\(\int (1-2 x) (3+5 x)^3 \, dx\) [1183]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int (1-2 x) (3+5 x)^3 \, dx=\frac {11}{100} (3+5 x)^4-\frac {2}{125} (3+5 x)^5 \]

[Out]

11/100*(3+5*x)^4-2/125*(3+5*x)^5

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int (1-2 x) (3+5 x)^3 \, dx=\frac {11}{100} (5 x+3)^4-\frac {2}{125} (5 x+3)^5 \]

[In]

Int[(1 - 2*x)*(3 + 5*x)^3,x]

[Out]

(11*(3 + 5*x)^4)/100 - (2*(3 + 5*x)^5)/125

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 (\frac {11}{5} (3+5 x)^3-\frac {2}{5} (3+5 x)^4\right ) \, dx \\ & = \frac {11}{100} (3+5 x)^4-\frac {2}{125} (3+5 x)^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int (1-2 x) (3+5 x)^3 \, dx=27 x+\frac {81 x^2}{2}-15 x^3-\frac {325 x^4}{4}-50 x^5 \]

[In]

Integrate[(1 - 2*x)*(3 + 5*x)^3,x]

[Out]

27*x + (81*x^2)/2 - 15*x^3 - (325*x^4)/4 - 50*x^5

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
gosper	\(-\frac {x \left (200 x^{4}+325 x^{3}+60 x^{2}-162 x -108\right )}{4}\)	\(24\)
default	\(-50 x^{5}-\frac {325}{4} x^{4}-15 x^{3}+\frac {81}{2} x^{2}+27 x\)	\(25\)
norman	\(-50 x^{5}-\frac {325}{4} x^{4}-15 x^{3}+\frac {81}{2} x^{2}+27 x\)	\(25\)
risch	\(-50 x^{5}-\frac {325}{4} x^{4}-15 x^{3}+\frac {81}{2} x^{2}+27 x\)	\(25\)
parallelrisch	\(-50 x^{5}-\frac {325}{4} x^{4}-15 x^{3}+\frac {81}{2} x^{2}+27 x\)	\(25\)

[In]

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

[Out]

-1/4*x*(200*x^4+325*x^3+60*x^2-162*x-108)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x) (3+5 x)^3 \, dx=-50 \, x^{5} - \frac {325}{4} \, x^{4} - 15 \, x^{3} + \frac {81}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

-50*x^5 - 325/4*x^4 - 15*x^3 + 81/2*x^2 + 27*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int (1-2 x) (3+5 x)^3 \, dx=- 50 x^{5} - \frac {325 x^{4}}{4} - 15 x^{3} + \frac {81 x^{2}}{2} + 27 x \]

[In]

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

[Out]

-50*x**5 - 325*x**4/4 - 15*x**3 + 81*x**2/2 + 27*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x) (3+5 x)^3 \, dx=-50 \, x^{5} - \frac {325}{4} \, x^{4} - 15 \, x^{3} + \frac {81}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

-50*x^5 - 325/4*x^4 - 15*x^3 + 81/2*x^2 + 27*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x) (3+5 x)^3 \, dx=-50 \, x^{5} - \frac {325}{4} \, x^{4} - 15 \, x^{3} + \frac {81}{2} \, x^{2} + 27 \, x \]

[In]

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

[Out]

-50*x^5 - 325/4*x^4 - 15*x^3 + 81/2*x^2 + 27*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int (1-2 x) (3+5 x)^3 \, dx=-50\,x^5-\frac {325\,x^4}{4}-15\,x^3+\frac {81\,x^2}{2}+27\,x \]

[In]

int(-(2*x - 1)*(5*x + 3)^3,x)

[Out]

27*x + (81*x^2)/2 - 15*x^3 - (325*x^4)/4 - 50*x^5