3.13.77 \(\int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx\) [1277]

3.13.77.1 Optimal result

Integrand size = 22, antiderivative size = 52 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=216 x+594 x^2+258 x^3-\frac {7145 x^4}{4}-\frac {15709 x^5}{5}+\frac {121 x^6}{6}+\frac {33255 x^7}{7}+4725 x^8+1500 x^9 \]

output

216*x+594*x^2+258*x^3-7145/4*x^4-15709/5*x^5+121/6*x^6+33255/7*x^7+4725*x^ 
8+1500*x^9
 

3.13.77.2 Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=216 x+594 x^2+258 x^3-\frac {7145 x^4}{4}-\frac {15709 x^5}{5}+\frac {121 x^6}{6}+\frac {33255 x^7}{7}+4725 x^8+1500 x^9 \]

input

Integrate[(1 - 2*x)^2*(2 + 3*x)^3*(3 + 5*x)^3,x]
 

output

216*x + 594*x^2 + 258*x^3 - (7145*x^4)/4 - (15709*x^5)/5 + (121*x^6)/6 + ( 
33255*x^7)/7 + 4725*x^8 + 1500*x^9
 

3.13.77.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 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.

\(\displaystyle \int (1-2 x)^2 (3 x+2)^3 (5 x+3)^3 \, dx\)

\(\Big \downarrow \) 99

\(\displaystyle \int \left (13500 x^8+37800 x^7+33255 x^6+121 x^5-15709 x^4-7145 x^3+774 x^2+1188 x+216\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 1500 x^9+4725 x^8+\frac {33255 x^7}{7}+\frac {121 x^6}{6}-\frac {15709 x^5}{5}-\frac {7145 x^4}{4}+258 x^3+594 x^2+216 x\)

input

Int[(1 - 2*x)^2*(2 + 3*x)^3*(3 + 5*x)^3,x]
 

output

216*x + 594*x^2 + 258*x^3 - (7145*x^4)/4 - (15709*x^5)/5 + (121*x^6)/6 + ( 
33255*x^7)/7 + 4725*x^8 + 1500*x^9
 

3.13.77.3.1 Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

3.13.77.4 Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {x \left (630000 x^{8}+1984500 x^{7}+1995300 x^{6}+8470 x^{5}-1319556 x^{4}-750225 x^{3}+108360 x^{2}+249480 x +90720\right )}{420}\)	\(44\)
default	\(216 x +594 x^{2}+258 x^{3}-\frac {7145}{4} x^{4}-\frac {15709}{5} x^{5}+\frac {121}{6} x^{6}+\frac {33255}{7} x^{7}+4725 x^{8}+1500 x^{9}\)	\(45\)
norman	\(216 x +594 x^{2}+258 x^{3}-\frac {7145}{4} x^{4}-\frac {15709}{5} x^{5}+\frac {121}{6} x^{6}+\frac {33255}{7} x^{7}+4725 x^{8}+1500 x^{9}\)	\(45\)
risch	\(216 x +594 x^{2}+258 x^{3}-\frac {7145}{4} x^{4}-\frac {15709}{5} x^{5}+\frac {121}{6} x^{6}+\frac {33255}{7} x^{7}+4725 x^{8}+1500 x^{9}\)	\(45\)
parallelrisch	\(216 x +594 x^{2}+258 x^{3}-\frac {7145}{4} x^{4}-\frac {15709}{5} x^{5}+\frac {121}{6} x^{6}+\frac {33255}{7} x^{7}+4725 x^{8}+1500 x^{9}\)	\(45\)

input

int((1-2*x)^2*(2+3*x)^3*(3+5*x)^3,x,method=_RETURNVERBOSE)
 

output

1/420*x*(630000*x^8+1984500*x^7+1995300*x^6+8470*x^5-1319556*x^4-750225*x^ 
3+108360*x^2+249480*x+90720)
 

3.13.77.5 Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=1500 \, x^{9} + 4725 \, x^{8} + \frac {33255}{7} \, x^{7} + \frac {121}{6} \, x^{6} - \frac {15709}{5} \, x^{5} - \frac {7145}{4} \, x^{4} + 258 \, x^{3} + 594 \, x^{2} + 216 \, x \]

input

integrate((1-2*x)^2*(2+3*x)^3*(3+5*x)^3,x, algorithm="fricas")
 

output

1500*x^9 + 4725*x^8 + 33255/7*x^7 + 121/6*x^6 - 15709/5*x^5 - 7145/4*x^4 + 
 258*x^3 + 594*x^2 + 216*x
 

3.13.77.6 Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=1500 x^{9} + 4725 x^{8} + \frac {33255 x^{7}}{7} + \frac {121 x^{6}}{6} - \frac {15709 x^{5}}{5} - \frac {7145 x^{4}}{4} + 258 x^{3} + 594 x^{2} + 216 x \]

input

integrate((1-2*x)**2*(2+3*x)**3*(3+5*x)**3,x)
 

output

1500*x**9 + 4725*x**8 + 33255*x**7/7 + 121*x**6/6 - 15709*x**5/5 - 7145*x* 
*4/4 + 258*x**3 + 594*x**2 + 216*x
 

3.13.77.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=1500 \, x^{9} + 4725 \, x^{8} + \frac {33255}{7} \, x^{7} + \frac {121}{6} \, x^{6} - \frac {15709}{5} \, x^{5} - \frac {7145}{4} \, x^{4} + 258 \, x^{3} + 594 \, x^{2} + 216 \, x \]

input

integrate((1-2*x)^2*(2+3*x)^3*(3+5*x)^3,x, algorithm="maxima")
 

output

1500*x^9 + 4725*x^8 + 33255/7*x^7 + 121/6*x^6 - 15709/5*x^5 - 7145/4*x^4 + 
 258*x^3 + 594*x^2 + 216*x
 

3.13.77.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=1500 \, x^{9} + 4725 \, x^{8} + \frac {33255}{7} \, x^{7} + \frac {121}{6} \, x^{6} - \frac {15709}{5} \, x^{5} - \frac {7145}{4} \, x^{4} + 258 \, x^{3} + 594 \, x^{2} + 216 \, x \]

input

integrate((1-2*x)^2*(2+3*x)^3*(3+5*x)^3,x, algorithm="giac")
 

output

1500*x^9 + 4725*x^8 + 33255/7*x^7 + 121/6*x^6 - 15709/5*x^5 - 7145/4*x^4 + 
 258*x^3 + 594*x^2 + 216*x
 

3.13.77.9 Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int (1-2 x)^2 (2+3 x)^3 (3+5 x)^3 \, dx=1500\,x^9+4725\,x^8+\frac {33255\,x^7}{7}+\frac {121\,x^6}{6}-\frac {15709\,x^5}{5}-\frac {7145\,x^4}{4}+258\,x^3+594\,x^2+216\,x \]

input

int((2*x - 1)^2*(3*x + 2)^3*(5*x + 3)^3,x)
 

output

216*x + 594*x^2 + 258*x^3 - (7145*x^4)/4 - (15709*x^5)/5 + (121*x^6)/6 + ( 
33255*x^7)/7 + 4725*x^8 + 1500*x^9