\(\int x^2 (2+x)^5 (2+3 x) \, dx\) [173]

Optimal result

Integrand size = 14, antiderivative size = 12 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {1}{3} x^3 (2+x)^6 \]

[Out]

1/3*x^3*(2+x)^6

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {75} \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {1}{3} x^3 (x+2)^6 \]

[In]

Int[x^2*(2 + x)^5*(2 + 3*x),x]

[Out]

(x^3*(2 + x)^6)/3

Rule 75

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 (2+x)^6 \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(42\) vs. \(2(12)=24\).

Time = 0.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.50 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {64 x^3}{3}+64 x^4+80 x^5+\frac {160 x^6}{3}+20 x^7+4 x^8+\frac {x^9}{3} \]

[In]

Integrate[x^2*(2 + x)^5*(2 + 3*x),x]

[Out]

(64*x^3)/3 + 64*x^4 + 80*x^5 + (160*x^6)/3 + 20*x^7 + 4*x^8 + x^9/3

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(36\) vs. \(2(10)=20\).

Time = 0.40 (sec) , antiderivative size = 37, normalized size of antiderivative = 3.08

method	result	size
gosper	\(\frac {1}{3} x^{9}+4 x^{8}+20 x^{7}+\frac {160}{3} x^{6}+80 x^{5}+64 x^{4}+\frac {64}{3} x^{3}\)	\(37\)
default	\(\frac {1}{3} x^{9}+4 x^{8}+20 x^{7}+\frac {160}{3} x^{6}+80 x^{5}+64 x^{4}+\frac {64}{3} x^{3}\)	\(37\)
norman	\(\frac {1}{3} x^{9}+4 x^{8}+20 x^{7}+\frac {160}{3} x^{6}+80 x^{5}+64 x^{4}+\frac {64}{3} x^{3}\)	\(37\)
risch	\(\frac {1}{3} x^{9}+4 x^{8}+20 x^{7}+\frac {160}{3} x^{6}+80 x^{5}+64 x^{4}+\frac {64}{3} x^{3}\)	\(37\)
parallelrisch	\(\frac {1}{3} x^{9}+4 x^{8}+20 x^{7}+\frac {160}{3} x^{6}+80 x^{5}+64 x^{4}+\frac {64}{3} x^{3}\)	\(37\)

[In]

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

[Out]

1/3*x^9+4*x^8+20*x^7+160/3*x^6+80*x^5+64*x^4+64/3*x^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (10) = 20\).

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 3.00 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {1}{3} \, x^{9} + 4 \, x^{8} + 20 \, x^{7} + \frac {160}{3} \, x^{6} + 80 \, x^{5} + 64 \, x^{4} + \frac {64}{3} \, x^{3} \]

[In]

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

[Out]

1/3*x^9 + 4*x^8 + 20*x^7 + 160/3*x^6 + 80*x^5 + 64*x^4 + 64/3*x^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (8) = 16\).

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 3.08 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {x^{9}}{3} + 4 x^{8} + 20 x^{7} + \frac {160 x^{6}}{3} + 80 x^{5} + 64 x^{4} + \frac {64 x^{3}}{3} \]

[In]

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

[Out]

x**9/3 + 4*x**8 + 20*x**7 + 160*x**6/3 + 80*x**5 + 64*x**4 + 64*x**3/3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (10) = 20\).

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 3.00 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {1}{3} \, x^{9} + 4 \, x^{8} + 20 \, x^{7} + \frac {160}{3} \, x^{6} + 80 \, x^{5} + 64 \, x^{4} + \frac {64}{3} \, x^{3} \]

[In]

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

[Out]

1/3*x^9 + 4*x^8 + 20*x^7 + 160/3*x^6 + 80*x^5 + 64*x^4 + 64/3*x^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (10) = 20\).

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 3.00 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {1}{3} \, x^{9} + 4 \, x^{8} + 20 \, x^{7} + \frac {160}{3} \, x^{6} + 80 \, x^{5} + 64 \, x^{4} + \frac {64}{3} \, x^{3} \]

[In]

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

[Out]

1/3*x^9 + 4*x^8 + 20*x^7 + 160/3*x^6 + 80*x^5 + 64*x^4 + 64/3*x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 3.00 \[ \int x^2 (2+x)^5 (2+3 x) \, dx=\frac {x^9}{3}+4\,x^8+20\,x^7+\frac {160\,x^6}{3}+80\,x^5+64\,x^4+\frac {64\,x^3}{3} \]

[In]

int(x^2*(3*x + 2)*(x + 2)^5,x)

[Out]

(64*x^3)/3 + 64*x^4 + 80*x^5 + (160*x^6)/3 + 20*x^7 + 4*x^8 + x^9/3