Integrand size = 13, antiderivative size = 109 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=\frac {2}{15 (3-4 x)^{5/2} x^{5/2}}+\frac {4}{27 (3-4 x)^{3/2} x^{5/2}}+\frac {32}{81 \sqrt {3-4 x} x^{5/2}}-\frac {64 \sqrt {3-4 x}}{405 x^{5/2}}-\frac {1024 \sqrt {3-4 x}}{3645 x^{3/2}}-\frac {8192 \sqrt {3-4 x}}{10935 \sqrt {x}} \] Output:
2/15/(3-4*x)^(5/2)/x^(5/2)+4/27/(3-4*x)^(3/2)/x^(5/2)+32/81/(3-4*x)^(1/2)/ x^(5/2)-64/405*(3-4*x)^(1/2)/x^(5/2)-1024/3645*(3-4*x)^(1/2)/x^(3/2)-8192/ 10935*(3-4*x)^(1/2)/x^(1/2)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=\frac {2 \left (-729-3240 x-34560 x^2+276480 x^3-491520 x^4+262144 x^5\right )}{10935 (-x (-3+4 x))^{5/2}} \] Input:
Integrate[(3*x - 4*x^2)^(-7/2),x]
Output:
(2*(-729 - 3240*x - 34560*x^2 + 276480*x^3 - 491520*x^4 + 262144*x^5))/(10 935*(-(x*(-3 + 4*x)))^(5/2))
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.66, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1089, 1089, 1088}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle \frac {64}{45} \int \frac {1}{\left (3 x-4 x^2\right )^{5/2}}dx-\frac {2 (3-8 x)}{45 \left (3 x-4 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 1089 |
\(\displaystyle \frac {64}{45} \left (\frac {32}{27} \int \frac {1}{\left (3 x-4 x^2\right )^{3/2}}dx-\frac {2 (3-8 x)}{27 \left (3 x-4 x^2\right )^{3/2}}\right )-\frac {2 (3-8 x)}{45 \left (3 x-4 x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle \frac {64}{45} \left (-\frac {64 (3-8 x)}{243 \sqrt {3 x-4 x^2}}-\frac {2 (3-8 x)}{27 \left (3 x-4 x^2\right )^{3/2}}\right )-\frac {2 (3-8 x)}{45 \left (3 x-4 x^2\right )^{5/2}}\) |
Input:
Int[(3*x - 4*x^2)^(-7/2),x]
Output:
(-2*(3 - 8*x))/(45*(3*x - 4*x^2)^(5/2)) + (64*((-2*(3 - 8*x))/(27*(3*x - 4 *x^2)^(3/2)) - (64*(3 - 8*x))/(243*Sqrt[3*x - 4*x^2])))/45
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]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.38
method | result | size |
meijerg | \(-\frac {2 \sqrt {3}\, \left (-\frac {262144}{243} x^{5}+\frac {163840}{81} x^{4}-\frac {10240}{9} x^{3}+\frac {1280}{9} x^{2}+\frac {40}{3} x +3\right )}{1215 x^{\frac {5}{2}} \left (-\frac {4 x}{3}+1\right )^{\frac {5}{2}}}\) | \(41\) |
gosper | \(-\frac {2 x \left (4 x -3\right ) \left (262144 x^{5}-491520 x^{4}+276480 x^{3}-34560 x^{2}-3240 x -729\right )}{10935 \left (-4 x^{2}+3 x \right )^{\frac {7}{2}}}\) | \(45\) |
orering | \(-\frac {2 x \left (4 x -3\right ) \left (262144 x^{5}-491520 x^{4}+276480 x^{3}-34560 x^{2}-3240 x -729\right )}{10935 \left (-4 x^{2}+3 x \right )^{\frac {7}{2}}}\) | \(45\) |
trager | \(-\frac {2 \left (262144 x^{5}-491520 x^{4}+276480 x^{3}-34560 x^{2}-3240 x -729\right ) \sqrt {-4 x^{2}+3 x}}{10935 x^{3} \left (4 x -3\right )^{3}}\) | \(49\) |
pseudoelliptic | \(\frac {\frac {524288}{10935} x^{5}-\frac {65536}{729} x^{4}+\frac {4096}{81} x^{3}-\frac {512}{81} x^{2}-\frac {16}{27} x -\frac {2}{15}}{x^{2} \left (4 x -3\right )^{2} \sqrt {-4 x^{2}+3 x}}\) | \(49\) |
default | \(-\frac {2 \left (-8 x +3\right )}{45 \left (-4 x^{2}+3 x \right )^{\frac {5}{2}}}-\frac {128 \left (-8 x +3\right )}{1215 \left (-4 x^{2}+3 x \right )^{\frac {3}{2}}}-\frac {4096 \left (-8 x +3\right )}{10935 \sqrt {-4 x^{2}+3 x}}\) | \(56\) |
Input:
int(1/(-4*x^2+3*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/1215/x^(5/2)*3^(1/2)*(-262144/243*x^5+163840/81*x^4-10240/9*x^3+1280/9* x^2+40/3*x+3)/(-4/3*x+1)^(5/2)
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (262144 \, x^{5} - 491520 \, x^{4} + 276480 \, x^{3} - 34560 \, x^{2} - 3240 \, x - 729\right )} \sqrt {-4 \, x^{2} + 3 \, x}}{10935 \, {\left (64 \, x^{6} - 144 \, x^{5} + 108 \, x^{4} - 27 \, x^{3}\right )}} \] Input:
integrate(1/(-4*x^2+3*x)^(7/2),x, algorithm="fricas")
Output:
-2/10935*(262144*x^5 - 491520*x^4 + 276480*x^3 - 34560*x^2 - 3240*x - 729) *sqrt(-4*x^2 + 3*x)/(64*x^6 - 144*x^5 + 108*x^4 - 27*x^3)
\[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=\int \frac {1}{\left (- 4 x^{2} + 3 x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate(1/(-4*x**2+3*x)**(7/2),x)
Output:
Integral((-4*x**2 + 3*x)**(-7/2), x)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=\frac {32768 \, x}{10935 \, \sqrt {-4 \, x^{2} + 3 \, x}} - \frac {4096}{3645 \, \sqrt {-4 \, x^{2} + 3 \, x}} + \frac {1024 \, x}{1215 \, {\left (-4 \, x^{2} + 3 \, x\right )}^{\frac {3}{2}}} - \frac {128}{405 \, {\left (-4 \, x^{2} + 3 \, x\right )}^{\frac {3}{2}}} + \frac {16 \, x}{45 \, {\left (-4 \, x^{2} + 3 \, x\right )}^{\frac {5}{2}}} - \frac {2}{15 \, {\left (-4 \, x^{2} + 3 \, x\right )}^{\frac {5}{2}}} \] Input:
integrate(1/(-4*x^2+3*x)^(7/2),x, algorithm="maxima")
Output:
32768/10935*x/sqrt(-4*x^2 + 3*x) - 4096/3645/sqrt(-4*x^2 + 3*x) + 1024/121 5*x/(-4*x^2 + 3*x)^(3/2) - 128/405/(-4*x^2 + 3*x)^(3/2) + 16/45*x/(-4*x^2 + 3*x)^(5/2) - 2/15/(-4*x^2 + 3*x)^(5/2)
Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (8 \, {\left (32 \, {\left (8 \, {\left (16 \, {\left (8 \, x - 15\right )} x + 135\right )} x - 135\right )} x - 405\right )} x - 729\right )} \sqrt {-4 \, x^{2} + 3 \, x}}{10935 \, {\left (4 \, x^{2} - 3 \, x\right )}^{3}} \] Input:
integrate(1/(-4*x^2+3*x)^(7/2),x, algorithm="giac")
Output:
-2/10935*(8*(32*(8*(16*(8*x - 15)*x + 135)*x - 135)*x - 405)*x - 729)*sqrt (-4*x^2 + 3*x)/(4*x^2 - 3*x)^3
Time = 9.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=-\frac {6480\,x-9216\,x\,\left (3\,x-4\,x^2\right )-32768\,x\,{\left (3\,x-4\,x^2\right )}^2+12288\,{\left (3\,x-4\,x^2\right )}^2-13824\,x^2+1458}{{\left (3\,x-4\,x^2\right )}^{3/2}\,\left (32805\,x-43740\,x^2\right )} \] Input:
int(1/(3*x - 4*x^2)^(7/2),x)
Output:
-(6480*x - 9216*x*(3*x - 4*x^2) - 32768*x*(3*x - 4*x^2)^2 + 12288*(3*x - 4 *x^2)^2 - 13824*x^2 + 1458)/((3*x - 4*x^2)^(3/2)*(32805*x - 43740*x^2))
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (3 x-4 x^2\right )^{7/2}} \, dx=\frac {\frac {262144 \sqrt {-4 x +3}\, i \,x^{5}}{10935}-\frac {131072 \sqrt {-4 x +3}\, i \,x^{4}}{3645}+\frac {16384 \sqrt {-4 x +3}\, i \,x^{3}}{1215}+\frac {524288 \sqrt {x}\, x^{5}}{10935}-\frac {65536 \sqrt {x}\, x^{4}}{729}+\frac {4096 \sqrt {x}\, x^{3}}{81}-\frac {512 \sqrt {x}\, x^{2}}{81}-\frac {16 \sqrt {x}\, x}{27}-\frac {2 \sqrt {x}}{15}}{\sqrt {-4 x +3}\, x^{3} \left (16 x^{2}-24 x +9\right )} \] Input:
int(1/(-4*x^2+3*x)^(7/2),x)
Output:
(2*(131072*sqrt( - 4*x + 3)*i*x**5 - 196608*sqrt( - 4*x + 3)*i*x**4 + 7372 8*sqrt( - 4*x + 3)*i*x**3 + 262144*sqrt(x)*x**5 - 491520*sqrt(x)*x**4 + 27 6480*sqrt(x)*x**3 - 34560*sqrt(x)*x**2 - 3240*sqrt(x)*x - 729*sqrt(x)))/(1 0935*sqrt( - 4*x + 3)*x**3*(16*x**2 - 24*x + 9))