\(\int (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4)^2 \, dx\) [35]

Optimal result

Integrand size = 29, antiderivative size = 92 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^2 c^2 x+\frac {32}{3} a c^3 x^3+8 a c^2 d x^4+\frac {8}{5} c \left (2 c^3+a d^2\right ) x^5+\frac {16}{3} c^3 d x^6+\frac {24}{7} c^2 d^2 x^7+c d^3 x^8+\frac {d^4 x^9}{9} \]

[Out]

16*a^2*c^2*x+32/3*a*c^3*x^3+8*a*c^2*d*x^4+8/5*c*(a*d^2+2*c^3)*x^5+16/3*c^3*d*x^6+24/7*c^2*d^2*x^7+c*d^3*x^8+1/
9*d^4*x^9

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2086} \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^2 c^2 x+\frac {8}{5} c x^5 \left (a d^2+2 c^3\right )+\frac {32}{3} a c^3 x^3+8 a c^2 d x^4+\frac {16}{3} c^3 d x^6+\frac {24}{7} c^2 d^2 x^7+c d^3 x^8+\frac {d^4 x^9}{9} \]

[In]

Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^2,x]

[Out]

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

Rule 2086

Int[(P_)^(p_), x_Symbol] :> Int[ExpandToSum[P^p, x], x] /; PolyQ[P, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (16 a^2 c^2+32 a c^3 x^2+32 a c^2 d x^3+8 c \left (2 c^3+a d^2\right ) x^4+32 c^3 d x^5+24 c^2 d^2 x^6+8 c d^3 x^7+d^4 x^8\right ) \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^2 c^2 x+\frac {32}{3} a c^3 x^3+8 a c^2 d x^4+\frac {8}{5} c \left (2 c^3+a d^2\right ) x^5+\frac {16}{3} c^3 d x^6+\frac {24}{7} c^2 d^2 x^7+c d^3 x^8+\frac {d^4 x^9}{9} \]

[In]

Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90

method	result	size
norman	\(\frac {d^{4} x^{9}}{9}+c \,d^{3} x^{8}+\frac {24 c^{2} d^{2} x^{7}}{7}+\frac {16 c^{3} d \,x^{6}}{3}+\left (\frac {8}{5} d^{2} a c +\frac {16}{5} c^{4}\right ) x^{5}+8 a \,c^{2} d \,x^{4}+\frac {32 a \,c^{3} x^{3}}{3}+16 a^{2} c^{2} x\)	\(83\)
gosper	\(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} d^{2} a c +\frac {16}{5} x^{5} c^{4}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\)	\(84\)
default	\(\frac {d^{4} x^{9}}{9}+c \,d^{3} x^{8}+\frac {24 c^{2} d^{2} x^{7}}{7}+\frac {16 c^{3} d \,x^{6}}{3}+\frac {\left (8 d^{2} a c +16 c^{4}\right ) x^{5}}{5}+8 a \,c^{2} d \,x^{4}+\frac {32 a \,c^{3} x^{3}}{3}+16 a^{2} c^{2} x\)	\(84\)
risch	\(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} d^{2} a c +\frac {16}{5} x^{5} c^{4}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\)	\(84\)
parallelrisch	\(\frac {1}{9} d^{4} x^{9}+c \,d^{3} x^{8}+\frac {24}{7} c^{2} d^{2} x^{7}+\frac {16}{3} c^{3} d \,x^{6}+\frac {8}{5} x^{5} d^{2} a c +\frac {16}{5} x^{5} c^{4}+8 a \,c^{2} d \,x^{4}+\frac {32}{3} a \,c^{3} x^{3}+16 a^{2} c^{2} x\)	\(84\)

[In]

int((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x,method=_RETURNVERBOSE)

[Out]

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*d^2*a*c+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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {24}{7} \, c^{2} d^{2} x^{7} + \frac {16}{3} \, c^{3} d x^{6} + 8 \, a c^{2} d x^{4} + \frac {32}{3} \, a c^{3} x^{3} + \frac {8}{5} \, {\left (2 \, c^{4} + a c d^{2}\right )} x^{5} + 16 \, a^{2} c^{2} x \]

[In]

integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=16 a^{2} c^{2} x + \frac {32 a c^{3} x^{3}}{3} + 8 a c^{2} d x^{4} + \frac {16 c^{3} d x^{6}}{3} + \frac {24 c^{2} d^{2} x^{7}}{7} + c d^{3} x^{8} + \frac {d^{4} x^{9}}{9} + x^{5} \cdot \left (\frac {8 a c d^{2}}{5} + \frac {16 c^{4}}{5}\right ) \]

[In]

integrate((d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**2,x)

[Out]

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 + 24*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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {16}{7} \, c^{2} d^{2} x^{7} + \frac {16}{5} \, c^{4} x^{5} + 16 \, a^{2} c^{2} x + \frac {8}{15} \, {\left (3 \, d^{2} x^{5} + 15 \, c d x^{4} + 20 \, c^{2} x^{3}\right )} a c + \frac {8}{21} \, {\left (3 \, d^{2} x^{7} + 14 \, c d x^{6}\right )} c^{2} \]

[In]

integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=\frac {1}{9} \, d^{4} x^{9} + c d^{3} x^{8} + \frac {24}{7} \, c^{2} d^{2} x^{7} + \frac {16}{3} \, c^{3} d x^{6} + \frac {16}{5} \, c^{4} x^{5} + \frac {8}{5} \, a c d^{2} x^{5} + 8 \, a c^{2} d x^{4} + \frac {32}{3} \, a c^{3} x^{3} + 16 \, a^{2} c^{2} x \]

[In]

integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^2 \, dx=x^5\,\left (\frac {16\,c^4}{5}+\frac {8\,a\,c\,d^2}{5}\right )+\frac {d^4\,x^9}{9}+16\,a^2\,c^2\,x+\frac {32\,a\,c^3\,x^3}{3}+\frac {16\,c^3\,d\,x^6}{3}+c\,d^3\,x^8+\frac {24\,c^2\,d^2\,x^7}{7}+8\,a\,c^2\,d\,x^4 \]

[In]

int((4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^2,x)

[Out]

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