Integrand size = 24, antiderivative size = 55 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=\frac {1}{12} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^3+\frac {1}{4} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^3 \] Output:
1/12*(-4*a*c+b^2)*d^3*(c*x^2+b*x+a)^3+1/4*d^3*(2*c*x+b)^2*(c*x^2+b*x+a)^3
Time = 0.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.76 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=\frac {1}{3} d^3 x (b+c x) \left (3 a^2 \left (b^2+2 b c x+2 c^2 x^2\right )+x^2 (b+c x)^2 \left (b^2+3 b c x+3 c^2 x^2\right )+a x \left (3 b^3+11 b^2 c x+16 b c^2 x^2+8 c^3 x^3\right )\right ) \] Input:
Integrate[(b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^2,x]
Output:
(d^3*x*(b + c*x)*(3*a^2*(b^2 + 2*b*c*x + 2*c^2*x^2) + x^2*(b + c*x)^2*(b^2 + 3*b*c*x + 3*c^2*x^2) + a*x*(3*b^3 + 11*b^2*c*x + 16*b*c^2*x^2 + 8*c^3*x ^3)))/3
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1116, 27, 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.
\(\displaystyle \int \left (a+b x+c x^2\right )^2 (b d+2 c d x)^3 \, dx\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle \frac {1}{4} d^2 \left (b^2-4 a c\right ) \int d (b+2 c x) \left (c x^2+b x+a\right )^2dx+\frac {1}{4} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} d^3 \left (b^2-4 a c\right ) \int (b+2 c x) \left (c x^2+b x+a\right )^2dx+\frac {1}{4} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^3\) |
\(\Big \downarrow \) 1104 |
\(\displaystyle \frac {1}{12} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3+\frac {1}{4} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^3\) |
Input:
Int[(b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^2,x]
Output:
((b^2 - 4*a*c)*d^3*(a + b*x + c*x^2)^3)/12 + (d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^3)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
Time = 0.75 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84
method | result | size |
default | \(-d^{3} \left (-c \left (c \,x^{2}+b x +a \right )^{4}+\frac {\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{3}}{3}\right )\) | \(46\) |
gosper | \(\frac {x \left (3 c^{5} x^{7}+12 b \,c^{4} x^{6}+8 x^{5} a \,c^{4}+19 x^{5} b^{2} c^{3}+24 x^{4} a b \,c^{3}+15 b^{3} c^{2} x^{4}+6 a^{2} c^{3} x^{3}+27 a \,b^{2} c^{2} x^{3}+6 b^{4} c \,x^{3}+12 a^{2} b \,c^{2} x^{2}+14 a \,b^{3} c \,x^{2}+x^{2} b^{5}+9 a^{2} b^{2} c x +3 a \,b^{4} x +3 a^{2} b^{3}\right ) d^{3}}{3}\) | \(152\) |
orering | \(\frac {x \left (3 c^{5} x^{7}+12 b \,c^{4} x^{6}+8 x^{5} a \,c^{4}+19 x^{5} b^{2} c^{3}+24 x^{4} a b \,c^{3}+15 b^{3} c^{2} x^{4}+6 a^{2} c^{3} x^{3}+27 a \,b^{2} c^{2} x^{3}+6 b^{4} c \,x^{3}+12 a^{2} b \,c^{2} x^{2}+14 a \,b^{3} c \,x^{2}+x^{2} b^{5}+9 a^{2} b^{2} c x +3 a \,b^{4} x +3 a^{2} b^{3}\right ) \left (2 c d x +b d \right )^{3}}{3 \left (2 c x +b \right )^{3}}\) | \(168\) |
norman | \(\left (\frac {8}{3} a \,c^{4} d^{3}+\frac {19}{3} b^{2} c^{3} d^{3}\right ) x^{6}+\left (4 a^{2} b \,c^{2} d^{3}+\frac {14}{3} a \,b^{3} c \,d^{3}+\frac {1}{3} b^{5} d^{3}\right ) x^{3}+\left (8 a b \,c^{3} d^{3}+5 b^{3} c^{2} d^{3}\right ) x^{5}+\left (3 a^{2} b^{2} c \,d^{3}+a \,b^{4} d^{3}\right ) x^{2}+\left (2 a^{2} c^{3} d^{3}+9 d^{3} a \,c^{2} b^{2}+2 b^{4} c \,d^{3}\right ) x^{4}+c^{5} d^{3} x^{8}+d^{3} a^{2} b^{3} x +4 b \,c^{4} d^{3} x^{7}\) | \(183\) |
parallelrisch | \(d^{3} a^{2} b^{3} x +3 x^{2} a^{2} b^{2} c \,d^{3}+d^{3} a \,b^{4} x^{2}+4 a^{2} b \,c^{2} d^{3} x^{3}+\frac {14}{3} a \,b^{3} c \,d^{3} x^{3}+\frac {1}{3} d^{3} b^{5} x^{3}+2 a^{2} c^{3} d^{3} x^{4}+9 d^{3} a \,b^{2} c^{2} x^{4}+2 b^{4} c \,x^{4} d^{3}+8 a b \,c^{3} d^{3} x^{5}+5 d^{3} b^{3} c^{2} x^{5}+\frac {8}{3} d^{3} a \,c^{4} x^{6}+\frac {19}{3} d^{3} b^{2} c^{3} x^{6}+4 b \,c^{4} d^{3} x^{7}+c^{5} d^{3} x^{8}\) | \(194\) |
risch | \(\frac {8}{3} d^{3} a \,c^{4} x^{6}+\frac {1}{3} d^{3} b^{5} x^{3}+\frac {1}{3} d^{3} a^{3} b^{2}+2 b^{4} c \,x^{4} d^{3}-\frac {1}{3} a^{4} c \,d^{3}+5 d^{3} b^{3} c^{2} x^{5}+4 b \,c^{4} d^{3} x^{7}+\frac {14}{3} a \,b^{3} c \,d^{3} x^{3}+3 x^{2} a^{2} b^{2} c \,d^{3}+2 a^{2} c^{3} d^{3} x^{4}+\frac {19}{3} d^{3} b^{2} c^{3} x^{6}+d^{3} a \,b^{4} x^{2}+9 d^{3} a \,b^{2} c^{2} x^{4}+8 a b \,c^{3} d^{3} x^{5}+4 a^{2} b \,c^{2} d^{3} x^{3}+c^{5} d^{3} x^{8}+d^{3} a^{2} b^{3} x\) | \(214\) |
Input:
int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-d^3*(-c*(c*x^2+b*x+a)^4+1/3*(4*a*c-b^2)*(c*x^2+b*x+a)^3)
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (51) = 102\).
Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.93 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=c^{5} d^{3} x^{8} + 4 \, b c^{4} d^{3} x^{7} + \frac {1}{3} \, {\left (19 \, b^{2} c^{3} + 8 \, a c^{4}\right )} d^{3} x^{6} + a^{2} b^{3} d^{3} x + {\left (5 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} d^{3} x^{5} + {\left (2 \, b^{4} c + 9 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} d^{3} x^{4} + \frac {1}{3} \, {\left (b^{5} + 14 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} d^{3} x^{3} + {\left (a b^{4} + 3 \, a^{2} b^{2} c\right )} d^{3} x^{2} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
c^5*d^3*x^8 + 4*b*c^4*d^3*x^7 + 1/3*(19*b^2*c^3 + 8*a*c^4)*d^3*x^6 + a^2*b ^3*d^3*x + (5*b^3*c^2 + 8*a*b*c^3)*d^3*x^5 + (2*b^4*c + 9*a*b^2*c^2 + 2*a^ 2*c^3)*d^3*x^4 + 1/3*(b^5 + 14*a*b^3*c + 12*a^2*b*c^2)*d^3*x^3 + (a*b^4 + 3*a^2*b^2*c)*d^3*x^2
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (49) = 98\).
Time = 0.03 (sec) , antiderivative size = 194, normalized size of antiderivative = 3.53 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=a^{2} b^{3} d^{3} x + 4 b c^{4} d^{3} x^{7} + c^{5} d^{3} x^{8} + x^{6} \cdot \left (\frac {8 a c^{4} d^{3}}{3} + \frac {19 b^{2} c^{3} d^{3}}{3}\right ) + x^{5} \cdot \left (8 a b c^{3} d^{3} + 5 b^{3} c^{2} d^{3}\right ) + x^{4} \cdot \left (2 a^{2} c^{3} d^{3} + 9 a b^{2} c^{2} d^{3} + 2 b^{4} c d^{3}\right ) + x^{3} \cdot \left (4 a^{2} b c^{2} d^{3} + \frac {14 a b^{3} c d^{3}}{3} + \frac {b^{5} d^{3}}{3}\right ) + x^{2} \cdot \left (3 a^{2} b^{2} c d^{3} + a b^{4} d^{3}\right ) \] Input:
integrate((2*c*d*x+b*d)**3*(c*x**2+b*x+a)**2,x)
Output:
a**2*b**3*d**3*x + 4*b*c**4*d**3*x**7 + c**5*d**3*x**8 + x**6*(8*a*c**4*d* *3/3 + 19*b**2*c**3*d**3/3) + x**5*(8*a*b*c**3*d**3 + 5*b**3*c**2*d**3) + x**4*(2*a**2*c**3*d**3 + 9*a*b**2*c**2*d**3 + 2*b**4*c*d**3) + x**3*(4*a** 2*b*c**2*d**3 + 14*a*b**3*c*d**3/3 + b**5*d**3/3) + x**2*(3*a**2*b**2*c*d* *3 + a*b**4*d**3)
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (51) = 102\).
Time = 0.04 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.93 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=c^{5} d^{3} x^{8} + 4 \, b c^{4} d^{3} x^{7} + \frac {1}{3} \, {\left (19 \, b^{2} c^{3} + 8 \, a c^{4}\right )} d^{3} x^{6} + a^{2} b^{3} d^{3} x + {\left (5 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} d^{3} x^{5} + {\left (2 \, b^{4} c + 9 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} d^{3} x^{4} + \frac {1}{3} \, {\left (b^{5} + 14 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} d^{3} x^{3} + {\left (a b^{4} + 3 \, a^{2} b^{2} c\right )} d^{3} x^{2} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
c^5*d^3*x^8 + 4*b*c^4*d^3*x^7 + 1/3*(19*b^2*c^3 + 8*a*c^4)*d^3*x^6 + a^2*b ^3*d^3*x + (5*b^3*c^2 + 8*a*b*c^3)*d^3*x^5 + (2*b^4*c + 9*a*b^2*c^2 + 2*a^ 2*c^3)*d^3*x^4 + 1/3*(b^5 + 14*a*b^3*c + 12*a^2*b*c^2)*d^3*x^3 + (a*b^4 + 3*a^2*b^2*c)*d^3*x^2
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (51) = 102\).
Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.25 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx={\left (c d x^{2} + b d x\right )} a^{2} b^{2} d^{2} + \frac {3 \, {\left (c d x^{2} + b d x\right )}^{2} a b^{2} d^{2} + 6 \, {\left (c d x^{2} + b d x\right )}^{2} a^{2} c d^{2} + {\left (c d x^{2} + b d x\right )}^{3} b^{2} d + 8 \, {\left (c d x^{2} + b d x\right )}^{3} a c d + 3 \, {\left (c d x^{2} + b d x\right )}^{4} c}{3 \, d} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
(c*d*x^2 + b*d*x)*a^2*b^2*d^2 + 1/3*(3*(c*d*x^2 + b*d*x)^2*a*b^2*d^2 + 6*( c*d*x^2 + b*d*x)^2*a^2*c*d^2 + (c*d*x^2 + b*d*x)^3*b^2*d + 8*(c*d*x^2 + b* d*x)^3*a*c*d + 3*(c*d*x^2 + b*d*x)^4*c)/d
Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.76 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=c^5\,d^3\,x^8+a^2\,b^3\,d^3\,x+4\,b\,c^4\,d^3\,x^7+\frac {b\,d^3\,x^3\,\left (12\,a^2\,c^2+14\,a\,b^2\,c+b^4\right )}{3}+\frac {c^3\,d^3\,x^6\,\left (19\,b^2+8\,a\,c\right )}{3}+c\,d^3\,x^4\,\left (2\,a^2\,c^2+9\,a\,b^2\,c+2\,b^4\right )+a\,b^2\,d^3\,x^2\,\left (b^2+3\,a\,c\right )+b\,c^2\,d^3\,x^5\,\left (5\,b^2+8\,a\,c\right ) \] Input:
int((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^2,x)
Output:
c^5*d^3*x^8 + a^2*b^3*d^3*x + 4*b*c^4*d^3*x^7 + (b*d^3*x^3*(b^4 + 12*a^2*c ^2 + 14*a*b^2*c))/3 + (c^3*d^3*x^6*(8*a*c + 19*b^2))/3 + c*d^3*x^4*(2*b^4 + 2*a^2*c^2 + 9*a*b^2*c) + a*b^2*d^3*x^2*(3*a*c + b^2) + b*c^2*d^3*x^5*(8* a*c + 5*b^2)
Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.75 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^2 \, dx=\frac {d^{3} x \left (3 c^{5} x^{7}+12 b \,c^{4} x^{6}+8 a \,c^{4} x^{5}+19 b^{2} c^{3} x^{5}+24 a b \,c^{3} x^{4}+15 b^{3} c^{2} x^{4}+6 a^{2} c^{3} x^{3}+27 a \,b^{2} c^{2} x^{3}+6 b^{4} c \,x^{3}+12 a^{2} b \,c^{2} x^{2}+14 a \,b^{3} c \,x^{2}+b^{5} x^{2}+9 a^{2} b^{2} c x +3 a \,b^{4} x +3 a^{2} b^{3}\right )}{3} \] Input:
int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^2,x)
Output:
(d**3*x*(3*a**2*b**3 + 9*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 6*a**2*c**3 *x**3 + 3*a*b**4*x + 14*a*b**3*c*x**2 + 27*a*b**2*c**2*x**3 + 24*a*b*c**3* x**4 + 8*a*c**4*x**5 + b**5*x**2 + 6*b**4*c*x**3 + 15*b**3*c**2*x**4 + 19* b**2*c**3*x**5 + 12*b*c**4*x**6 + 3*c**5*x**7))/3