Integrand size = 24, antiderivative size = 116 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {5929 \sqrt {1-2 x} \sqrt {3+5 x}}{16000}+\frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {65219 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{16000 \sqrt {10}} \]
65219/160000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+539/4800*(1-2*x) ^(3/2)*(3+5*x)^(1/2)+49/1200*(1-2*x)^(5/2)*(3+5*x)^(1/2)-3/40*(1-2*x)^(7/2 )*(3+5*x)^(1/2)+5929/16000*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {10 \sqrt {1-2 x} \left (64611+116625 x-91180 x^2-90400 x^3+144000 x^4\right )-195657 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{480000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(64611 + 116625*x - 91180*x^2 - 90400*x^3 + 144000*x^4) - 195657*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(480000*Sq rt[3 + 5*x])
Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 60, 60, 60, 64, 223}
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.
\(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)}{\sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {49}{80} \int \frac {(1-2 x)^{5/2}}{\sqrt {5 x+3}}dx-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {49}{80} \left (\frac {11}{6} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {49}{80} \left (\frac {11}{6} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {49}{80} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {49}{80} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {49}{80} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {3}{40} (1-2 x)^{7/2} \sqrt {5 x+3}\) |
(-3*(1 - 2*x)^(7/2)*Sqrt[3 + 5*x])/40 + (49*(((1 - 2*x)^(5/2)*Sqrt[3 + 5*x ])/15 + (11*(((1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/10 + (33*((Sqrt[1 - 2*x]*Sqrt [3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/20))/6 ))/80
3.25.27.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Time = 3.87 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {\left (28800 x^{3}-35360 x^{2}+2980 x +21537\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{48000 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {65219 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{320000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(103\) |
default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (576000 x^{3} \sqrt {-10 x^{2}-x +3}-707200 x^{2} \sqrt {-10 x^{2}-x +3}+195657 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+59600 x \sqrt {-10 x^{2}-x +3}+430740 \sqrt {-10 x^{2}-x +3}\right )}{960000 \sqrt {-10 x^{2}-x +3}}\) | \(104\) |
-1/48000*(28800*x^3-35360*x^2+2980*x+21537)*(-1+2*x)*(3+5*x)^(1/2)/(-(-1+2 *x)*(3+5*x))^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+65219/320000*10^( 1/2)*arcsin(20/11*x+1/11)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1 /2)
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {1}{48000} \, {\left (28800 \, x^{3} - 35360 \, x^{2} + 2980 \, x + 21537\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {65219}{320000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
1/48000*(28800*x^3 - 35360*x^2 + 2980*x + 21537)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 65219/320000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)* sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \cdot \left (3 x + 2\right )}{\sqrt {5 x + 3}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {3}{5} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {221}{300} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + \frac {149}{2400} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {65219}{320000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {7179}{16000} \, \sqrt {-10 \, x^{2} - x + 3} \]
3/5*sqrt(-10*x^2 - x + 3)*x^3 - 221/300*sqrt(-10*x^2 - x + 3)*x^2 + 149/24 00*sqrt(-10*x^2 - x + 3)*x - 65219/320000*sqrt(10)*arcsin(-20/11*x - 1/11) + 7179/16000*sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (83) = 166\).
Time = 0.31 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.75 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {1}{800000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (60 \, x - 119\right )} {\left (5 \, x + 3\right )} + 6163\right )} {\left (5 \, x + 3\right )} - 66189\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 184305 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{30000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{400} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \]
1/800000*sqrt(5)*(2*(4*(8*(60*x - 119)*(5*x + 3) + 6163)*(5*x + 3) - 66189 )*sqrt(5*x + 3)*sqrt(-10*x + 5) - 184305*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt (5*x + 3))) - 1/30000*sqrt(5)*(2*(4*(40*x - 59)*(5*x + 3) + 1293)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 4785*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) - 1/400*sqrt(5)*(2*(20*x - 23)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 143*sqrt(2) *arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) + 1/25*sqrt(5)*(11*sqrt(2)*arcsin(1/ 11*sqrt(22)*sqrt(5*x + 3)) + 2*sqrt(5*x + 3)*sqrt(-10*x + 5))
Timed out. \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )}{\sqrt {5\,x+3}} \,d x \]