\(\int (1-2 x) (3+5 x) \, dx\) [1151]

Optimal result

Integrand size = 11, antiderivative size = 18 \[ \int (1-2 x) (3+5 x) \, dx=3 x-\frac {x^2}{2}-\frac {10 x^3}{3} \]

[Out]

3*x-1/2*x^2-10/3*x^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int (1-2 x) (3+5 x) \, dx=-\frac {10 x^3}{3}-\frac {x^2}{2}+3 x \]

[In]

Int[(1 - 2*x)*(3 + 5*x),x]

[Out]

3*x - x^2/2 - (10*x^3)/3

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (3-x-10 x^2\right ) \, dx \\ & = 3 x-\frac {x^2}{2}-\frac {10 x^3}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int (1-2 x) (3+5 x) \, dx=3 x-\frac {x^2}{2}-\frac {10 x^3}{3} \]

[In]

Integrate[(1 - 2*x)*(3 + 5*x),x]

[Out]

3*x - x^2/2 - (10*x^3)/3

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78

method	result	size
gosper	\(-\frac {x \left (20 x^{2}+3 x -18\right )}{6}\)	\(14\)
default	\(3 x -\frac {1}{2} x^{2}-\frac {10}{3} x^{3}\)	\(15\)
norman	\(3 x -\frac {1}{2} x^{2}-\frac {10}{3} x^{3}\)	\(15\)
risch	\(3 x -\frac {1}{2} x^{2}-\frac {10}{3} x^{3}\)	\(15\)
parallelrisch	\(3 x -\frac {1}{2} x^{2}-\frac {10}{3} x^{3}\)	\(15\)

[In]

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

[Out]

-1/6*x*(20*x^2+3*x-18)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int (1-2 x) (3+5 x) \, dx=-\frac {10}{3} \, x^{3} - \frac {1}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-10/3*x^3 - 1/2*x^2 + 3*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int (1-2 x) (3+5 x) \, dx=- \frac {10 x^{3}}{3} - \frac {x^{2}}{2} + 3 x \]

[In]

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

[Out]

-10*x**3/3 - x**2/2 + 3*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int (1-2 x) (3+5 x) \, dx=-\frac {10}{3} \, x^{3} - \frac {1}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-10/3*x^3 - 1/2*x^2 + 3*x

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int (1-2 x) (3+5 x) \, dx=-\frac {10}{3} \, x^{3} - \frac {1}{2} \, x^{2} + 3 \, x \]

[In]

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

[Out]

-10/3*x^3 - 1/2*x^2 + 3*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int (1-2 x) (3+5 x) \, dx=-\frac {x\,\left (20\,x^2+3\,x-18\right )}{6} \]

[In]

int(-(2*x - 1)*(5*x + 3),x)

[Out]

-(x*(3*x + 20*x^2 - 18))/6