\(\int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx\) [1240]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=-\frac {49}{405} (2+3 x)^5+\frac {91}{162} (2+3 x)^6-\frac {16}{63} (2+3 x)^7+\frac {5}{162} (2+3 x)^8 \]

[Out]

-49/405*(2+3*x)^5+91/162*(2+3*x)^6-16/63*(2+3*x)^7+5/162*(2+3*x)^8

Rubi [A] (verified)

Time = 0.01 (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} \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {5}{162} (3 x+2)^8-\frac {16}{63} (3 x+2)^7+\frac {91}{162} (3 x+2)^6-\frac {49}{405} (3 x+2)^5 \]

[In]

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

[Out]

(-49*(2 + 3*x)^5)/405 + (91*(2 + 3*x)^6)/162 - (16*(2 + 3*x)^7)/63 + (5*(2 + 3*x)^8)/162

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 {49}{27} (2+3 x)^4+\frac {91}{9} (2+3 x)^5-\frac {16}{3} (2+3 x)^6+\frac {20}{27} (2+3 x)^7\right ) \, dx \\ & = -\frac {49}{405} (2+3 x)^5+\frac {91}{162} (2+3 x)^6-\frac {16}{63} (2+3 x)^7+\frac {5}{162} (2+3 x)^8 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=48 x+88 x^2-\frac {152 x^3}{3}-328 x^4-\frac {1077 x^5}{5}+\frac {675 x^6}{2}+\frac {3672 x^7}{7}+\frac {405 x^8}{2} \]

[In]

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

[Out]

48*x + 88*x^2 - (152*x^3)/3 - 328*x^4 - (1077*x^5)/5 + (675*x^6)/2 + (3672*x^7)/7 + (405*x^8)/2

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87

method	result	size
gosper	\(\frac {x \left (42525 x^{7}+110160 x^{6}+70875 x^{5}-45234 x^{4}-68880 x^{3}-10640 x^{2}+18480 x +10080\right )}{210}\)	\(39\)
default	\(\frac {405}{2} x^{8}+\frac {3672}{7} x^{7}+\frac {675}{2} x^{6}-\frac {1077}{5} x^{5}-328 x^{4}-\frac {152}{3} x^{3}+88 x^{2}+48 x\)	\(40\)
norman	\(\frac {405}{2} x^{8}+\frac {3672}{7} x^{7}+\frac {675}{2} x^{6}-\frac {1077}{5} x^{5}-328 x^{4}-\frac {152}{3} x^{3}+88 x^{2}+48 x\)	\(40\)
risch	\(\frac {405}{2} x^{8}+\frac {3672}{7} x^{7}+\frac {675}{2} x^{6}-\frac {1077}{5} x^{5}-328 x^{4}-\frac {152}{3} x^{3}+88 x^{2}+48 x\)	\(40\)
parallelrisch	\(\frac {405}{2} x^{8}+\frac {3672}{7} x^{7}+\frac {675}{2} x^{6}-\frac {1077}{5} x^{5}-328 x^{4}-\frac {152}{3} x^{3}+88 x^{2}+48 x\)	\(40\)

[In]

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

[Out]

1/210*x*(42525*x^7+110160*x^6+70875*x^5-45234*x^4-68880*x^3-10640*x^2+18480*x+10080)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {405}{2} \, x^{8} + \frac {3672}{7} \, x^{7} + \frac {675}{2} \, x^{6} - \frac {1077}{5} \, x^{5} - 328 \, x^{4} - \frac {152}{3} \, x^{3} + 88 \, x^{2} + 48 \, x \]

[In]

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

[Out]

405/2*x^8 + 3672/7*x^7 + 675/2*x^6 - 1077/5*x^5 - 328*x^4 - 152/3*x^3 + 88*x^2 + 48*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {405 x^{8}}{2} + \frac {3672 x^{7}}{7} + \frac {675 x^{6}}{2} - \frac {1077 x^{5}}{5} - 328 x^{4} - \frac {152 x^{3}}{3} + 88 x^{2} + 48 x \]

[In]

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

[Out]

405*x**8/2 + 3672*x**7/7 + 675*x**6/2 - 1077*x**5/5 - 328*x**4 - 152*x**3/3 + 88*x**2 + 48*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {405}{2} \, x^{8} + \frac {3672}{7} \, x^{7} + \frac {675}{2} \, x^{6} - \frac {1077}{5} \, x^{5} - 328 \, x^{4} - \frac {152}{3} \, x^{3} + 88 \, x^{2} + 48 \, x \]

[In]

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

[Out]

405/2*x^8 + 3672/7*x^7 + 675/2*x^6 - 1077/5*x^5 - 328*x^4 - 152/3*x^3 + 88*x^2 + 48*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {405}{2} \, x^{8} + \frac {3672}{7} \, x^{7} + \frac {675}{2} \, x^{6} - \frac {1077}{5} \, x^{5} - 328 \, x^{4} - \frac {152}{3} \, x^{3} + 88 \, x^{2} + 48 \, x \]

[In]

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

[Out]

405/2*x^8 + 3672/7*x^7 + 675/2*x^6 - 1077/5*x^5 - 328*x^4 - 152/3*x^3 + 88*x^2 + 48*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int (1-2 x)^2 (2+3 x)^4 (3+5 x) \, dx=\frac {405\,x^8}{2}+\frac {3672\,x^7}{7}+\frac {675\,x^6}{2}-\frac {1077\,x^5}{5}-328\,x^4-\frac {152\,x^3}{3}+88\,x^2+48\,x \]

[In]

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

[Out]

48*x + 88*x^2 - (152*x^3)/3 - 328*x^4 - (1077*x^5)/5 + (675*x^6)/2 + (3672*x^7)/7 + (405*x^8)/2