\(\int (b d+2 c d x)^5 (a+b x+c x^2)^p \, dx\) [332]

Optimal result

Integrand size = 24, antiderivative size = 121 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {2 \left (b^2-4 a c\right )^2 d^5 \left (a+b x+c x^2\right )^{1+p}}{(1+p) (2+p) (3+p)}+\frac {2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^2 \left (a+b x+c x^2\right )^{1+p}}{(2+p) (3+p)}+\frac {d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{1+p}}{3+p} \] Output:

2*(-4*a*c+b^2)^2*d^5*(c*x^2+b*x+a)^(p+1)/(p+1)/(2+p)/(3+p)+2*(-4*a*c+b^2)* 
d^5*(2*c*x+b)^2*(c*x^2+b*x+a)^(p+1)/(2+p)/(3+p)+d^5*(2*c*x+b)^4*(c*x^2+b*x 
+a)^(p+1)/(3+p)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.20 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {d^5 (a+x (b+c x))^{1+p} \left (b^4 \left (6+5 p+p^2\right )+8 b^3 c \left (3+4 p+p^2\right ) x-32 b c^2 (1+p) x \left (a-c (2+p) x^2\right )-8 b^2 c \left (a (3+p)-c \left (7+10 p+3 p^2\right ) x^2\right )+16 c^2 \left (2 a^2-2 a c (1+p) x^2+c^2 \left (2+3 p+p^2\right ) x^4\right )\right )}{(1+p) (2+p) (3+p)} \] Input:

Integrate[(b*d + 2*c*d*x)^5*(a + b*x + c*x^2)^p,x]
 

Output:

(d^5*(a + x*(b + c*x))^(1 + p)*(b^4*(6 + 5*p + p^2) + 8*b^3*c*(3 + 4*p + p 
^2)*x - 32*b*c^2*(1 + p)*x*(a - c*(2 + p)*x^2) - 8*b^2*c*(a*(3 + p) - c*(7 
 + 10*p + 3*p^2)*x^2) + 16*c^2*(2*a^2 - 2*a*c*(1 + p)*x^2 + c^2*(2 + 3*p + 
 p^2)*x^4)))/((1 + p)*(2 + p)*(3 + p))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1116, 27, 1116, 1104}

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.

\(\displaystyle \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {2 d^2 \left (b^2-4 a c\right ) \int d^3 (b+2 c x)^3 \left (c x^2+b x+a\right )^pdx}{p+3}+\frac {d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 d^5 \left (b^2-4 a c\right ) \int (b+2 c x)^3 \left (c x^2+b x+a\right )^pdx}{p+3}+\frac {d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}\)

\(\Big \downarrow \) 1116

\(\displaystyle \frac {2 d^5 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \int (b+2 c x) \left (c x^2+b x+a\right )^pdx}{p+2}+\frac {(b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{p+2}\right )}{p+3}+\frac {d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}\)

\(\Big \downarrow \) 1104

\(\displaystyle \frac {2 d^5 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{p+1}}{(p+1) (p+2)}+\frac {(b+2 c x)^2 \left (a+b x+c x^2\right )^{p+1}}{p+2}\right )}{p+3}+\frac {d^5 (b+2 c x)^4 \left (a+b x+c x^2\right )^{p+1}}{p+3}\)

Input:

Int[(b*d + 2*c*d*x)^5*(a + b*x + c*x^2)^p,x]
 

Output:

(d^5*(b + 2*c*x)^4*(a + b*x + c*x^2)^(1 + p))/(3 + p) + (2*(b^2 - 4*a*c)*d 
^5*(((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1 + p))/((1 + p)*(2 + p)) + ((b + 2* 
c*x)^2*(a + b*x + c*x^2)^(1 + p))/(2 + p)))/(3 + p)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1104

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol 
] :> Simp[d*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, 
d, e, p}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.93

method	result	size
gosper	\(\frac {d^{5} \left (c \,x^{2}+b x +a \right )^{p +1} \left (16 c^{4} p^{2} x^{4}+32 b \,c^{3} p^{2} x^{3}+48 c^{4} p \,x^{4}+24 b^{2} c^{2} p^{2} x^{2}+96 b \,c^{3} p \,x^{3}+32 c^{4} x^{4}-32 a \,c^{3} p \,x^{2}+8 b^{3} c \,p^{2} x +80 b^{2} c^{2} p \,x^{2}+64 b \,c^{3} x^{3}-32 a b \,c^{2} p x -32 a \,c^{3} x^{2}+b^{4} p^{2}+32 b^{3} c p x +56 b^{2} c^{2} x^{2}-8 a \,b^{2} c p -32 a b \,c^{2} x +5 b^{4} p +24 b^{3} c x +32 a^{2} c^{2}-24 c a \,b^{2}+6 b^{4}\right )}{p^{3}+6 p^{2}+11 p +6}\)	\(233\)
orering	\(\frac {\left (16 c^{4} p^{2} x^{4}+32 b \,c^{3} p^{2} x^{3}+48 c^{4} p \,x^{4}+24 b^{2} c^{2} p^{2} x^{2}+96 b \,c^{3} p \,x^{3}+32 c^{4} x^{4}-32 a \,c^{3} p \,x^{2}+8 b^{3} c \,p^{2} x +80 b^{2} c^{2} p \,x^{2}+64 b \,c^{3} x^{3}-32 a b \,c^{2} p x -32 a \,c^{3} x^{2}+b^{4} p^{2}+32 b^{3} c p x +56 b^{2} c^{2} x^{2}-8 a \,b^{2} c p -32 a b \,c^{2} x +5 b^{4} p +24 b^{3} c x +32 a^{2} c^{2}-24 c a \,b^{2}+6 b^{4}\right ) \left (c \,x^{2}+b x +a \right ) \left (2 c d x +b d \right )^{5} \left (c \,x^{2}+b x +a \right )^{p}}{\left (p^{3}+6 p^{2}+11 p +6\right ) \left (2 c x +b \right )^{5}}\)	\(257\)
risch	\(\frac {d^{5} \left (16 c^{5} p^{2} x^{6}+48 b \,c^{4} p^{2} x^{5}+48 p \,c^{5} x^{6}+16 a \,c^{4} p^{2} x^{4}+56 b^{2} c^{3} p^{2} x^{4}+144 p b \,c^{4} x^{5}+32 c^{5} x^{6}+32 a b \,c^{3} p^{2} x^{3}+16 a \,c^{4} p \,x^{4}+32 b^{3} c^{2} p^{2} x^{3}+176 b^{2} c^{3} p \,x^{4}+96 b \,x^{5} c^{4}+24 a \,b^{2} c^{2} p^{2} x^{2}+32 a b \,c^{3} p \,x^{3}+9 b^{4} c \,p^{2} x^{2}+112 b^{3} c^{2} p \,x^{3}+120 b^{2} x^{4} c^{3}-32 a^{2} c^{3} p \,x^{2}+8 a \,b^{3} c \,p^{2} x +40 a \,b^{2} c^{2} p \,x^{2}+b^{5} p^{2} x +37 b^{4} c p \,x^{2}+80 b^{3} c^{2} x^{3}-32 a^{2} b \,c^{2} p x +a \,b^{4} p^{2}+24 a \,b^{3} c p x +5 b^{5} p x +30 b^{4} c \,x^{2}-8 a^{2} b^{2} c p +5 a \,b^{4} p +6 b^{5} x +32 c^{2} a^{3}-24 a^{2} b^{2} c +6 a \,b^{4}\right ) \left (c \,x^{2}+b x +a \right )^{p}}{\left (2+p \right ) \left (3+p \right ) \left (p +1\right )}\)	\(383\)
norman	\(\frac {a \,d^{5} \left (b^{4} p^{2}-8 a \,b^{2} c p +5 b^{4} p +32 a^{2} c^{2}-24 c a \,b^{2}+6 b^{4}\right ) {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{3}+6 p^{2}+11 p +6}+\frac {16 d^{5} c^{5} x^{6} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{3+p}+\frac {48 b \,c^{4} d^{5} x^{5} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{3+p}-\frac {b \,d^{5} \left (-8 a \,b^{2} c \,p^{2}-b^{4} p^{2}+32 a^{2} c^{2} p -24 a \,b^{2} c p -5 b^{4} p -6 b^{4}\right ) x \,{\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{3}+6 p^{2}+11 p +6}-\frac {d^{5} c \left (-24 a \,b^{2} c \,p^{2}-9 b^{4} p^{2}+32 a^{2} c^{2} p -40 a \,b^{2} c p -37 b^{4} p -30 b^{4}\right ) x^{2} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{3}+6 p^{2}+11 p +6}+\frac {8 d^{5} c^{3} \left (2 a c p +7 b^{2} p +15 b^{2}\right ) x^{4} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{2}+5 p +6}+\frac {16 b \,c^{2} d^{5} \left (2 a c p +2 b^{2} p +5 b^{2}\right ) x^{3} {\mathrm e}^{p \ln \left (c \,x^{2}+b x +a \right )}}{p^{2}+5 p +6}\)	\(412\)
parallelrisch	\(\frac {9 x^{2} \left (c \,x^{2}+b x +a \right )^{p} b^{4} c^{2} d^{5} p^{2}+16 x^{6} \left (c \,x^{2}+b x +a \right )^{p} c^{6} d^{5} p^{2}+48 x^{6} \left (c \,x^{2}+b x +a \right )^{p} c^{6} d^{5} p +96 x^{5} \left (c \,x^{2}+b x +a \right )^{p} b \,c^{5} d^{5}+32 x^{3} \left (c \,x^{2}+b x +a \right )^{p} a b \,c^{4} d^{5} p^{2}+32 x^{3} \left (c \,x^{2}+b x +a \right )^{p} a b \,c^{4} d^{5} p +24 x^{2} \left (c \,x^{2}+b x +a \right )^{p} a \,b^{2} c^{3} d^{5} p^{2}+40 x^{2} \left (c \,x^{2}+b x +a \right )^{p} a \,b^{2} c^{3} d^{5} p +8 x \left (c \,x^{2}+b x +a \right )^{p} a \,b^{3} c^{2} d^{5} p^{2}-32 x \left (c \,x^{2}+b x +a \right )^{p} a^{2} b \,c^{3} d^{5} p +24 x \left (c \,x^{2}+b x +a \right )^{p} a \,b^{3} c^{2} d^{5} p +48 x^{5} \left (c \,x^{2}+b x +a \right )^{p} b \,c^{5} d^{5} p^{2}+144 x^{5} \left (c \,x^{2}+b x +a \right )^{p} b \,c^{5} d^{5} p +16 x^{4} \left (c \,x^{2}+b x +a \right )^{p} a \,c^{5} d^{5} p^{2}+37 x^{2} \left (c \,x^{2}+b x +a \right )^{p} b^{4} c^{2} d^{5} p +x \left (c \,x^{2}+b x +a \right )^{p} b^{5} c \,d^{5} p^{2}+5 x \left (c \,x^{2}+b x +a \right )^{p} b^{5} c \,d^{5} p +\left (c \,x^{2}+b x +a \right )^{p} a \,b^{4} c \,d^{5} p^{2}-8 \left (c \,x^{2}+b x +a \right )^{p} a^{2} b^{2} c^{2} d^{5} p +5 \left (c \,x^{2}+b x +a \right )^{p} a \,b^{4} c \,d^{5} p +112 x^{3} \left (c \,x^{2}+b x +a \right )^{p} b^{3} c^{3} d^{5} p +120 x^{4} \left (c \,x^{2}+b x +a \right )^{p} b^{2} c^{4} d^{5}+80 x^{3} \left (c \,x^{2}+b x +a \right )^{p} b^{3} c^{3} d^{5}+30 x^{2} \left (c \,x^{2}+b x +a \right )^{p} b^{4} c^{2} d^{5}+6 x \left (c \,x^{2}+b x +a \right )^{p} b^{5} c \,d^{5}-24 \left (c \,x^{2}+b x +a \right )^{p} a^{2} b^{2} c^{2} d^{5}+6 \left (c \,x^{2}+b x +a \right )^{p} a \,b^{4} c \,d^{5}+32 x^{3} \left (c \,x^{2}+b x +a \right )^{p} b^{3} c^{3} d^{5} p^{2}+16 x^{4} \left (c \,x^{2}+b x +a \right )^{p} a \,c^{5} d^{5} p +176 x^{4} \left (c \,x^{2}+b x +a \right )^{p} b^{2} c^{4} d^{5} p +56 x^{4} \left (c \,x^{2}+b x +a \right )^{p} b^{2} c^{4} d^{5} p^{2}+32 x^{6} \left (c \,x^{2}+b x +a \right )^{p} c^{6} d^{5}+32 \left (c \,x^{2}+b x +a \right )^{p} a^{3} c^{3} d^{5}-32 x^{2} \left (c \,x^{2}+b x +a \right )^{p} a^{2} c^{4} d^{5} p}{c \left (p^{3}+6 p^{2}+11 p +6\right )}\)	\(901\)

Input:

int((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x,method=_RETURNVERBOSE)
 

Output:

d^5*(c*x^2+b*x+a)^(p+1)/(p^3+6*p^2+11*p+6)*(16*c^4*p^2*x^4+32*b*c^3*p^2*x^ 
3+48*c^4*p*x^4+24*b^2*c^2*p^2*x^2+96*b*c^3*p*x^3+32*c^4*x^4-32*a*c^3*p*x^2 
+8*b^3*c*p^2*x+80*b^2*c^2*p*x^2+64*b*c^3*x^3-32*a*b*c^2*p*x-32*a*c^3*x^2+b 
^4*p^2+32*b^3*c*p*x+56*b^2*c^2*x^2-8*a*b^2*c*p-32*a*b*c^2*x+5*b^4*p+24*b^3 
*c*x+32*a^2*c^2-24*a*b^2*c+6*b^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (121) = 242\).

Time = 0.10 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.31 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {{\left (a b^{4} d^{5} p^{2} + {\left (5 \, a b^{4} - 8 \, a^{2} b^{2} c\right )} d^{5} p + 16 \, {\left (c^{5} d^{5} p^{2} + 3 \, c^{5} d^{5} p + 2 \, c^{5} d^{5}\right )} x^{6} + 2 \, {\left (3 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d^{5} + 48 \, {\left (b c^{4} d^{5} p^{2} + 3 \, b c^{4} d^{5} p + 2 \, b c^{4} d^{5}\right )} x^{5} + 8 \, {\left (15 \, b^{2} c^{3} d^{5} + {\left (7 \, b^{2} c^{3} + 2 \, a c^{4}\right )} d^{5} p^{2} + 2 \, {\left (11 \, b^{2} c^{3} + a c^{4}\right )} d^{5} p\right )} x^{4} + 16 \, {\left (5 \, b^{3} c^{2} d^{5} + 2 \, {\left (b^{3} c^{2} + a b c^{3}\right )} d^{5} p^{2} + {\left (7 \, b^{3} c^{2} + 2 \, a b c^{3}\right )} d^{5} p\right )} x^{3} + {\left (30 \, b^{4} c d^{5} + 3 \, {\left (3 \, b^{4} c + 8 \, a b^{2} c^{2}\right )} d^{5} p^{2} + {\left (37 \, b^{4} c + 40 \, a b^{2} c^{2} - 32 \, a^{2} c^{3}\right )} d^{5} p\right )} x^{2} + {\left (6 \, b^{5} d^{5} + {\left (b^{5} + 8 \, a b^{3} c\right )} d^{5} p^{2} + {\left (5 \, b^{5} + 24 \, a b^{3} c - 32 \, a^{2} b c^{2}\right )} d^{5} p\right )} x\right )} {\left (c x^{2} + b x + a\right )}^{p}}{p^{3} + 6 \, p^{2} + 11 \, p + 6} \] Input:

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

Output:

(a*b^4*d^5*p^2 + (5*a*b^4 - 8*a^2*b^2*c)*d^5*p + 16*(c^5*d^5*p^2 + 3*c^5*d 
^5*p + 2*c^5*d^5)*x^6 + 2*(3*a*b^4 - 12*a^2*b^2*c + 16*a^3*c^2)*d^5 + 48*( 
b*c^4*d^5*p^2 + 3*b*c^4*d^5*p + 2*b*c^4*d^5)*x^5 + 8*(15*b^2*c^3*d^5 + (7* 
b^2*c^3 + 2*a*c^4)*d^5*p^2 + 2*(11*b^2*c^3 + a*c^4)*d^5*p)*x^4 + 16*(5*b^3 
*c^2*d^5 + 2*(b^3*c^2 + a*b*c^3)*d^5*p^2 + (7*b^3*c^2 + 2*a*b*c^3)*d^5*p)* 
x^3 + (30*b^4*c*d^5 + 3*(3*b^4*c + 8*a*b^2*c^2)*d^5*p^2 + (37*b^4*c + 40*a 
*b^2*c^2 - 32*a^2*c^3)*d^5*p)*x^2 + (6*b^5*d^5 + (b^5 + 8*a*b^3*c)*d^5*p^2 
 + (5*b^5 + 24*a*b^3*c - 32*a^2*b*c^2)*d^5*p)*x)*(c*x^2 + b*x + a)^p/(p^3 
+ 6*p^2 + 11*p + 6)
 

Sympy [F(-1)]

Timed out. \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**5*(c*x**2+b*x+a)**p,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (121) = 242\).

Time = 0.08 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.44 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {{\left (16 \, {\left (p^{2} + 3 \, p + 2\right )} c^{5} d^{5} x^{6} + 48 \, {\left (p^{2} + 3 \, p + 2\right )} b c^{4} d^{5} x^{5} + {\left (p^{2} + 5 \, p + 6\right )} a b^{4} d^{5} - 8 \, a^{2} b^{2} c d^{5} {\left (p + 3\right )} + 32 \, a^{3} c^{2} d^{5} + 8 \, {\left ({\left (7 \, p^{2} + 22 \, p + 15\right )} b^{2} c^{3} d^{5} + 2 \, {\left (p^{2} + p\right )} a c^{4} d^{5}\right )} x^{4} + 16 \, {\left ({\left (2 \, p^{2} + 7 \, p + 5\right )} b^{3} c^{2} d^{5} + 2 \, {\left (p^{2} + p\right )} a b c^{3} d^{5}\right )} x^{3} + {\left ({\left (9 \, p^{2} + 37 \, p + 30\right )} b^{4} c d^{5} + 8 \, {\left (3 \, p^{2} + 5 \, p\right )} a b^{2} c^{2} d^{5} - 32 \, a^{2} c^{3} d^{5} p\right )} x^{2} + {\left ({\left (p^{2} + 5 \, p + 6\right )} b^{5} d^{5} + 8 \, {\left (p^{2} + 3 \, p\right )} a b^{3} c d^{5} - 32 \, a^{2} b c^{2} d^{5} p\right )} x\right )} {\left (c x^{2} + b x + a\right )}^{p}}{p^{3} + 6 \, p^{2} + 11 \, p + 6} \] Input:

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

Output:

(16*(p^2 + 3*p + 2)*c^5*d^5*x^6 + 48*(p^2 + 3*p + 2)*b*c^4*d^5*x^5 + (p^2 
+ 5*p + 6)*a*b^4*d^5 - 8*a^2*b^2*c*d^5*(p + 3) + 32*a^3*c^2*d^5 + 8*((7*p^ 
2 + 22*p + 15)*b^2*c^3*d^5 + 2*(p^2 + p)*a*c^4*d^5)*x^4 + 16*((2*p^2 + 7*p 
 + 5)*b^3*c^2*d^5 + 2*(p^2 + p)*a*b*c^3*d^5)*x^3 + ((9*p^2 + 37*p + 30)*b^ 
4*c*d^5 + 8*(3*p^2 + 5*p)*a*b^2*c^2*d^5 - 32*a^2*c^3*d^5*p)*x^2 + ((p^2 + 
5*p + 6)*b^5*d^5 + 8*(p^2 + 3*p)*a*b^3*c*d^5 - 32*a^2*b*c^2*d^5*p)*x)*(c*x 
^2 + b*x + a)^p/(p^3 + 6*p^2 + 11*p + 6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (121) = 242\).

Time = 0.37 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.45 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {{\left (c x^{2} + b x + a\right )}^{p + 1} b^{4} d^{5}}{p + 1} - \frac {8 \, {\left (c x^{2} + b x + a\right )}^{p + 1} a b^{2} c d^{5}}{p + 1} + \frac {16 \, {\left (c x^{2} + b x + a\right )}^{p + 1} a^{2} c^{2} d^{5}}{p + 1} + \frac {8 \, {\left ({\left (c x^{2} + b x + a\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} b^{2} c d^{5} p + 2 \, {\left (c x^{2} + b x + a\right )}^{3} {\left (c x^{2} + b x + a\right )}^{p} c^{2} d^{5} p - 4 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} a c^{2} d^{5} p + 3 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} b^{2} c d^{5} + 4 \, {\left (c x^{2} + b x + a\right )}^{3} {\left (c x^{2} + b x + a\right )}^{p} c^{2} d^{5} - 12 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (c x^{2} + b x + a\right )}^{p} a c^{2} d^{5}\right )}}{p^{2} + 5 \, p + 6} \] Input:

integrate((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

Output:

(c*x^2 + b*x + a)^(p + 1)*b^4*d^5/(p + 1) - 8*(c*x^2 + b*x + a)^(p + 1)*a* 
b^2*c*d^5/(p + 1) + 16*(c*x^2 + b*x + a)^(p + 1)*a^2*c^2*d^5/(p + 1) + 8*( 
(c*x^2 + b*x + a)^2*(c*x^2 + b*x + a)^p*b^2*c*d^5*p + 2*(c*x^2 + b*x + a)^ 
3*(c*x^2 + b*x + a)^p*c^2*d^5*p - 4*(c*x^2 + b*x + a)^2*(c*x^2 + b*x + a)^ 
p*a*c^2*d^5*p + 3*(c*x^2 + b*x + a)^2*(c*x^2 + b*x + a)^p*b^2*c*d^5 + 4*(c 
*x^2 + b*x + a)^3*(c*x^2 + b*x + a)^p*c^2*d^5 - 12*(c*x^2 + b*x + a)^2*(c* 
x^2 + b*x + a)^p*a*c^2*d^5)/(p^2 + 5*p + 6)
 

Mupad [B] (verification not implemented)

Time = 5.97 (sec) , antiderivative size = 375, normalized size of antiderivative = 3.10 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx={\left (c\,x^2+b\,x+a\right )}^p\,\left (\frac {a\,d^5\,\left (32\,a^2\,c^2-8\,a\,b^2\,c\,p-24\,a\,b^2\,c+b^4\,p^2+5\,b^4\,p+6\,b^4\right )}{p^3+6\,p^2+11\,p+6}+\frac {c\,d^5\,x^2\,\left (-32\,a^2\,c^2\,p+24\,a\,b^2\,c\,p^2+40\,a\,b^2\,c\,p+9\,b^4\,p^2+37\,b^4\,p+30\,b^4\right )}{p^3+6\,p^2+11\,p+6}+\frac {16\,c^5\,d^5\,x^6\,\left (p^2+3\,p+2\right )}{p^3+6\,p^2+11\,p+6}+\frac {b\,d^5\,x\,\left (-32\,a^2\,c^2\,p+8\,a\,b^2\,c\,p^2+24\,a\,b^2\,c\,p+b^4\,p^2+5\,b^4\,p+6\,b^4\right )}{p^3+6\,p^2+11\,p+6}+\frac {48\,b\,c^4\,d^5\,x^5\,\left (p^2+3\,p+2\right )}{p^3+6\,p^2+11\,p+6}+\frac {8\,c^3\,d^5\,x^4\,\left (p+1\right )\,\left (7\,b^2\,p+15\,b^2+2\,a\,c\,p\right )}{p^3+6\,p^2+11\,p+6}+\frac {16\,b\,c^2\,d^5\,x^3\,\left (p+1\right )\,\left (2\,b^2\,p+5\,b^2+2\,a\,c\,p\right )}{p^3+6\,p^2+11\,p+6}\right ) \] Input:

int((b*d + 2*c*d*x)^5*(a + b*x + c*x^2)^p,x)
 

Output:

(a + b*x + c*x^2)^p*((a*d^5*(5*b^4*p + 6*b^4 + 32*a^2*c^2 + b^4*p^2 - 24*a 
*b^2*c - 8*a*b^2*c*p))/(11*p + 6*p^2 + p^3 + 6) + (c*d^5*x^2*(37*b^4*p + 3 
0*b^4 + 9*b^4*p^2 - 32*a^2*c^2*p + 40*a*b^2*c*p + 24*a*b^2*c*p^2))/(11*p + 
 6*p^2 + p^3 + 6) + (16*c^5*d^5*x^6*(3*p + p^2 + 2))/(11*p + 6*p^2 + p^3 + 
 6) + (b*d^5*x*(5*b^4*p + 6*b^4 + b^4*p^2 - 32*a^2*c^2*p + 24*a*b^2*c*p + 
8*a*b^2*c*p^2))/(11*p + 6*p^2 + p^3 + 6) + (48*b*c^4*d^5*x^5*(3*p + p^2 + 
2))/(11*p + 6*p^2 + p^3 + 6) + (8*c^3*d^5*x^4*(p + 1)*(7*b^2*p + 15*b^2 + 
2*a*c*p))/(11*p + 6*p^2 + p^3 + 6) + (16*b*c^2*d^5*x^3*(p + 1)*(2*b^2*p + 
5*b^2 + 2*a*c*p))/(11*p + 6*p^2 + p^3 + 6))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 382, normalized size of antiderivative = 3.16 \[ \int (b d+2 c d x)^5 \left (a+b x+c x^2\right )^p \, dx=\frac {\left (c \,x^{2}+b x +a \right )^{p} d^{5} \left (16 c^{5} p^{2} x^{6}+48 b \,c^{4} p^{2} x^{5}+48 c^{5} p \,x^{6}+16 a \,c^{4} p^{2} x^{4}+56 b^{2} c^{3} p^{2} x^{4}+144 b \,c^{4} p \,x^{5}+32 c^{5} x^{6}+32 a b \,c^{3} p^{2} x^{3}+16 a \,c^{4} p \,x^{4}+32 b^{3} c^{2} p^{2} x^{3}+176 b^{2} c^{3} p \,x^{4}+96 b \,c^{4} x^{5}+24 a \,b^{2} c^{2} p^{2} x^{2}+32 a b \,c^{3} p \,x^{3}+9 b^{4} c \,p^{2} x^{2}+112 b^{3} c^{2} p \,x^{3}+120 b^{2} c^{3} x^{4}-32 a^{2} c^{3} p \,x^{2}+8 a \,b^{3} c \,p^{2} x +40 a \,b^{2} c^{2} p \,x^{2}+b^{5} p^{2} x +37 b^{4} c p \,x^{2}+80 b^{3} c^{2} x^{3}-32 a^{2} b \,c^{2} p x +a \,b^{4} p^{2}+24 a \,b^{3} c p x +5 b^{5} p x +30 b^{4} c \,x^{2}-8 a^{2} b^{2} c p +5 a \,b^{4} p +6 b^{5} x +32 a^{3} c^{2}-24 a^{2} b^{2} c +6 a \,b^{4}\right )}{p^{3}+6 p^{2}+11 p +6} \] Input:

int((2*c*d*x+b*d)^5*(c*x^2+b*x+a)^p,x)
 

Output:

((a + b*x + c*x**2)**p*d**5*(32*a**3*c**2 - 8*a**2*b**2*c*p - 24*a**2*b**2 
*c - 32*a**2*b*c**2*p*x - 32*a**2*c**3*p*x**2 + a*b**4*p**2 + 5*a*b**4*p + 
 6*a*b**4 + 8*a*b**3*c*p**2*x + 24*a*b**3*c*p*x + 24*a*b**2*c**2*p**2*x**2 
 + 40*a*b**2*c**2*p*x**2 + 32*a*b*c**3*p**2*x**3 + 32*a*b*c**3*p*x**3 + 16 
*a*c**4*p**2*x**4 + 16*a*c**4*p*x**4 + b**5*p**2*x + 5*b**5*p*x + 6*b**5*x 
 + 9*b**4*c*p**2*x**2 + 37*b**4*c*p*x**2 + 30*b**4*c*x**2 + 32*b**3*c**2*p 
**2*x**3 + 112*b**3*c**2*p*x**3 + 80*b**3*c**2*x**3 + 56*b**2*c**3*p**2*x* 
*4 + 176*b**2*c**3*p*x**4 + 120*b**2*c**3*x**4 + 48*b*c**4*p**2*x**5 + 144 
*b*c**4*p*x**5 + 96*b*c**4*x**5 + 16*c**5*p**2*x**6 + 48*c**5*p*x**6 + 32* 
c**5*x**6))/(p**3 + 6*p**2 + 11*p + 6)