Integrand size = 24, antiderivative size = 72 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {107}{80} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1177 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{80 \sqrt {10}} \]
1177/800*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-3/20*(3+5*x)^(3/2)*( 1-2*x)^(1/2)-107/80*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {-\sqrt {5-10 x} \sqrt {3+5 x} (143+60 x)-1177 \sqrt {2} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{80 \sqrt {5}} \]
(-(Sqrt[5 - 10*x]*Sqrt[3 + 5*x]*(143 + 60*x)) - 1177*Sqrt[2]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/(80*Sqrt[5])
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {90, 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.
\(\displaystyle \int \frac {(3 x+2) \sqrt {5 x+3}}{\sqrt {1-2 x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {107}{40} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {3}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {107}{40} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {107}{40} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {107}{40} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\) |
(-3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/20 + (107*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/40
3.25.61.3.1 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 = 70, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (1177 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1200 x \sqrt {-10 x^{2}-x +3}-2860 \sqrt {-10 x^{2}-x +3}\right )}{1600 \sqrt {-10 x^{2}-x +3}}\) | \(70\) |
risch | \(\frac {\left (143+60 x \right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{80 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {1177 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{1600 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(93\) |
1/1600*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(1177*10^(1/2)*arcsin(20/11*x+1/11)-120 0*x*(-10*x^2-x+3)^(1/2)-2860*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{80} \, {\left (60 \, x + 143\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {1177}{1600} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/80*(60*x + 143)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 1177/1600*sqrt(10)*arcta n(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right ) \sqrt {5 x + 3}}{\sqrt {1 - 2 x}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {1177}{1600} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {3}{4} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {143}{80} \, \sqrt {-10 \, x^{2} - x + 3} \]
1177/1600*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 3/4*sqrt(-10*x^2 - x + 3)*x - 143/80*sqrt(-10*x^2 - x + 3)
Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{800} \, \sqrt {5} {\left (2 \, {\left (60 \, x + 143\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 1177 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]
-1/800*sqrt(5)*(2*(60*x + 143)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1177*sqrt(2 )*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)))
Time = 13.49 (sec) , antiderivative size = 509, normalized size of antiderivative = 7.07 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {1177\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\left (\sqrt {1-2\,x}-1\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{400}-\frac {\frac {297\,\left (\sqrt {1-2\,x}-1\right )}{3125\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {2559\,{\left (\sqrt {1-2\,x}-1\right )}^3}{1250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}+\frac {2559\,{\left (\sqrt {1-2\,x}-1\right )}^5}{500\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^5}-\frac {297\,{\left (\sqrt {1-2\,x}-1\right )}^7}{200\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^7}+\frac {288\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {192\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^4}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {72\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^6}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}}{\frac {32\,{\left (\sqrt {1-2\,x}-1\right )}^2}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {24\,{\left (\sqrt {1-2\,x}-1\right )}^4}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {8\,{\left (\sqrt {1-2\,x}-1\right )}^6}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^8}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {16}{625}}-\frac {\frac {2\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {4\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {16\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}}{\frac {4\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {4}{25}} \]
(1177*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/400 - ((297*((1 - 2*x)^(1/2) - 1))/(3125*(3^(1/2) - (5*x + 3) ^(1/2))) - (2559*((1 - 2*x)^(1/2) - 1)^3)/(1250*(3^(1/2) - (5*x + 3)^(1/2) )^3) + (2559*((1 - 2*x)^(1/2) - 1)^5)/(500*(3^(1/2) - (5*x + 3)^(1/2))^5) - (297*((1 - 2*x)^(1/2) - 1)^7)/(200*(3^(1/2) - (5*x + 3)^(1/2))^7) + (288 *3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (1 92*3^(1/2)*((1 - 2*x)^(1/2) - 1)^4)/(125*(3^(1/2) - (5*x + 3)^(1/2))^4) + (72*3^(1/2)*((1 - 2*x)^(1/2) - 1)^6)/(25*(3^(1/2) - (5*x + 3)^(1/2))^6))/( (32*((1 - 2*x)^(1/2) - 1)^2)/(125*(3^(1/2) - (5*x + 3)^(1/2))^2) + (24*((1 - 2*x)^(1/2) - 1)^4)/(25*(3^(1/2) - (5*x + 3)^(1/2))^4) + (8*((1 - 2*x)^( 1/2) - 1)^6)/(5*(3^(1/2) - (5*x + 3)^(1/2))^6) + ((1 - 2*x)^(1/2) - 1)^8/( 3^(1/2) - (5*x + 3)^(1/2))^8 + 16/625) - ((2*((1 - 2*x)^(1/2) - 1)^3)/(5*( 3^(1/2) - (5*x + 3)^(1/2))^3) - (4*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - ( 5*x + 3)^(1/2))) + (16*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x + 3)^(1/2))^2))/((4*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x + 3)^(1/2 ))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + 4/25)