\(\int (1-2 x) (2+3 x) (3+5 x) \, dx\) [1150]

Optimal result

Integrand size = 16, antiderivative size = 25 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=6 x+\frac {7 x^2}{2}-\frac {23 x^3}{3}-\frac {15 x^4}{2} \]

[Out]

6*x+7/2*x^2-23/3*x^3-15/2*x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=-\frac {15 x^4}{2}-\frac {23 x^3}{3}+\frac {7 x^2}{2}+6 x \]

[In]

Int[(1 - 2*x)*(2 + 3*x)*(3 + 5*x),x]

[Out]

6*x + (7*x^2)/2 - (23*x^3)/3 - (15*x^4)/2

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 (6+7 x-23 x^2-30 x^3\right ) \, dx \\ & = 6 x+\frac {7 x^2}{2}-\frac {23 x^3}{3}-\frac {15 x^4}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=6 x+\frac {7 x^2}{2}-\frac {23 x^3}{3}-\frac {15 x^4}{2} \]

[In]

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

[Out]

6*x + (7*x^2)/2 - (23*x^3)/3 - (15*x^4)/2

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
gosper	\(-\frac {x \left (45 x^{3}+46 x^{2}-21 x -36\right )}{6}\)	\(19\)
default	\(6 x +\frac {7}{2} x^{2}-\frac {23}{3} x^{3}-\frac {15}{2} x^{4}\)	\(20\)
norman	\(6 x +\frac {7}{2} x^{2}-\frac {23}{3} x^{3}-\frac {15}{2} x^{4}\)	\(20\)
risch	\(6 x +\frac {7}{2} x^{2}-\frac {23}{3} x^{3}-\frac {15}{2} x^{4}\)	\(20\)
parallelrisch	\(6 x +\frac {7}{2} x^{2}-\frac {23}{3} x^{3}-\frac {15}{2} x^{4}\)	\(20\)

[In]

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

[Out]

-1/6*x*(45*x^3+46*x^2-21*x-36)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=-\frac {15}{2} \, x^{4} - \frac {23}{3} \, x^{3} + \frac {7}{2} \, x^{2} + 6 \, x \]

[In]

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

[Out]

-15/2*x^4 - 23/3*x^3 + 7/2*x^2 + 6*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=- \frac {15 x^{4}}{2} - \frac {23 x^{3}}{3} + \frac {7 x^{2}}{2} + 6 x \]

[In]

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

[Out]

-15*x**4/2 - 23*x**3/3 + 7*x**2/2 + 6*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=-\frac {15}{2} \, x^{4} - \frac {23}{3} \, x^{3} + \frac {7}{2} \, x^{2} + 6 \, x \]

[In]

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

[Out]

-15/2*x^4 - 23/3*x^3 + 7/2*x^2 + 6*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=-\frac {15}{2} \, x^{4} - \frac {23}{3} \, x^{3} + \frac {7}{2} \, x^{2} + 6 \, x \]

[In]

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

[Out]

-15/2*x^4 - 23/3*x^3 + 7/2*x^2 + 6*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int (1-2 x) (2+3 x) (3+5 x) \, dx=-\frac {15\,x^4}{2}-\frac {23\,x^3}{3}+\frac {7\,x^2}{2}+6\,x \]

[In]

int(-(2*x - 1)*(3*x + 2)*(5*x + 3),x)

[Out]

6*x + (7*x^2)/2 - (23*x^3)/3 - (15*x^4)/2