Integrand size = 29, antiderivative size = 106 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {c^2 \left (c^3+4 a d^2\right )^2 x}{d^4}-\frac {4 c^3 \left (c^3+4 a d^2\right ) (c+d x)^3}{3 d^5}+\frac {2 c \left (3 c^3+4 a d^2\right ) (c+d x)^5}{5 d^5}-\frac {4 c^2 (c+d x)^7}{7 d^5}+\frac {(c+d x)^9}{9 d^5} \] Output:
c^2*(4*a*d^2+c^3)^2*x/d^4-4/3*c^3*(4*a*d^2+c^3)*(d*x+c)^3/d^5+2/5*c*(4*a*d ^2+3*c^3)*(d*x+c)^5/d^5-4/7*c^2*(d*x+c)^7/d^5+1/9*(d*x+c)^9/d^5
Time = 0.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^2 c^2 x+\frac {32}{3} a c^3 x^3+8 a c^2 d x^4+\frac {8}{5} c \left (2 c^3+a d^2\right ) x^5+\frac {16}{3} c^3 d x^6+\frac {24}{7} c^2 d^2 x^7+c d^3 x^8+\frac {d^4 x^9}{9} \] Input:
Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^2,x]
Output:
16*a^2*c^2*x + (32*a*c^3*x^3)/3 + 8*a*c^2*d*x^4 + (8*c*(2*c^3 + a*d^2)*x^5 )/5 + (16*c^3*d*x^6)/3 + (24*c^2*d^2*x^7)/7 + c*d^3*x^8 + (d^4*x^9)/9
Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2458, 1403, 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.
\(\displaystyle \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx\) |
\(\Big \downarrow \) 2458 |
\(\displaystyle \int \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )^2d\left (\frac {c}{d}+x\right )\) |
\(\Big \downarrow \) 1403 |
\(\displaystyle \int \left (\frac {\left (4 a c d^2+c^4\right )^2}{d^4}-\frac {4 c^3 \left (4 a d^2+c^3\right ) \left (\frac {c}{d}+x\right )^2}{d^2}+4 c^4 \left (\frac {2 a d^2}{c^3}+\frac {3}{2}\right ) \left (\frac {c}{d}+x\right )^4-4 c^2 d^2 \left (\frac {c}{d}+x\right )^6+d^4 \left (\frac {c}{d}+x\right )^8\right )d\left (\frac {c}{d}+x\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} c \left (4 a d^2+3 c^3\right ) \left (\frac {c}{d}+x\right )^5-\frac {4 c^3 \left (4 a d^2+c^3\right ) \left (\frac {c}{d}+x\right )^3}{3 d^2}+\frac {c^2 \left (4 a d^2+c^3\right )^2 \left (\frac {c}{d}+x\right )}{d^4}-\frac {4}{7} c^2 d^2 \left (\frac {c}{d}+x\right )^7+\frac {1}{9} d^4 \left (\frac {c}{d}+x\right )^9\) |
Input:
Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^2,x]
Output:
(c^2*(c^3 + 4*a*d^2)^2*(c/d + x))/d^4 - (4*c^3*(c^3 + 4*a*d^2)*(c/d + x)^3 )/(3*d^2) + (2*c*(3*c^3 + 4*a*d^2)*(c/d + x)^5)/5 - (4*c^2*d^2*(c/d + x)^7 )/7 + (d^4*(c/d + x)^9)/9
Int[((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandInte grand[(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a *c, 0] && IGtQ[p, 0]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78
method | result | size |
norman | \(\frac {d^{4} x^{9}}{9}+c \,d^{3} x^{8}+\frac {24 c^{2} d^{2} x^{7}}{7}+\frac {16 c^{3} d \,x^{6}}{3}+\left (\frac {8}{5} a c \,d^{2}+\frac {16}{5} c^{4}\right ) x^{5}+8 a \,c^{2} d \,x^{4}+\frac {32 a \,c^{3} x^{3}}{3}+16 a^{2} c^{2} x\) | \(83\) |
gosper | \(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} a c \,d^{2}+\frac {16}{5} c^{4} x^{5}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\) | \(84\) |
default | \(\frac {d^{4} x^{9}}{9}+c \,d^{3} x^{8}+\frac {24 c^{2} d^{2} x^{7}}{7}+\frac {16 c^{3} d \,x^{6}}{3}+\frac {\left (8 a c \,d^{2}+16 c^{4}\right ) x^{5}}{5}+8 a \,c^{2} d \,x^{4}+\frac {32 a \,c^{3} x^{3}}{3}+16 a^{2} c^{2} x\) | \(84\) |
risch | \(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} a c \,d^{2}+\frac {16}{5} c^{4} x^{5}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\) | \(84\) |
parallelrisch | \(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} a c \,d^{2}+\frac {16}{5} c^{4} x^{5}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\) | \(84\) |
orering | \(\frac {x \left (35 d^{4} x^{8}+315 c \,d^{3} x^{7}+1080 c^{2} d^{2} x^{6}+1680 c^{3} d \,x^{5}+504 a c \,d^{2} x^{4}+1008 c^{4} x^{4}+2520 a \,c^{2} d \,x^{3}+3360 a \,c^{3} x^{2}+5040 a^{2} c^{2}\right )}{315}\) | \(87\) |
Input:
int((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x,method=_RETURNVERBOSE)
Output:
1/9*d^4*x^9+c*d^3*x^8+24/7*c^2*d^2*x^7+16/3*c^3*d*x^6+(8/5*a*c*d^2+16/5*c^ 4)*x^5+8*a*c^2*d*x^4+32/3*a*c^3*x^3+16*a^2*c^2*x
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {24}{7} \, c^{2} d^{2} x^{7} + \frac {16}{3} \, c^{3} d x^{6} + 8 \, a c^{2} d x^{4} + \frac {32}{3} \, a c^{3} x^{3} + \frac {8}{5} \, {\left (2 \, c^{4} + a c d^{2}\right )} x^{5} + 16 \, a^{2} c^{2} x \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="fricas")
Output:
1/9*d^4*x^9 + c*d^3*x^8 + 24/7*c^2*d^2*x^7 + 16/3*c^3*d*x^6 + 8*a*c^2*d*x^ 4 + 32/3*a*c^3*x^3 + 8/5*(2*c^4 + a*c*d^2)*x^5 + 16*a^2*c^2*x
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^{2} c^{2} x + \frac {32 a c^{3} x^{3}}{3} + 8 a c^{2} d x^{4} + \frac {16 c^{3} d x^{6}}{3} + \frac {24 c^{2} d^{2} x^{7}}{7} + c d^{3} x^{8} + \frac {d^{4} x^{9}}{9} + x^{5} \cdot \left (\frac {8 a c d^{2}}{5} + \frac {16 c^{4}}{5}\right ) \] Input:
integrate((d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**2,x)
Output:
16*a**2*c**2*x + 32*a*c**3*x**3/3 + 8*a*c**2*d*x**4 + 16*c**3*d*x**6/3 + 2 4*c**2*d**2*x**7/7 + c*d**3*x**8 + d**4*x**9/9 + x**5*(8*a*c*d**2/5 + 16*c **4/5)
Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {16}{7} \, c^{2} d^{2} x^{7} + \frac {16}{5} \, c^{4} x^{5} + 16 \, a^{2} c^{2} x + \frac {8}{15} \, {\left (3 \, d^{2} x^{5} + 15 \, c d x^{4} + 20 \, c^{2} x^{3}\right )} a c + \frac {8}{21} \, {\left (3 \, d^{2} x^{7} + 14 \, c d x^{6}\right )} c^{2} \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="maxima")
Output:
1/9*d^4*x^9 + c*d^3*x^8 + 16/7*c^2*d^2*x^7 + 16/5*c^4*x^5 + 16*a^2*c^2*x + 8/15*(3*d^2*x^5 + 15*c*d*x^4 + 20*c^2*x^3)*a*c + 8/21*(3*d^2*x^7 + 14*c*d *x^6)*c^2
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {24}{7} \, c^{2} d^{2} x^{7} + \frac {16}{3} \, c^{3} d x^{6} + \frac {16}{5} \, c^{4} x^{5} + \frac {8}{5} \, a c d^{2} x^{5} + 8 \, a c^{2} d x^{4} + \frac {32}{3} \, a c^{3} x^{3} + 16 \, a^{2} c^{2} x \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="giac")
Output:
1/9*d^4*x^9 + c*d^3*x^8 + 24/7*c^2*d^2*x^7 + 16/3*c^3*d*x^6 + 16/5*c^4*x^5 + 8/5*a*c*d^2*x^5 + 8*a*c^2*d*x^4 + 32/3*a*c^3*x^3 + 16*a^2*c^2*x
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=x^5\,\left (\frac {16\,c^4}{5}+\frac {8\,a\,c\,d^2}{5}\right )+\frac {d^4\,x^9}{9}+16\,a^2\,c^2\,x+\frac {32\,a\,c^3\,x^3}{3}+\frac {16\,c^3\,d\,x^6}{3}+c\,d^3\,x^8+\frac {24\,c^2\,d^2\,x^7}{7}+8\,a\,c^2\,d\,x^4 \] Input:
int((4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^2,x)
Output:
x^5*((16*c^4)/5 + (8*a*c*d^2)/5) + (d^4*x^9)/9 + 16*a^2*c^2*x + (32*a*c^3* x^3)/3 + (16*c^3*d*x^6)/3 + c*d^3*x^8 + (24*c^2*d^2*x^7)/7 + 8*a*c^2*d*x^4
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {x \left (35 d^{4} x^{8}+315 c \,d^{3} x^{7}+1080 c^{2} d^{2} x^{6}+1680 c^{3} d \,x^{5}+504 a c \,d^{2} x^{4}+1008 c^{4} x^{4}+2520 a \,c^{2} d \,x^{3}+3360 a \,c^{3} x^{2}+5040 a^{2} c^{2}\right )}{315} \] Input:
int((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x)
Output:
(x*(5040*a**2*c**2 + 3360*a*c**3*x**2 + 2520*a*c**2*d*x**3 + 504*a*c*d**2* x**4 + 1008*c**4*x**4 + 1680*c**3*d*x**5 + 1080*c**2*d**2*x**6 + 315*c*d** 3*x**7 + 35*d**4*x**8))/315