Integrand size = 24, antiderivative size = 94 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {1947}{320} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {59}{80} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {1}{10} \sqrt {1-2 x} (3+5 x)^{5/2}+\frac {21417 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{320 \sqrt {10}} \]
21417/3200*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-59/80*(3+5*x)^(3/2 )*(1-2*x)^(1/2)-1/10*(3+5*x)^(5/2)*(1-2*x)^(1/2)-1947/320*(1-2*x)^(1/2)*(3 +5*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\frac {-10 \sqrt {1-2 x} \left (8829+21135 x+13100 x^2+4000 x^3\right )-21417 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{3200 \sqrt {3+5 x}} \]
(-10*Sqrt[1 - 2*x]*(8829 + 21135*x + 13100*x^2 + 4000*x^3) - 21417*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(3200*Sqrt[3 + 5*x])
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {90, 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.
\(\displaystyle \int \frac {(3 x+2) (5 x+3)^{3/2}}{\sqrt {1-2 x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {59}{20} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {1}{10} \sqrt {1-2 x} (5 x+3)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {59}{20} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (5 x+3)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {59}{20} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (5 x+3)^{5/2}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {59}{20} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (5 x+3)^{5/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {59}{20} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (5 x+3)^{5/2}\) |
-1/10*(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)) + (59*(-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))/20
3.25.70.3.1 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 = 1.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-16000 x^{2} \sqrt {-10 x^{2}-x +3}+21417 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-42800 x \sqrt {-10 x^{2}-x +3}-58860 \sqrt {-10 x^{2}-x +3}\right )}{6400 \sqrt {-10 x^{2}-x +3}}\) | \(87\) |
risch | \(\frac {\left (800 x^{2}+2140 x +2943\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{320 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {21417 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{6400 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(98\) |
1/6400*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(-16000*x^2*(-10*x^2-x+3)^(1/2)+21417*1 0^(1/2)*arcsin(20/11*x+1/11)-42800*x*(-10*x^2-x+3)^(1/2)-58860*(-10*x^2-x+ 3)^(1/2))/(-10*x^2-x+3)^(1/2)
Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{320} \, {\left (800 \, x^{2} + 2140 \, x + 2943\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {21417}{6400} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/320*(800*x^2 + 2140*x + 2943)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 21417/6400 *sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10 *x^2 + x - 3))
\[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {3}{2}}}{\sqrt {1 - 2 x}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.62 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {5}{2} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {107}{16} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {21417}{6400} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {2943}{320} \, \sqrt {-10 \, x^{2} - x + 3} \]
-5/2*sqrt(-10*x^2 - x + 3)*x^2 - 107/16*sqrt(-10*x^2 - x + 3)*x - 21417/64 00*sqrt(10)*arcsin(-20/11*x - 1/11) - 2943/320*sqrt(-10*x^2 - x + 3)
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57 \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{3200} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x + 83\right )} {\left (5 \, x + 3\right )} + 1947\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 21417 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]
-1/3200*sqrt(5)*(2*(4*(40*x + 83)*(5*x + 3) + 1947)*sqrt(5*x + 3)*sqrt(-10 *x + 5) - 21417*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)))
Timed out. \[ \int \frac {(2+3 x) (3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{3/2}}{\sqrt {1-2\,x}} \,d x \]