Integrand size = 22, antiderivative size = 79 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=\frac {26411}{96 (1-2 x)^{3/2}}-\frac {57281}{32 \sqrt {1-2 x}}-\frac {24843}{16} \sqrt {1-2 x}+\frac {3591}{16} (1-2 x)^{3/2}-\frac {4671}{160} (1-2 x)^{5/2}+\frac {405}{224} (1-2 x)^{7/2} \] Output:
26411/96/(1-2*x)^(3/2)-57281/32/(1-2*x)^(1/2)-24843/16*(1-2*x)^(1/2)+3591/ 16*(1-2*x)^(3/2)-4671/160*(1-2*x)^(5/2)+405/224*(1-2*x)^(7/2)
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.48 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=-\frac {301408-909264 x+435312 x^2+105624 x^3+33858 x^4+6075 x^5}{105 (1-2 x)^{3/2}} \] Input:
Integrate[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^(5/2),x]
Output:
-1/105*(301408 - 909264*x + 435312*x^2 + 105624*x^3 + 33858*x^4 + 6075*x^5 )/(1 - 2*x)^(3/2)
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {86, 2009}
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.
\(\displaystyle \int \frac {(3 x+2)^4 (5 x+3)}{(1-2 x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {405}{32} (1-2 x)^{5/2}+\frac {4671}{32} (1-2 x)^{3/2}-\frac {10773}{16} \sqrt {1-2 x}+\frac {24843}{16 \sqrt {1-2 x}}-\frac {57281}{32 (1-2 x)^{3/2}}+\frac {26411}{32 (1-2 x)^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {405}{224} (1-2 x)^{7/2}-\frac {4671}{160} (1-2 x)^{5/2}+\frac {3591}{16} (1-2 x)^{3/2}-\frac {24843}{16} \sqrt {1-2 x}-\frac {57281}{32 \sqrt {1-2 x}}+\frac {26411}{96 (1-2 x)^{3/2}}\) |
Input:
Int[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x)^(5/2),x]
Output:
26411/(96*(1 - 2*x)^(3/2)) - 57281/(32*Sqrt[1 - 2*x]) - (24843*Sqrt[1 - 2* x])/16 + (3591*(1 - 2*x)^(3/2))/16 - (4671*(1 - 2*x)^(5/2))/160 + (405*(1 - 2*x)^(7/2))/224
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((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]))))
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {6075 x^{5}+33858 x^{4}+105624 x^{3}+435312 x^{2}-909264 x +301408}{105 \left (1-2 x \right )^{\frac {3}{2}}}\) | \(35\) |
pseudoelliptic | \(\frac {-6075 x^{5}-33858 x^{4}-105624 x^{3}-435312 x^{2}+909264 x -301408}{105 \left (1-2 x \right )^{\frac {3}{2}}}\) | \(35\) |
orering | \(\frac {\left (-1+2 x \right ) \left (6075 x^{5}+33858 x^{4}+105624 x^{3}+435312 x^{2}-909264 x +301408\right )}{105 \left (1-2 x \right )^{\frac {5}{2}}}\) | \(40\) |
trager | \(-\frac {\left (6075 x^{5}+33858 x^{4}+105624 x^{3}+435312 x^{2}-909264 x +301408\right ) \sqrt {1-2 x}}{105 \left (-1+2 x \right )^{2}}\) | \(42\) |
risch | \(\frac {6075 x^{5}+33858 x^{4}+105624 x^{3}+435312 x^{2}-909264 x +301408}{105 \left (-1+2 x \right ) \sqrt {1-2 x}}\) | \(42\) |
derivativedivides | \(\frac {26411}{96 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {57281}{32 \sqrt {1-2 x}}-\frac {24843 \sqrt {1-2 x}}{16}+\frac {3591 \left (1-2 x \right )^{\frac {3}{2}}}{16}-\frac {4671 \left (1-2 x \right )^{\frac {5}{2}}}{160}+\frac {405 \left (1-2 x \right )^{\frac {7}{2}}}{224}\) | \(56\) |
default | \(\frac {26411}{96 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {57281}{32 \sqrt {1-2 x}}-\frac {24843 \sqrt {1-2 x}}{16}+\frac {3591 \left (1-2 x \right )^{\frac {3}{2}}}{16}-\frac {4671 \left (1-2 x \right )^{\frac {5}{2}}}{160}+\frac {405 \left (1-2 x \right )^{\frac {7}{2}}}{224}\) | \(56\) |
meijerg | \(-\frac {32 \left (\frac {\sqrt {\pi }}{2}-\frac {\sqrt {\pi }}{2 \left (1-2 x \right )^{\frac {3}{2}}}\right )}{\sqrt {\pi }}+\frac {\frac {368 \sqrt {\pi }}{3}-\frac {46 \sqrt {\pi }\, \left (-24 x +8\right )}{3 \left (1-2 x \right )^{\frac {3}{2}}}}{\sqrt {\pi }}-\frac {188 \left (-4 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (24 x^{2}-48 x +16\right )}{4 \left (1-2 x \right )^{\frac {3}{2}}}\right )}{\sqrt {\pi }}+\frac {1152 \sqrt {\pi }-\frac {9 \sqrt {\pi }\, \left (64 x^{3}+192 x^{2}-384 x +128\right )}{\left (1-2 x \right )^{\frac {3}{2}}}}{\sqrt {\pi }}-\frac {441 \left (-\frac {64 \sqrt {\pi }}{5}+\frac {\sqrt {\pi }\, \left (96 x^{4}+128 x^{3}+384 x^{2}-768 x +256\right )}{20 \left (1-2 x \right )^{\frac {3}{2}}}\right )}{8 \sqrt {\pi }}+\frac {\frac {1080 \sqrt {\pi }}{7}-\frac {135 \sqrt {\pi }\, \left (384 x^{5}+384 x^{4}+512 x^{3}+1536 x^{2}-3072 x +1024\right )}{896 \left (1-2 x \right )^{\frac {3}{2}}}}{\sqrt {\pi }}\) | \(213\) |
Input:
int((2+3*x)^4*(3+5*x)/(1-2*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/105/(1-2*x)^(3/2)*(6075*x^5+33858*x^4+105624*x^3+435312*x^2-909264*x+30 1408)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=-\frac {{\left (6075 \, x^{5} + 33858 \, x^{4} + 105624 \, x^{3} + 435312 \, x^{2} - 909264 \, x + 301408\right )} \sqrt {-2 \, x + 1}}{105 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] Input:
integrate((2+3*x)^4*(3+5*x)/(1-2*x)^(5/2),x, algorithm="fricas")
Output:
-1/105*(6075*x^5 + 33858*x^4 + 105624*x^3 + 435312*x^2 - 909264*x + 301408 )*sqrt(-2*x + 1)/(4*x^2 - 4*x + 1)
Time = 0.88 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=\frac {405 \left (1 - 2 x\right )^{\frac {7}{2}}}{224} - \frac {4671 \left (1 - 2 x\right )^{\frac {5}{2}}}{160} + \frac {3591 \left (1 - 2 x\right )^{\frac {3}{2}}}{16} - \frac {24843 \sqrt {1 - 2 x}}{16} - \frac {57281}{32 \sqrt {1 - 2 x}} + \frac {26411}{96 \left (1 - 2 x\right )^{\frac {3}{2}}} \] Input:
integrate((2+3*x)**4*(3+5*x)/(1-2*x)**(5/2),x)
Output:
405*(1 - 2*x)**(7/2)/224 - 4671*(1 - 2*x)**(5/2)/160 + 3591*(1 - 2*x)**(3/ 2)/16 - 24843*sqrt(1 - 2*x)/16 - 57281/(32*sqrt(1 - 2*x)) + 26411/(96*(1 - 2*x)**(3/2))
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=\frac {405}{224} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {4671}{160} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {3591}{16} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {24843}{16} \, \sqrt {-2 \, x + 1} + \frac {343 \, {\left (501 \, x - 212\right )}}{48 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \] Input:
integrate((2+3*x)^4*(3+5*x)/(1-2*x)^(5/2),x, algorithm="maxima")
Output:
405/224*(-2*x + 1)^(7/2) - 4671/160*(-2*x + 1)^(5/2) + 3591/16*(-2*x + 1)^ (3/2) - 24843/16*sqrt(-2*x + 1) + 343/48*(501*x - 212)/(-2*x + 1)^(3/2)
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=-\frac {405}{224} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {4671}{160} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {3591}{16} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {24843}{16} \, \sqrt {-2 \, x + 1} - \frac {343 \, {\left (501 \, x - 212\right )}}{48 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \] Input:
integrate((2+3*x)^4*(3+5*x)/(1-2*x)^(5/2),x, algorithm="giac")
Output:
-405/224*(2*x - 1)^3*sqrt(-2*x + 1) - 4671/160*(2*x - 1)^2*sqrt(-2*x + 1) + 3591/16*(-2*x + 1)^(3/2) - 24843/16*sqrt(-2*x + 1) - 343/48*(501*x - 212 )/((2*x - 1)*sqrt(-2*x + 1))
Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.63 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=\frac {\frac {57281\,x}{16}-\frac {18179}{12}}{{\left (1-2\,x\right )}^{3/2}}-\frac {24843\,\sqrt {1-2\,x}}{16}+\frac {3591\,{\left (1-2\,x\right )}^{3/2}}{16}-\frac {4671\,{\left (1-2\,x\right )}^{5/2}}{160}+\frac {405\,{\left (1-2\,x\right )}^{7/2}}{224} \] Input:
int(((3*x + 2)^4*(5*x + 3))/(1 - 2*x)^(5/2),x)
Output:
((57281*x)/16 - 18179/12)/(1 - 2*x)^(3/2) - (24843*(1 - 2*x)^(1/2))/16 + ( 3591*(1 - 2*x)^(3/2))/16 - (4671*(1 - 2*x)^(5/2))/160 + (405*(1 - 2*x)^(7/ 2))/224
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.53 \[ \int \frac {(2+3 x)^4 (3+5 x)}{(1-2 x)^{5/2}} \, dx=\frac {6075 x^{5}+33858 x^{4}+105624 x^{3}+435312 x^{2}-909264 x +301408}{105 \sqrt {-2 x +1}\, \left (2 x -1\right )} \] Input:
int((2+3*x)^4*(3+5*x)/(1-2*x)^(5/2),x)
Output:
(6075*x**5 + 33858*x**4 + 105624*x**3 + 435312*x**2 - 909264*x + 301408)/( 105*sqrt( - 2*x + 1)*(2*x - 1))