3 + 86*c^2*m^2 + 176*c^2*m + 105*c^2)*x^2 + 86*m^2 + 176*m + 105), x))/ 
(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)
 

Giac [F]

\[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(e*x^2+d)^3*(a+b*arctan(c*x)),x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)^3*(b*arctan(c*x) + a)*x^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^3 \,d x \] Input:

int(x^m*(a + b*atan(c*x))*(d + e*x^2)^3,x)
 

Output:

int(x^m*(a + b*atan(c*x))*(d + e*x^2)^3, x)
 

Reduce [F]

\[ \int x^m \left (d+e x^2\right )^3 (a+b \arctan (c x)) \, dx=\text {too large to display} \] Input:

int(x^m*(e*x^2+d)^3*(a+b*atan(c*x)),x)
 

Output:

(x**m*atan(c*x)*b*c**7*d**3*m**7*x + 27*x**m*atan(c*x)*b*c**7*d**3*m**6*x 
+ 295*x**m*atan(c*x)*b*c**7*d**3*m**5*x + 1665*x**m*atan(c*x)*b*c**7*d**3* 
m**4*x + 5104*x**m*atan(c*x)*b*c**7*d**3*m**3*x + 8028*x**m*atan(c*x)*b*c* 
*7*d**3*m**2*x + 5040*x**m*atan(c*x)*b*c**7*d**3*m*x + 3*x**m*atan(c*x)*b* 
c**7*d**2*e*m**7*x**3 + 75*x**m*atan(c*x)*b*c**7*d**2*e*m**6*x**3 + 741*x* 
*m*atan(c*x)*b*c**7*d**2*e*m**5*x**3 + 3657*x**m*atan(c*x)*b*c**7*d**2*e*m 
**4*x**3 + 9336*x**m*atan(c*x)*b*c**7*d**2*e*m**3*x**3 + 11388*x**m*atan(c 
*x)*b*c**7*d**2*e*m**2*x**3 + 5040*x**m*atan(c*x)*b*c**7*d**2*e*m*x**3 + 3 
*x**m*atan(c*x)*b*c**7*d*e**2*m**7*x**5 + 69*x**m*atan(c*x)*b*c**7*d*e**2* 
m**6*x**5 + 621*x**m*atan(c*x)*b*c**7*d*e**2*m**5*x**5 + 2775*x**m*atan(c* 
x)*b*c**7*d*e**2*m**4*x**5 + 6432*x**m*atan(c*x)*b*c**7*d*e**2*m**3*x**5 + 
 7236*x**m*atan(c*x)*b*c**7*d*e**2*m**2*x**5 + 3024*x**m*atan(c*x)*b*c**7* 
d*e**2*m*x**5 + x**m*atan(c*x)*b*c**7*e**3*m**7*x**7 + 21*x**m*atan(c*x)*b 
*c**7*e**3*m**6*x**7 + 175*x**m*atan(c*x)*b*c**7*e**3*m**5*x**7 + 735*x**m 
*atan(c*x)*b*c**7*e**3*m**4*x**7 + 1624*x**m*atan(c*x)*b*c**7*e**3*m**3*x* 
*7 + 1764*x**m*atan(c*x)*b*c**7*e**3*m**2*x**7 + 720*x**m*atan(c*x)*b*c**7 
*e**3*m*x**7 + x**m*a*c**7*d**3*m**7*x + 27*x**m*a*c**7*d**3*m**6*x + 295* 
x**m*a*c**7*d**3*m**5*x + 1665*x**m*a*c**7*d**3*m**4*x + 5104*x**m*a*c**7* 
d**3*m**3*x + 8028*x**m*a*c**7*d**3*m**2*x + 5040*x**m*a*c**7*d**3*m*x + 3 
*x**m*a*c**7*d**2*e*m**7*x**3 + 75*x**m*a*c**7*d**2*e*m**6*x**3 + 741*x...