\(\int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx\) [1155]

Optimal result

Integrand size = 18, antiderivative size = 34 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {7}{81 (2+3 x)^3}-\frac {37}{54 (2+3 x)^2}+\frac {10}{27 (2+3 x)} \]

[Out]

7/81/(2+3*x)^3-37/54/(2+3*x)^2+10/27/(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {10}{27 (3 x+2)}-\frac {37}{54 (3 x+2)^2}+\frac {7}{81 (3 x+2)^3} \]

[In]

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

[Out]

7/(81*(2 + 3*x)^3) - 37/(54*(2 + 3*x)^2) + 10/(27*(2 + 3*x))

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}{9 (2+3 x)^4}+\frac {37}{9 (2+3 x)^3}-\frac {10}{9 (2+3 x)^2}\right ) \, dx \\ & = \frac {7}{81 (2+3 x)^3}-\frac {37}{54 (2+3 x)^2}+\frac {10}{27 (2+3 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {32+387 x+540 x^2}{162 (2+3 x)^3} \]

[In]

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

[Out]

(32 + 387*x + 540*x^2)/(162*(2 + 3*x)^3)

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59

method	result	size
gosper	\(\frac {540 x^{2}+387 x +32}{162 \left (2+3 x \right )^{3}}\)	\(20\)
risch	\(\frac {\frac {10}{3} x^{2}+\frac {43}{18} x +\frac {16}{81}}{\left (2+3 x \right )^{3}}\)	\(20\)
norman	\(\frac {-\frac {2}{3} x^{3}+\frac {3}{2} x +2 x^{2}}{\left (2+3 x \right )^{3}}\)	\(23\)
parallelrisch	\(\frac {-16 x^{3}+48 x^{2}+36 x}{24 \left (2+3 x \right )^{3}}\)	\(24\)
default	\(\frac {7}{81 \left (2+3 x \right )^{3}}-\frac {37}{54 \left (2+3 x \right )^{2}}+\frac {10}{27 \left (2+3 x \right )}\)	\(29\)
meijerg	\(\frac {x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {x^{2} \left (3+\frac {3 x}{2}\right )}{96 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {5 x^{3}}{24 \left (1+\frac {3 x}{2}\right )^{3}}\)	\(51\)

[In]

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

[Out]

1/162*(540*x^2+387*x+32)/(2+3*x)^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {540 \, x^{2} + 387 \, x + 32}{162 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

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

[Out]

1/162*(540*x^2 + 387*x + 32)/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=- \frac {- 540 x^{2} - 387 x - 32}{4374 x^{3} + 8748 x^{2} + 5832 x + 1296} \]

[In]

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

[Out]

-(-540*x**2 - 387*x - 32)/(4374*x**3 + 8748*x**2 + 5832*x + 1296)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {540 \, x^{2} + 387 \, x + 32}{162 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

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

[Out]

1/162*(540*x^2 + 387*x + 32)/(27*x^3 + 54*x^2 + 36*x + 8)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {540 \, x^{2} + 387 \, x + 32}{162 \, {\left (3 \, x + 2\right )}^{3}} \]

[In]

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

[Out]

1/162*(540*x^2 + 387*x + 32)/(3*x + 2)^3

Mupad [B] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx=\frac {10}{27\,\left (3\,x+2\right )}-\frac {37}{54\,{\left (3\,x+2\right )}^2}+\frac {7}{81\,{\left (3\,x+2\right )}^3} \]

[In]

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

[Out]

10/(27*(3*x + 2)) - 37/(54*(3*x + 2)^2) + 7/(81*(3*x + 2)^3)