3.1.34 \(\int (d+e x) (a+c x^2)^3 (A+B x+C x^2) \, dx\) [34]

3.1.34.1 Optimal result
3.1.34.2 Mathematica [A] (verified)
3.1.34.3 Rubi [A] (verified)
3.1.34.4 Maple [A] (verified)
3.1.34.5 Fricas [A] (verification not implemented)
3.1.34.6 Sympy [A] (verification not implemented)
3.1.34.7 Maxima [A] (verification not implemented)
3.1.34.8 Giac [A] (verification not implemented)
3.1.34.9 Mupad [B] (verification not implemented)

3.1.34.1 Optimal result

Integrand size = 25, antiderivative size = 169 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=a^3 A d x+\frac {1}{3} a^2 (3 A c d+a C d+a B e) x^3+\frac {1}{4} a^3 C e x^4+\frac {3}{5} a c (A c d+a C d+a B e) x^5+\frac {1}{2} a^2 c C e x^6+\frac {1}{7} c^2 (A c d+3 a (C d+B e)) x^7+\frac {3}{8} a c^2 C e x^8+\frac {1}{9} c^3 (C d+B e) x^9+\frac {1}{10} c^3 C e x^{10}+\frac {(B d+A e) \left (a+c x^2\right )^4}{8 c} \]

output
a^3*A*d*x+1/3*a^2*(3*A*c*d+B*a*e+C*a*d)*x^3+1/4*a^3*C*e*x^4+3/5*a*c*(A*c*d 
+B*a*e+C*a*d)*x^5+1/2*a^2*c*C*e*x^6+1/7*c^2*(A*c*d+3*a*(B*e+C*d))*x^7+3/8* 
a*c^2*C*e*x^8+1/9*c^3*(B*e+C*d)*x^9+1/10*c^3*C*e*x^10+1/8*(A*e+B*d)*(c*x^2 
+a)^4/c
 
3.1.34.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.16 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=a^3 A d x+\frac {1}{2} a^3 (B d+A e) x^2+\frac {1}{3} a^2 (3 A c d+a C d+a B e) x^3+\frac {1}{4} a^2 (3 B c d+3 A c e+a C e) x^4+\frac {3}{5} a c (A c d+a C d+a B e) x^5+\frac {1}{2} a c (B c d+A c e+a C e) x^6+\frac {1}{7} c^2 (A c d+3 a C d+3 a B e) x^7+\frac {1}{8} c^2 (B c d+A c e+3 a C e) x^8+\frac {1}{9} c^3 (C d+B e) x^9+\frac {1}{10} c^3 C e x^{10} \]

input
Integrate[(d + e*x)*(a + c*x^2)^3*(A + B*x + C*x^2),x]
 
output
a^3*A*d*x + (a^3*(B*d + A*e)*x^2)/2 + (a^2*(3*A*c*d + a*C*d + a*B*e)*x^3)/ 
3 + (a^2*(3*B*c*d + 3*A*c*e + a*C*e)*x^4)/4 + (3*a*c*(A*c*d + a*C*d + a*B* 
e)*x^5)/5 + (a*c*(B*c*d + A*c*e + a*C*e)*x^6)/2 + (c^2*(A*c*d + 3*a*C*d + 
3*a*B*e)*x^7)/7 + (c^2*(B*c*d + A*c*e + 3*a*C*e)*x^8)/8 + (c^3*(C*d + B*e) 
*x^9)/9 + (c^3*C*e*x^10)/10
 
3.1.34.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2017, 2341, 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.

\(\displaystyle \int \left (a+c x^2\right )^3 (d+e x) \left (A+B x+C x^2\right ) \, dx\)

\(\Big \downarrow \) 2017

\(\displaystyle \int \left (c x^2+a\right )^3 \left ((d+e x) \left (C x^2+B x+A\right )-(B d+A e) x\right )dx+\frac {\left (a+c x^2\right )^4 (A e+B d)}{8 c}\)

\(\Big \downarrow \) 2341

\(\displaystyle \int \left (c^3 C e x^9+c^3 (C d+B e) x^8+3 a c^2 C e x^7+c^2 (A c d+3 a (C d+B e)) x^6+3 a^2 c C e x^5+3 a c (A c d+a C d+a B e) x^4+a^3 C e x^3+a^2 (3 A c d+a C d+a B e) x^2+a^3 A d\right )dx+\frac {\left (a+c x^2\right )^4 (A e+B d)}{8 c}\)

\(\Big \downarrow \) 2009

\(\displaystyle a^3 A d x+\frac {1}{4} a^3 C e x^4+\frac {1}{3} a^2 x^3 (a B e+a C d+3 A c d)+\frac {1}{2} a^2 c C e x^6+\frac {1}{7} c^2 x^7 (3 a (B e+C d)+A c d)+\frac {3}{5} a c x^5 (a B e+a C d+A c d)+\frac {\left (a+c x^2\right )^4 (A e+B d)}{8 c}+\frac {3}{8} a c^2 C e x^8+\frac {1}{9} c^3 x^9 (B e+C d)+\frac {1}{10} c^3 C e x^{10}\)

input
Int[(d + e*x)*(a + c*x^2)^3*(A + B*x + C*x^2),x]
 
output
a^3*A*d*x + (a^2*(3*A*c*d + a*C*d + a*B*e)*x^3)/3 + (a^3*C*e*x^4)/4 + (3*a 
*c*(A*c*d + a*C*d + a*B*e)*x^5)/5 + (a^2*c*C*e*x^6)/2 + (c^2*(A*c*d + 3*a* 
(C*d + B*e))*x^7)/7 + (3*a*c^2*C*e*x^8)/8 + (c^3*(C*d + B*e)*x^9)/9 + (c^3 
*C*e*x^10)/10 + ((B*d + A*e)*(a + c*x^2)^4)/(8*c)
 

3.1.34.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2017
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 
 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] 
*x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p 
, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n 
 - 1]*x^(n - 1)] &&  !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ 
[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a 
+ b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
 

rule 2341
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 
3.1.34.4 Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.32

method result size
default \(\frac {c^{3} C e \,x^{10}}{10}+\frac {\left (e \,c^{3} B +c^{3} d C \right ) x^{9}}{9}+\frac {\left (e \,c^{3} A +c^{3} d B +3 a \,c^{2} e C \right ) x^{8}}{8}+\frac {\left (A d \,c^{3}+3 a \,c^{2} e B +3 a \,c^{2} d C \right ) x^{7}}{7}+\frac {\left (3 A a \,c^{2} e +3 B a \,c^{2} d +3 a^{2} c e C \right ) x^{6}}{6}+\frac {\left (3 A a \,c^{2} d +3 a^{2} c e B +3 a^{2} c d C \right ) x^{5}}{5}+\frac {\left (3 A \,a^{2} c e +3 B \,a^{2} c d +a^{3} e C \right ) x^{4}}{4}+\frac {\left (3 a^{2} c d A +B e \,a^{3}+d \,a^{3} C \right ) x^{3}}{3}+\frac {\left (A \,a^{3} e +B \,a^{3} d \right ) x^{2}}{2}+a^{3} A d x\) \(223\)
norman \(\frac {c^{3} C e \,x^{10}}{10}+\left (\frac {1}{9} e \,c^{3} B +\frac {1}{9} c^{3} d C \right ) x^{9}+\left (\frac {1}{8} e \,c^{3} A +\frac {1}{8} c^{3} d B +\frac {3}{8} a \,c^{2} e C \right ) x^{8}+\left (\frac {1}{7} A d \,c^{3}+\frac {3}{7} a \,c^{2} e B +\frac {3}{7} a \,c^{2} d C \right ) x^{7}+\left (\frac {1}{2} A a \,c^{2} e +\frac {1}{2} B a \,c^{2} d +\frac {1}{2} a^{2} c e C \right ) x^{6}+\left (\frac {3}{5} A a \,c^{2} d +\frac {3}{5} a^{2} c e B +\frac {3}{5} a^{2} c d C \right ) x^{5}+\left (\frac {3}{4} A \,a^{2} c e +\frac {3}{4} B \,a^{2} c d +\frac {1}{4} a^{3} e C \right ) x^{4}+\left (a^{2} c d A +\frac {1}{3} B e \,a^{3}+\frac {1}{3} d \,a^{3} C \right ) x^{3}+\left (\frac {1}{2} A \,a^{3} e +\frac {1}{2} B \,a^{3} d \right ) x^{2}+a^{3} A d x\) \(224\)
gosper \(\frac {1}{10} c^{3} C e \,x^{10}+\frac {1}{9} B \,c^{3} e \,x^{9}+\frac {1}{9} x^{9} c^{3} d C +\frac {1}{8} x^{8} A \,c^{3} e +\frac {1}{8} x^{8} B \,c^{3} d +\frac {3}{8} a \,c^{2} C e \,x^{8}+\frac {1}{7} x^{7} A d \,c^{3}+\frac {3}{7} x^{7} B e a \,c^{2}+\frac {3}{7} x^{7} a \,c^{2} d C +\frac {1}{2} x^{6} A a \,c^{2} e +\frac {1}{2} x^{6} B a \,c^{2} d +\frac {1}{2} a^{2} c C e \,x^{6}+\frac {3}{5} x^{5} d A a \,c^{2}+\frac {3}{5} x^{5} B e c \,a^{2}+\frac {3}{5} x^{5} a^{2} c d C +\frac {3}{4} x^{4} A \,a^{2} c e +\frac {3}{4} x^{4} B \,a^{2} c d +\frac {1}{4} a^{3} C e \,x^{4}+x^{3} d A c \,a^{2}+\frac {1}{3} x^{3} B e \,a^{3}+\frac {1}{3} x^{3} d \,a^{3} C +\frac {1}{2} x^{2} A \,a^{3} e +\frac {1}{2} x^{2} B \,a^{3} d +a^{3} A d x\) \(250\)
risch \(\frac {1}{10} c^{3} C e \,x^{10}+\frac {1}{9} B \,c^{3} e \,x^{9}+\frac {1}{9} x^{9} c^{3} d C +\frac {1}{8} x^{8} A \,c^{3} e +\frac {1}{8} x^{8} B \,c^{3} d +\frac {3}{8} a \,c^{2} C e \,x^{8}+\frac {1}{7} x^{7} A d \,c^{3}+\frac {3}{7} x^{7} B e a \,c^{2}+\frac {3}{7} x^{7} a \,c^{2} d C +\frac {1}{2} x^{6} A a \,c^{2} e +\frac {1}{2} x^{6} B a \,c^{2} d +\frac {1}{2} a^{2} c C e \,x^{6}+\frac {3}{5} x^{5} d A a \,c^{2}+\frac {3}{5} x^{5} B e c \,a^{2}+\frac {3}{5} x^{5} a^{2} c d C +\frac {3}{4} x^{4} A \,a^{2} c e +\frac {3}{4} x^{4} B \,a^{2} c d +\frac {1}{4} a^{3} C e \,x^{4}+x^{3} d A c \,a^{2}+\frac {1}{3} x^{3} B e \,a^{3}+\frac {1}{3} x^{3} d \,a^{3} C +\frac {1}{2} x^{2} A \,a^{3} e +\frac {1}{2} x^{2} B \,a^{3} d +a^{3} A d x\) \(250\)
parallelrisch \(\frac {1}{10} c^{3} C e \,x^{10}+\frac {1}{9} B \,c^{3} e \,x^{9}+\frac {1}{9} x^{9} c^{3} d C +\frac {1}{8} x^{8} A \,c^{3} e +\frac {1}{8} x^{8} B \,c^{3} d +\frac {3}{8} a \,c^{2} C e \,x^{8}+\frac {1}{7} x^{7} A d \,c^{3}+\frac {3}{7} x^{7} B e a \,c^{2}+\frac {3}{7} x^{7} a \,c^{2} d C +\frac {1}{2} x^{6} A a \,c^{2} e +\frac {1}{2} x^{6} B a \,c^{2} d +\frac {1}{2} a^{2} c C e \,x^{6}+\frac {3}{5} x^{5} d A a \,c^{2}+\frac {3}{5} x^{5} B e c \,a^{2}+\frac {3}{5} x^{5} a^{2} c d C +\frac {3}{4} x^{4} A \,a^{2} c e +\frac {3}{4} x^{4} B \,a^{2} c d +\frac {1}{4} a^{3} C e \,x^{4}+x^{3} d A c \,a^{2}+\frac {1}{3} x^{3} B e \,a^{3}+\frac {1}{3} x^{3} d \,a^{3} C +\frac {1}{2} x^{2} A \,a^{3} e +\frac {1}{2} x^{2} B \,a^{3} d +a^{3} A d x\) \(250\)

input
int((e*x+d)*(c*x^2+a)^3*(C*x^2+B*x+A),x,method=_RETURNVERBOSE)
 
output
1/10*c^3*C*e*x^10+1/9*(B*c^3*e+C*c^3*d)*x^9+1/8*(A*c^3*e+B*c^3*d+3*C*a*c^2 
*e)*x^8+1/7*(A*c^3*d+3*B*a*c^2*e+3*C*a*c^2*d)*x^7+1/6*(3*A*a*c^2*e+3*B*a*c 
^2*d+3*C*a^2*c*e)*x^6+1/5*(3*A*a*c^2*d+3*B*a^2*c*e+3*C*a^2*c*d)*x^5+1/4*(3 
*A*a^2*c*e+3*B*a^2*c*d+C*a^3*e)*x^4+1/3*(3*A*a^2*c*d+B*a^3*e+C*a^3*d)*x^3+ 
1/2*(A*a^3*e+B*a^3*d)*x^2+a^3*A*d*x
 
3.1.34.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.31 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{10} \, C c^{3} e x^{10} + \frac {1}{9} \, {\left (C c^{3} d + B c^{3} e\right )} x^{9} + \frac {1}{8} \, {\left (B c^{3} d + {\left (3 \, C a c^{2} + A c^{3}\right )} e\right )} x^{8} + \frac {1}{7} \, {\left (3 \, B a c^{2} e + {\left (3 \, C a c^{2} + A c^{3}\right )} d\right )} x^{7} + \frac {1}{2} \, {\left (B a c^{2} d + {\left (C a^{2} c + A a c^{2}\right )} e\right )} x^{6} + A a^{3} d x + \frac {3}{5} \, {\left (B a^{2} c e + {\left (C a^{2} c + A a c^{2}\right )} d\right )} x^{5} + \frac {1}{4} \, {\left (3 \, B a^{2} c d + {\left (C a^{3} + 3 \, A a^{2} c\right )} e\right )} x^{4} + \frac {1}{3} \, {\left (B a^{3} e + {\left (C a^{3} + 3 \, A a^{2} c\right )} d\right )} x^{3} + \frac {1}{2} \, {\left (B a^{3} d + A a^{3} e\right )} x^{2} \]

input
integrate((e*x+d)*(c*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="fricas")
 
output
1/10*C*c^3*e*x^10 + 1/9*(C*c^3*d + B*c^3*e)*x^9 + 1/8*(B*c^3*d + (3*C*a*c^ 
2 + A*c^3)*e)*x^8 + 1/7*(3*B*a*c^2*e + (3*C*a*c^2 + A*c^3)*d)*x^7 + 1/2*(B 
*a*c^2*d + (C*a^2*c + A*a*c^2)*e)*x^6 + A*a^3*d*x + 3/5*(B*a^2*c*e + (C*a^ 
2*c + A*a*c^2)*d)*x^5 + 1/4*(3*B*a^2*c*d + (C*a^3 + 3*A*a^2*c)*e)*x^4 + 1/ 
3*(B*a^3*e + (C*a^3 + 3*A*a^2*c)*d)*x^3 + 1/2*(B*a^3*d + A*a^3*e)*x^2
 
3.1.34.6 Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.57 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=A a^{3} d x + \frac {C c^{3} e x^{10}}{10} + x^{9} \left (\frac {B c^{3} e}{9} + \frac {C c^{3} d}{9}\right ) + x^{8} \left (\frac {A c^{3} e}{8} + \frac {B c^{3} d}{8} + \frac {3 C a c^{2} e}{8}\right ) + x^{7} \left (\frac {A c^{3} d}{7} + \frac {3 B a c^{2} e}{7} + \frac {3 C a c^{2} d}{7}\right ) + x^{6} \left (\frac {A a c^{2} e}{2} + \frac {B a c^{2} d}{2} + \frac {C a^{2} c e}{2}\right ) + x^{5} \cdot \left (\frac {3 A a c^{2} d}{5} + \frac {3 B a^{2} c e}{5} + \frac {3 C a^{2} c d}{5}\right ) + x^{4} \cdot \left (\frac {3 A a^{2} c e}{4} + \frac {3 B a^{2} c d}{4} + \frac {C a^{3} e}{4}\right ) + x^{3} \left (A a^{2} c d + \frac {B a^{3} e}{3} + \frac {C a^{3} d}{3}\right ) + x^{2} \left (\frac {A a^{3} e}{2} + \frac {B a^{3} d}{2}\right ) \]

input
integrate((e*x+d)*(c*x**2+a)**3*(C*x**2+B*x+A),x)
 
output
A*a**3*d*x + C*c**3*e*x**10/10 + x**9*(B*c**3*e/9 + C*c**3*d/9) + x**8*(A* 
c**3*e/8 + B*c**3*d/8 + 3*C*a*c**2*e/8) + x**7*(A*c**3*d/7 + 3*B*a*c**2*e/ 
7 + 3*C*a*c**2*d/7) + x**6*(A*a*c**2*e/2 + B*a*c**2*d/2 + C*a**2*c*e/2) + 
x**5*(3*A*a*c**2*d/5 + 3*B*a**2*c*e/5 + 3*C*a**2*c*d/5) + x**4*(3*A*a**2*c 
*e/4 + 3*B*a**2*c*d/4 + C*a**3*e/4) + x**3*(A*a**2*c*d + B*a**3*e/3 + C*a* 
*3*d/3) + x**2*(A*a**3*e/2 + B*a**3*d/2)
 
3.1.34.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.31 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{10} \, C c^{3} e x^{10} + \frac {1}{9} \, {\left (C c^{3} d + B c^{3} e\right )} x^{9} + \frac {1}{8} \, {\left (B c^{3} d + {\left (3 \, C a c^{2} + A c^{3}\right )} e\right )} x^{8} + \frac {1}{7} \, {\left (3 \, B a c^{2} e + {\left (3 \, C a c^{2} + A c^{3}\right )} d\right )} x^{7} + \frac {1}{2} \, {\left (B a c^{2} d + {\left (C a^{2} c + A a c^{2}\right )} e\right )} x^{6} + A a^{3} d x + \frac {3}{5} \, {\left (B a^{2} c e + {\left (C a^{2} c + A a c^{2}\right )} d\right )} x^{5} + \frac {1}{4} \, {\left (3 \, B a^{2} c d + {\left (C a^{3} + 3 \, A a^{2} c\right )} e\right )} x^{4} + \frac {1}{3} \, {\left (B a^{3} e + {\left (C a^{3} + 3 \, A a^{2} c\right )} d\right )} x^{3} + \frac {1}{2} \, {\left (B a^{3} d + A a^{3} e\right )} x^{2} \]

input
integrate((e*x+d)*(c*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="maxima")
 
output
1/10*C*c^3*e*x^10 + 1/9*(C*c^3*d + B*c^3*e)*x^9 + 1/8*(B*c^3*d + (3*C*a*c^ 
2 + A*c^3)*e)*x^8 + 1/7*(3*B*a*c^2*e + (3*C*a*c^2 + A*c^3)*d)*x^7 + 1/2*(B 
*a*c^2*d + (C*a^2*c + A*a*c^2)*e)*x^6 + A*a^3*d*x + 3/5*(B*a^2*c*e + (C*a^ 
2*c + A*a*c^2)*d)*x^5 + 1/4*(3*B*a^2*c*d + (C*a^3 + 3*A*a^2*c)*e)*x^4 + 1/ 
3*(B*a^3*e + (C*a^3 + 3*A*a^2*c)*d)*x^3 + 1/2*(B*a^3*d + A*a^3*e)*x^2
 
3.1.34.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.47 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{10} \, C c^{3} e x^{10} + \frac {1}{9} \, C c^{3} d x^{9} + \frac {1}{9} \, B c^{3} e x^{9} + \frac {1}{8} \, B c^{3} d x^{8} + \frac {3}{8} \, C a c^{2} e x^{8} + \frac {1}{8} \, A c^{3} e x^{8} + \frac {3}{7} \, C a c^{2} d x^{7} + \frac {1}{7} \, A c^{3} d x^{7} + \frac {3}{7} \, B a c^{2} e x^{7} + \frac {1}{2} \, B a c^{2} d x^{6} + \frac {1}{2} \, C a^{2} c e x^{6} + \frac {1}{2} \, A a c^{2} e x^{6} + \frac {3}{5} \, C a^{2} c d x^{5} + \frac {3}{5} \, A a c^{2} d x^{5} + \frac {3}{5} \, B a^{2} c e x^{5} + \frac {3}{4} \, B a^{2} c d x^{4} + \frac {1}{4} \, C a^{3} e x^{4} + \frac {3}{4} \, A a^{2} c e x^{4} + \frac {1}{3} \, C a^{3} d x^{3} + A a^{2} c d x^{3} + \frac {1}{3} \, B a^{3} e x^{3} + \frac {1}{2} \, B a^{3} d x^{2} + \frac {1}{2} \, A a^{3} e x^{2} + A a^{3} d x \]

input
integrate((e*x+d)*(c*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="giac")
 
output
1/10*C*c^3*e*x^10 + 1/9*C*c^3*d*x^9 + 1/9*B*c^3*e*x^9 + 1/8*B*c^3*d*x^8 + 
3/8*C*a*c^2*e*x^8 + 1/8*A*c^3*e*x^8 + 3/7*C*a*c^2*d*x^7 + 1/7*A*c^3*d*x^7 
+ 3/7*B*a*c^2*e*x^7 + 1/2*B*a*c^2*d*x^6 + 1/2*C*a^2*c*e*x^6 + 1/2*A*a*c^2* 
e*x^6 + 3/5*C*a^2*c*d*x^5 + 3/5*A*a*c^2*d*x^5 + 3/5*B*a^2*c*e*x^5 + 3/4*B* 
a^2*c*d*x^4 + 1/4*C*a^3*e*x^4 + 3/4*A*a^2*c*e*x^4 + 1/3*C*a^3*d*x^3 + A*a^ 
2*c*d*x^3 + 1/3*B*a^3*e*x^3 + 1/2*B*a^3*d*x^2 + 1/2*A*a^3*e*x^2 + A*a^3*d* 
x
 
3.1.34.9 Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=x^3\,\left (\frac {B\,a^3\,e}{3}+\frac {C\,a^3\,d}{3}+A\,a^2\,c\,d\right )+x^8\,\left (\frac {A\,c^3\,e}{8}+\frac {B\,c^3\,d}{8}+\frac {3\,C\,a\,c^2\,e}{8}\right )+\frac {a^3\,x^2\,\left (A\,e+B\,d\right )}{2}+\frac {c^3\,x^9\,\left (B\,e+C\,d\right )}{9}+\frac {c^2\,x^7\,\left (A\,c\,d+3\,B\,a\,e+3\,C\,a\,d\right )}{7}+\frac {a^2\,x^4\,\left (3\,A\,c\,e+3\,B\,c\,d+C\,a\,e\right )}{4}+A\,a^3\,d\,x+\frac {3\,a\,c\,x^5\,\left (A\,c\,d+B\,a\,e+C\,a\,d\right )}{5}+\frac {a\,c\,x^6\,\left (A\,c\,e+B\,c\,d+C\,a\,e\right )}{2}+\frac {C\,c^3\,e\,x^{10}}{10} \]

input
int((a + c*x^2)^3*(d + e*x)*(A + B*x + C*x^2),x)
 
output
x^3*((B*a^3*e)/3 + (C*a^3*d)/3 + A*a^2*c*d) + x^8*((A*c^3*e)/8 + (B*c^3*d) 
/8 + (3*C*a*c^2*e)/8) + (a^3*x^2*(A*e + B*d))/2 + (c^3*x^9*(B*e + C*d))/9 
+ (c^2*x^7*(A*c*d + 3*B*a*e + 3*C*a*d))/7 + (a^2*x^4*(3*A*c*e + 3*B*c*d + 
C*a*e))/4 + A*a^3*d*x + (3*a*c*x^5*(A*c*d + B*a*e + C*a*d))/5 + (a*c*x^6*( 
A*c*e + B*c*d + C*a*e))/2 + (C*c^3*e*x^10)/10