Integrand size = 26, antiderivative size = 99 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {6269 \sqrt {1-2 x} \sqrt {3+5 x}}{1600}-\frac {181}{400} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {1}{10} \sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}+\frac {68959 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1600 \sqrt {10}} \]
68959/16000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-181/400*(3+5*x)^( 3/2)*(1-2*x)^(1/2)-1/10*(2+3*x)*(3+5*x)^(3/2)*(1-2*x)^(1/2)-6269/1600*(1-2 *x)^(1/2)*(3+5*x)^(1/2)
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {-\sqrt {5-10 x} \sqrt {3+5 x} \left (9401+6660 x+2400 x^2\right )-68959 \sqrt {2} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{1600 \sqrt {5}} \]
(-(Sqrt[5 - 10*x]*Sqrt[3 + 5*x]*(9401 + 6660*x + 2400*x^2)) - 68959*Sqrt[2 ]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/(1600*Sqrt[5])
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {101, 27, 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.
| \(\displaystyle \int \frac {(3 x+2)^2 \sqrt {5 x+3}}{\sqrt {1-2 x}} \, dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle -\frac {1}{30} \int -\frac {3 \sqrt {5 x+3} (181 x+116)}{2 \sqrt {1-2 x}}dx-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{20} \int \frac {\sqrt {5 x+3} (181 x+116)}{\sqrt {1-2 x}}dx-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{20} \left (\frac {6269}{40} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {181}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{20} \left (\frac {6269}{40} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {181}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{20} \left (\frac {6269}{40} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {181}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{20} \left (\frac {6269}{40} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {181}{20} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{10} \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}\) |
-1/10*(Sqrt[1 - 2*x]*(2 + 3*x)*(3 + 5*x)^(3/2)) + ((-181*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/20 + (6269*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[ Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/40)/20
3.25.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 = 1.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88
| method | result | size |
| default | \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-48000 x^{2} \sqrt {-10 x^{2}-x +3}+68959 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-133200 x \sqrt {-10 x^{2}-x +3}-188020 \sqrt {-10 x^{2}-x +3}\right )}{32000 \sqrt {-10 x^{2}-x +3}}\) | \(87\) |
| risch | \(\frac {\left (2400 x^{2}+6660 x +9401\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{1600 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {68959 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{32000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(98\) |
1/32000*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(-48000*x^2*(-10*x^2-x+3)^(1/2)+68959* 10^(1/2)*arcsin(20/11*x+1/11)-133200*x*(-10*x^2-x+3)^(1/2)-188020*(-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 = 0.68 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{1600} \, {\left (2400 \, x^{2} + 6660 \, x + 9401\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {68959}{32000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/1600*(2400*x^2 + 6660*x + 9401)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 68959/32 000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/ (10*x^2 + x - 3))
\[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right )^{2} \sqrt {5 x + 3}}{\sqrt {1 - 2 x}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {68959}{32000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3}{20} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {321}{80} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {10121}{1600} \, \sqrt {-10 \, x^{2} - x + 3} \]
68959/32000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 3/20*(-10*x^2 - x + 3 )^(3/2) - 321/80*sqrt(-10*x^2 - x + 3)*x - 10121/1600*sqrt(-10*x^2 - x + 3 )
Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.55 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{16000} \, \sqrt {5} {\left (2 \, {\left (12 \, {\left (40 \, x + 87\right )} {\left (5 \, x + 3\right )} + 6269\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 68959 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]
-1/16000*sqrt(5)*(2*(12*(40*x + 87)*(5*x + 3) + 6269)*sqrt(5*x + 3)*sqrt(- 10*x + 5) - 68959*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)))
Time = 8.09 (sec) , antiderivative size = 534, normalized size of antiderivative = 5.39 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {68959\,\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 )}{8000}-\frac {\frac {30559\,\left (\sqrt {1-2\,x}-1\right )}{390625\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {248379\,{\left (\sqrt {1-2\,x}-1\right )}^3}{156250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {70541\,{\left (\sqrt {1-2\,x}-1\right )}^5}{31250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^5}+\frac {70541\,{\left (\sqrt {1-2\,x}-1\right )}^7}{12500\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^7}+\frac {248379\,{\left (\sqrt {1-2\,x}-1\right )}^9}{10000\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^9}-\frac {30559\,{\left (\sqrt {1-2\,x}-1\right )}^{11}}{4000\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^{11}}+\frac {7168\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{15625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {95104\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^4}{15625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {32256\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^6}{3125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}+\frac {23776\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^8}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {448\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^{10}}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^{10}}}{\frac {192\,{\left (\sqrt {1-2\,x}-1\right )}^2}{3125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {48\,{\left (\sqrt {1-2\,x}-1\right )}^4}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {32\,{\left (\sqrt {1-2\,x}-1\right )}^6}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}+\frac {12\,{\left (\sqrt {1-2\,x}-1\right )}^8}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {12\,{\left (\sqrt {1-2\,x}-1\right )}^{10}}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^{10}}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^{12}}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^{12}}+\frac {64}{15625}} \]
(68959*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/8000 - ((30559*((1 - 2*x)^(1/2) - 1))/(390625*(3^(1/2) - (5* x + 3)^(1/2))) - (248379*((1 - 2*x)^(1/2) - 1)^3)/(156250*(3^(1/2) - (5*x + 3)^(1/2))^3) - (70541*((1 - 2*x)^(1/2) - 1)^5)/(31250*(3^(1/2) - (5*x + 3)^(1/2))^5) + (70541*((1 - 2*x)^(1/2) - 1)^7)/(12500*(3^(1/2) - (5*x + 3) ^(1/2))^7) + (248379*((1 - 2*x)^(1/2) - 1)^9)/(10000*(3^(1/2) - (5*x + 3)^ (1/2))^9) - (30559*((1 - 2*x)^(1/2) - 1)^11)/(4000*(3^(1/2) - (5*x + 3)^(1 /2))^11) + (7168*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(15625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (95104*3^(1/2)*((1 - 2*x)^(1/2) - 1)^4)/(15625*(3^(1/2) - (5*x + 3)^(1/2))^4) + (32256*3^(1/2)*((1 - 2*x)^(1/2) - 1)^6)/(3125*(3^(1/ 2) - (5*x + 3)^(1/2))^6) + (23776*3^(1/2)*((1 - 2*x)^(1/2) - 1)^8)/(625*(3 ^(1/2) - (5*x + 3)^(1/2))^8) + (448*3^(1/2)*((1 - 2*x)^(1/2) - 1)^10)/(25* (3^(1/2) - (5*x + 3)^(1/2))^10))/((192*((1 - 2*x)^(1/2) - 1)^2)/(3125*(3^( 1/2) - (5*x + 3)^(1/2))^2) + (48*((1 - 2*x)^(1/2) - 1)^4)/(125*(3^(1/2) - (5*x + 3)^(1/2))^4) + (32*((1 - 2*x)^(1/2) - 1)^6)/(25*(3^(1/2) - (5*x + 3 )^(1/2))^6) + (12*((1 - 2*x)^(1/2) - 1)^8)/(5*(3^(1/2) - (5*x + 3)^(1/2))^ 8) + (12*((1 - 2*x)^(1/2) - 1)^10)/(5*(3^(1/2) - (5*x + 3)^(1/2))^10) + (( 1 - 2*x)^(1/2) - 1)^12/(3^(1/2) - (5*x + 3)^(1/2))^12 + 64/15625)