\(\int \frac {1}{(3 x-4 x^2)^{3/2}} \, dx\) [22]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=-\frac {2 (3-8 x)}{9 \sqrt {3 x-4 x^2}} \]

[Out]

-2/9*(3-8*x)/(-4*x^2+3*x)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {627} \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=-\frac {2 (3-8 x)}{9 \sqrt {3 x-4 x^2}} \]

[In]

Int[(3*x - 4*x^2)^(-3/2),x]

[Out]

(-2*(3 - 8*x))/(9*Sqrt[3*x - 4*x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (3-8 x)}{9 \sqrt {3 x-4 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=\frac {2 (-3+8 x)}{9 \sqrt {-x (-3+4 x)}} \]

[In]

Integrate[(3*x - 4*x^2)^(-3/2),x]

[Out]

(2*(-3 + 8*x))/(9*Sqrt[-(x*(-3 + 4*x))])

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {2 \left (3-8 x \right )}{9 \sqrt {-4 x^{2}+3 x}}\)	\(19\)
pseudoelliptic	\(\frac {-\frac {2}{3}+\frac {16 x}{9}}{\sqrt {-4 x^{2}+3 x}}\)	\(19\)
meijerg	\(-\frac {2 \sqrt {3}\, \left (1-\frac {8 x}{3}\right )}{9 \sqrt {x}\, \sqrt {-\frac {4 x}{3}+1}}\)	\(21\)
gosper	\(-\frac {2 x \left (4 x -3\right ) \left (-3+8 x \right )}{9 \left (-4 x^{2}+3 x \right )^{\frac {3}{2}}}\)	\(25\)
trager	\(-\frac {2 \left (-3+8 x \right ) \sqrt {-4 x^{2}+3 x}}{9 x \left (4 x -3\right )}\)	\(29\)

[In]

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

[Out]

-2/9*(3-8*x)/(-4*x^2+3*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {-4 \, x^{2} + 3 \, x} {\left (8 \, x - 3\right )}}{9 \, {\left (4 \, x^{2} - 3 \, x\right )}} \]

[In]

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

[Out]

-2/9*sqrt(-4*x^2 + 3*x)*(8*x - 3)/(4*x^2 - 3*x)

Sympy [F]

\[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (- 4 x^{2} + 3 x\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral((-4*x**2 + 3*x)**(-3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=\frac {16 \, x}{9 \, \sqrt {-4 \, x^{2} + 3 \, x}} - \frac {2}{3 \, \sqrt {-4 \, x^{2} + 3 \, x}} \]

[In]

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

[Out]

16/9*x/sqrt(-4*x^2 + 3*x) - 2/3/sqrt(-4*x^2 + 3*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {-4 \, x^{2} + 3 \, x} {\left (8 \, x - 3\right )}}{9 \, {\left (4 \, x^{2} - 3 \, x\right )}} \]

[In]

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

[Out]

-2/9*sqrt(-4*x^2 + 3*x)*(8*x - 3)/(4*x^2 - 3*x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}} \, dx=\frac {16\,x-6}{9\,\sqrt {3\,x-4\,x^2}} \]

[In]

int(1/(3*x - 4*x^2)^(3/2),x)

[Out]

(16*x - 6)/(9*(3*x - 4*x^2)^(1/2))