\(\int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx\) [1158]

Optimal result

Integrand size = 19, antiderivative size = 57 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=\frac {x}{108 \sqrt {6} (1-2 x)^{3/2} (1+2 x)^{3/2}}+\frac {x}{54 \sqrt {6} \sqrt {1-2 x} \sqrt {1+2 x}} \]

[Out]

1/648*x/(1-2*x)^(3/2)/(1+2*x)^(3/2)*6^(1/2)+1/324*x*6^(1/2)/(1-2*x)^(1/2)/(1+2*x)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {40, 39} \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=\frac {x}{54 \sqrt {6} \sqrt {1-2 x} \sqrt {2 x+1}}+\frac {x}{108 \sqrt {6} (1-2 x)^{3/2} (2 x+1)^{3/2}} \]

[In]

Int[1/((3 - 6*x)^(5/2)*(2 + 4*x)^(5/2)),x]

[Out]

x/(108*Sqrt[6]*(1 - 2*x)^(3/2)*(1 + 2*x)^(3/2)) + x/(54*Sqrt[6]*Sqrt[1 - 2*x]*Sqrt[1 + 2*x])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /;
FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{108 \sqrt {6} (1-2 x)^{3/2} (1+2 x)^{3/2}}+\frac {1}{9} \int \frac {1}{(3-6 x)^{3/2} (2+4 x)^{3/2}} \, dx \\ & = \frac {x}{108 \sqrt {6} (1-2 x)^{3/2} (1+2 x)^{3/2}}+\frac {x}{54 \sqrt {6} \sqrt {1-2 x} \sqrt {1+2 x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=-\frac {x \left (-3+8 x^2\right )}{108 \sqrt {6} \left (1-4 x^2\right )^{3/2}} \]

[In]

Integrate[1/((3 - 6*x)^(5/2)*(2 + 4*x)^(5/2)),x]

[Out]

-1/108*(x*(-3 + 8*x^2))/(Sqrt[6]*(1 - 4*x^2)^(3/2))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61

method	result	size
gosper	\(\frac {\left (-1+2 x \right ) \left (1+2 x \right ) x \left (8 x^{2}-3\right )}{3 \left (3-6 x \right )^{\frac {5}{2}} \left (2+4 x \right )^{\frac {5}{2}}}\)	\(35\)
default	\(\frac {1}{36 \left (3-6 x \right )^{\frac {3}{2}} \left (2+4 x \right )^{\frac {3}{2}}}+\frac {1}{36 \sqrt {3-6 x}\, \left (2+4 x \right )^{\frac {3}{2}}}-\frac {\sqrt {3-6 x}}{162 \left (2+4 x \right )^{\frac {3}{2}}}-\frac {\sqrt {3-6 x}}{324 \sqrt {2+4 x}}\)	\(66\)

[In]

int(1/(3-6*x)^(5/2)/(2+4*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(-1+2*x)*(1+2*x)*x*(8*x^2-3)/(3-6*x)^(5/2)/(2+4*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=-\frac {{\left (8 \, x^{3} - 3 \, x\right )} \sqrt {4 \, x + 2} \sqrt {-6 \, x + 3}}{648 \, {\left (16 \, x^{4} - 8 \, x^{2} + 1\right )}} \]

[In]

integrate(1/(3-6*x)^(5/2)/(2+4*x)^(5/2),x, algorithm="fricas")

[Out]

-1/648*(8*x^3 - 3*x)*sqrt(4*x + 2)*sqrt(-6*x + 3)/(16*x^4 - 8*x^2 + 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(3-6*x)**(5/2)/(2+4*x)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=\frac {x}{54 \, \sqrt {-24 \, x^{2} + 6}} + \frac {x}{18 \, {\left (-24 \, x^{2} + 6\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(3-6*x)^(5/2)/(2+4*x)^(5/2),x, algorithm="maxima")

[Out]

1/54*x/sqrt(-24*x^2 + 6) + 1/18*x/(-24*x^2 + 6)^(3/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (41) = 82\).

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.54 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=\frac {\sqrt {6} {\left (\sqrt {2} - \sqrt {-2 \, x + 1}\right )}^{3}}{82944 \, {\left (2 \, x + 1\right )}^{\frac {3}{2}}} + \frac {11 \, \sqrt {6} {\left (\sqrt {2} - \sqrt {-2 \, x + 1}\right )}}{27648 \, \sqrt {2 \, x + 1}} - \frac {{\left (4 \, \sqrt {6} {\left (2 \, x + 1\right )} - 9 \, \sqrt {6}\right )} \sqrt {2 \, x + 1} \sqrt {-2 \, x + 1}}{5184 \, {\left (2 \, x - 1\right )}^{2}} - \frac {\sqrt {6} {\left (2 \, x + 1\right )}^{\frac {3}{2}} {\left (\frac {33 \, {\left (\sqrt {2} - \sqrt {-2 \, x + 1}\right )}^{2}}{2 \, x + 1} + 1\right )}}{82944 \, {\left (\sqrt {2} - \sqrt {-2 \, x + 1}\right )}^{3}} \]

[In]

integrate(1/(3-6*x)^(5/2)/(2+4*x)^(5/2),x, algorithm="giac")

[Out]

1/82944*sqrt(6)*(sqrt(2) - sqrt(-2*x + 1))^3/(2*x + 1)^(3/2) + 11/27648*sqrt(6)*(sqrt(2) - sqrt(-2*x + 1))/sqr
t(2*x + 1) - 1/5184*(4*sqrt(6)*(2*x + 1) - 9*sqrt(6))*sqrt(2*x + 1)*sqrt(-2*x + 1)/(2*x - 1)^2 - 1/82944*sqrt(
6)*(2*x + 1)^(3/2)*(33*(sqrt(2) - sqrt(-2*x + 1))^2/(2*x + 1) + 1)/(sqrt(2) - sqrt(-2*x + 1))^3

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(3-6 x)^{5/2} (2+4 x)^{5/2}} \, dx=-\frac {3\,x\,\sqrt {3-6\,x}-8\,x^3\,\sqrt {3-6\,x}}{\sqrt {4\,x+2}\,\left (-2592\,x^3+1296\,x^2+648\,x-324\right )} \]

[In]

int(1/((4*x + 2)^(5/2)*(3 - 6*x)^(5/2)),x)

[Out]

-(3*x*(3 - 6*x)^(1/2) - 8*x^3*(3 - 6*x)^(1/2))/((4*x + 2)^(1/2)*(648*x + 1296*x^2 - 2592*x^3 - 324))