Integrand size = 24, antiderivative size = 116 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {29887 \sqrt {1-2 x} \sqrt {3+5 x}}{1024}-\frac {2717}{768} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {247}{480} \sqrt {1-2 x} (3+5 x)^{5/2}-\frac {3}{40} \sqrt {1-2 x} (3+5 x)^{7/2}+\frac {328757 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1024 \sqrt {10}} \]
328757/10240*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2717/768*(3+5*x) ^(3/2)*(1-2*x)^(1/2)-247/480*(3+5*x)^(5/2)*(1-2*x)^(1/2)-3/40*(3+5*x)^(7/2 )*(1-2*x)^(1/2)-29887/1024*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=\frac {-10 \sqrt {1-2 x} \left (428139+1112169 x+938420 x^2+543200 x^3+144000 x^4\right )-986271 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{30720 \sqrt {3+5 x}} \]
(-10*Sqrt[1 - 2*x]*(428139 + 1112169*x + 938420*x^2 + 543200*x^3 + 144000* x^4) - 986271*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(3072 0*Sqrt[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.
\(\displaystyle \int \frac {(3 x+2) (5 x+3)^{5/2}}{\sqrt {1-2 x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {247}{80} \int \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}dx-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {247}{80} \left (\frac {55}{12} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {247}{80} \left (\frac {55}{12} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {247}{80} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {247}{80} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {247}{80} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {3}{40} \sqrt {1-2 x} (5 x+3)^{7/2}\) |
(-3*Sqrt[1 - 2*x]*(3 + 5*x)^(7/2))/40 + (247*(-1/6*(Sqrt[1 - 2*x]*(3 + 5*x )^(5/2)) + (55*(-1/4*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (33*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])) )/8))/12))/80
3.25.80.3.1 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 = 1.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {\left (28800 x^{3}+91360 x^{2}+132868 x +142713\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{3072 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {328757 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{20480 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(103\) |
default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-576000 x^{3} \sqrt {-10 x^{2}-x +3}-1827200 x^{2} \sqrt {-10 x^{2}-x +3}+986271 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-2657360 x \sqrt {-10 x^{2}-x +3}-2854260 \sqrt {-10 x^{2}-x +3}\right )}{61440 \sqrt {-10 x^{2}-x +3}}\) | \(104\) |
1/3072*(28800*x^3+91360*x^2+132868*x+142713)*(-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)+328757/20480*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 = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{3072} \, {\left (28800 \, x^{3} + 91360 \, x^{2} + 132868 \, x + 142713\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {328757}{20480} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/3072*(28800*x^3 + 91360*x^2 + 132868*x + 142713)*sqrt(5*x + 3)*sqrt(-2* x + 1) - 328757/20480*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {5}{2}}}{\sqrt {1 - 2 x}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {75}{8} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {2855}{96} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {33217}{768} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {328757}{20480} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {47571}{1024} \, \sqrt {-10 \, x^{2} - x + 3} \]
-75/8*sqrt(-10*x^2 - x + 3)*x^3 - 2855/96*sqrt(-10*x^2 - x + 3)*x^2 - 3321 7/768*sqrt(-10*x^2 - x + 3)*x - 328757/20480*sqrt(10)*arcsin(-20/11*x - 1/ 11) - 47571/1024*sqrt(-10*x^2 - x + 3)
Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.54 \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{30720} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (36 \, x + 71\right )} {\left (5 \, x + 3\right )} + 2717\right )} {\left (5 \, x + 3\right )} + 89661\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 986271 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]
-1/30720*sqrt(5)*(2*(4*(8*(36*x + 71)*(5*x + 3) + 2717)*(5*x + 3) + 89661) *sqrt(5*x + 3)*sqrt(-10*x + 5) - 986271*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt( 5*x + 3)))
Timed out. \[ \int \frac {(2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{5/2}}{\sqrt {1-2\,x}} \,d x \]