\(\int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx\) [1162]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=\frac {7}{405} (2+3 x)^5-\frac {4}{27} (2+3 x)^6+\frac {65}{189} (2+3 x)^7-\frac {25}{324} (2+3 x)^8 \]

[Out]

7/405*(2+3*x)^5-4/27*(2+3*x)^6+65/189*(2+3*x)^7-25/324*(2+3*x)^8

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=-\frac {25}{324} (3 x+2)^8+\frac {65}{189} (3 x+2)^7-\frac {4}{27} (3 x+2)^6+\frac {7}{405} (3 x+2)^5 \]

[In]

Int[(1 - 2*x)*(2 + 3*x)^4*(3 + 5*x)^2,x]

[Out]

(7*(2 + 3*x)^5)/405 - (4*(2 + 3*x)^6)/27 + (65*(2 + 3*x)^7)/189 - (25*(2 + 3*x)^8)/324

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {7}{27} (2+3 x)^4-\frac {8}{3} (2+3 x)^5+\frac {65}{9} (2+3 x)^6-\frac {50}{27} (2+3 x)^7\right ) \, dx \\ & = \frac {7}{405} (2+3 x)^5-\frac {4}{27} (2+3 x)^6+\frac {65}{189} (2+3 x)^7-\frac {25}{324} (2+3 x)^8 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=144 x+528 x^2+\frac {2536 x^3}{3}+94 x^4-\frac {9039 x^5}{5}-2898 x^6-\frac {13635 x^7}{7}-\frac {2025 x^8}{4} \]

[In]

Integrate[(1 - 2*x)*(2 + 3*x)^4*(3 + 5*x)^2,x]

[Out]

144*x + 528*x^2 + (2536*x^3)/3 + 94*x^4 - (9039*x^5)/5 - 2898*x^6 - (13635*x^7)/7 - (2025*x^8)/4

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87

method	result	size
gosper	\(-\frac {x \left (212625 x^{7}+818100 x^{6}+1217160 x^{5}+759276 x^{4}-39480 x^{3}-355040 x^{2}-221760 x -60480\right )}{420}\)	\(39\)
default	\(-\frac {2025}{4} x^{8}-\frac {13635}{7} x^{7}-2898 x^{6}-\frac {9039}{5} x^{5}+94 x^{4}+\frac {2536}{3} x^{3}+528 x^{2}+144 x\)	\(40\)
norman	\(-\frac {2025}{4} x^{8}-\frac {13635}{7} x^{7}-2898 x^{6}-\frac {9039}{5} x^{5}+94 x^{4}+\frac {2536}{3} x^{3}+528 x^{2}+144 x\)	\(40\)
risch	\(-\frac {2025}{4} x^{8}-\frac {13635}{7} x^{7}-2898 x^{6}-\frac {9039}{5} x^{5}+94 x^{4}+\frac {2536}{3} x^{3}+528 x^{2}+144 x\)	\(40\)
parallelrisch	\(-\frac {2025}{4} x^{8}-\frac {13635}{7} x^{7}-2898 x^{6}-\frac {9039}{5} x^{5}+94 x^{4}+\frac {2536}{3} x^{3}+528 x^{2}+144 x\)	\(40\)

[In]

int((1-2*x)*(2+3*x)^4*(3+5*x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/420*x*(212625*x^7+818100*x^6+1217160*x^5+759276*x^4-39480*x^3-355040*x^2-221760*x-60480)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=-\frac {2025}{4} \, x^{8} - \frac {13635}{7} \, x^{7} - 2898 \, x^{6} - \frac {9039}{5} \, x^{5} + 94 \, x^{4} + \frac {2536}{3} \, x^{3} + 528 \, x^{2} + 144 \, x \]

[In]

integrate((1-2*x)*(2+3*x)^4*(3+5*x)^2,x, algorithm="fricas")

[Out]

-2025/4*x^8 - 13635/7*x^7 - 2898*x^6 - 9039/5*x^5 + 94*x^4 + 2536/3*x^3 + 528*x^2 + 144*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=- \frac {2025 x^{8}}{4} - \frac {13635 x^{7}}{7} - 2898 x^{6} - \frac {9039 x^{5}}{5} + 94 x^{4} + \frac {2536 x^{3}}{3} + 528 x^{2} + 144 x \]

[In]

integrate((1-2*x)*(2+3*x)**4*(3+5*x)**2,x)

[Out]

-2025*x**8/4 - 13635*x**7/7 - 2898*x**6 - 9039*x**5/5 + 94*x**4 + 2536*x**3/3 + 528*x**2 + 144*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=-\frac {2025}{4} \, x^{8} - \frac {13635}{7} \, x^{7} - 2898 \, x^{6} - \frac {9039}{5} \, x^{5} + 94 \, x^{4} + \frac {2536}{3} \, x^{3} + 528 \, x^{2} + 144 \, x \]

[In]

integrate((1-2*x)*(2+3*x)^4*(3+5*x)^2,x, algorithm="maxima")

[Out]

-2025/4*x^8 - 13635/7*x^7 - 2898*x^6 - 9039/5*x^5 + 94*x^4 + 2536/3*x^3 + 528*x^2 + 144*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=-\frac {2025}{4} \, x^{8} - \frac {13635}{7} \, x^{7} - 2898 \, x^{6} - \frac {9039}{5} \, x^{5} + 94 \, x^{4} + \frac {2536}{3} \, x^{3} + 528 \, x^{2} + 144 \, x \]

[In]

integrate((1-2*x)*(2+3*x)^4*(3+5*x)^2,x, algorithm="giac")

[Out]

-2025/4*x^8 - 13635/7*x^7 - 2898*x^6 - 9039/5*x^5 + 94*x^4 + 2536/3*x^3 + 528*x^2 + 144*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x) (2+3 x)^4 (3+5 x)^2 \, dx=-\frac {2025\,x^8}{4}-\frac {13635\,x^7}{7}-2898\,x^6-\frac {9039\,x^5}{5}+94\,x^4+\frac {2536\,x^3}{3}+528\,x^2+144\,x \]

[In]

int(-(2*x - 1)*(3*x + 2)^4*(5*x + 3)^2,x)

[Out]

144*x + 528*x^2 + (2536*x^3)/3 + 94*x^4 - (9039*x^5)/5 - 2898*x^6 - (13635*x^7)/7 - (2025*x^8)/4