Integrand size = 24, antiderivative size = 94 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=\frac {407}{800} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {37}{80} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{3/2} (3+5 x)^{3/2}+\frac {4477 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{800 \sqrt {10}} \]
-1/10*(1-2*x)^(3/2)*(3+5*x)^(3/2)+4477/8000*arcsin(1/11*22^(1/2)*(3+5*x)^( 1/2))*10^(1/2)-37/80*(1-2*x)^(3/2)*(3+5*x)^(1/2)+407/800*(1-2*x)^(1/2)*(3+ 5*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=\frac {10 \sqrt {1-2 x} \left (-609+1445 x+6500 x^2+4000 x^3\right )-4477 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{8000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(-609 + 1445*x + 6500*x^2 + 4000*x^3) - 4477*Sqrt[30 + 5 0*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(8000*Sqrt[3 + 5*x])
Time = 0.18 (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.
| \(\displaystyle \int \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {37}{20} \int \sqrt {1-2 x} \sqrt {5 x+3}dx-\frac {1}{10} (1-2 x)^{3/2} (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {37}{20} \left (\frac {11}{8} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {1}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {1}{10} (1-2 x)^{3/2} (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {37}{20} \left (\frac {11}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {1}{10} (1-2 x)^{3/2} (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {37}{20} \left (\frac {11}{8} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {1}{10} (1-2 x)^{3/2} (5 x+3)^{3/2}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {37}{20} \left (\frac {11}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {1}{10} (1-2 x)^{3/2} (5 x+3)^{3/2}\) |
-1/10*((1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) + (37*(-1/4*((1 - 2*x)^(3/2)*Sqrt[ 3 + 5*x]) + (11*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*S qrt[3 + 5*x]])/(5*Sqrt[10])))/8))/20
3.23.65.3.1 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 = 1.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
| method | result | size |
| default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (16000 x^{2} \sqrt {-10 x^{2}-x +3}+4477 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+16400 x \sqrt {-10 x^{2}-x +3}-4060 \sqrt {-10 x^{2}-x +3}\right )}{16000 \sqrt {-10 x^{2}-x +3}}\) | \(87\) |
| risch | \(-\frac {\left (800 x^{2}+820 x -203\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{800 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {4477 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{16000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(98\) |
1/16000*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(16000*x^2*(-10*x^2-x+3)^(1/2)+4477*10 ^(1/2)*arcsin(20/11*x+1/11)+16400*x*(-10*x^2-x+3)^(1/2)-4060*(-10*x^2-x+3) ^(1/2))/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=\frac {1}{800} \, {\left (800 \, x^{2} + 820 \, x - 203\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {4477}{16000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
1/800*(800*x^2 + 820*x - 203)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 4477/16000*sq rt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^ 2 + x - 3))
\[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=\int \sqrt {1 - 2 x} \left (3 x + 2\right ) \sqrt {5 x + 3}\, dx \]
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.59 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=-\frac {1}{10} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {37}{40} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {4477}{16000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {37}{800} \, \sqrt {-10 \, x^{2} - x + 3} \]
-1/10*(-10*x^2 - x + 3)^(3/2) + 37/40*sqrt(-10*x^2 - x + 3)*x - 4477/16000 *sqrt(10)*arcsin(-20/11*x - 1/11) + 37/800*sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (67) = 134\).
Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=\frac {1}{8000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {19}{2000} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {3}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \]
1/8000*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))) + 19/2000*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))) + 3/25*sqrt(5)*(11*sqrt(2)*arcsin(1/11*sqrt(22)*s qrt(5*x + 3)) + 2*sqrt(5*x + 3)*sqrt(-10*x + 5))
Time = 8.13 (sec) , antiderivative size = 588, normalized size of antiderivative = 6.26 \[ \int \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x} \, dx=2\,\sqrt {1-2\,x}\,\sqrt {5\,x+3}\,\left (\frac {x}{2}+\frac {1}{40}\right )-\frac {363\,\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 )}{4000}-\frac {\frac {7543\,{\left (\sqrt {1-2\,x}-1\right )}^3}{78125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {726\,\left (\sqrt {1-2\,x}-1\right )}{390625\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {19023\,{\left (\sqrt {1-2\,x}-1\right )}^5}{15625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^5}+\frac {19023\,{\left (\sqrt {1-2\,x}-1\right )}^7}{6250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^7}-\frac {7543\,{\left (\sqrt {1-2\,x}-1\right )}^9}{5000\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^9}+\frac {363\,{\left (\sqrt {1-2\,x}-1\right )}^{11}}{2000\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^{11}}+\frac {1152\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{15625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}-\frac {11136\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^4}{15625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {15936\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^6}{3125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}-\frac {2784\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^8}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {72\,\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}}-\frac {\sqrt {2}\,\sqrt {5}\,\ln \left (x+\frac {1}{20}-\frac {\sqrt {10}\,\sqrt {1-2\,x}\,\sqrt {5\,x+3}\,1{}\mathrm {i}}{10}\right )\,121{}\mathrm {i}}{400} \]
2*(1 - 2*x)^(1/2)*(5*x + 3)^(1/2)*(x/2 + 1/40) - (363*10^(1/2)*atan((10^(1 /2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/4000 - (2^(1/ 2)*5^(1/2)*log(x - (10^(1/2)*(1 - 2*x)^(1/2)*(5*x + 3)^(1/2)*1i)/10 + 1/20 )*121i)/400 - ((7543*((1 - 2*x)^(1/2) - 1)^3)/(78125*(3^(1/2) - (5*x + 3)^ (1/2))^3) - (726*((1 - 2*x)^(1/2) - 1))/(390625*(3^(1/2) - (5*x + 3)^(1/2) )) - (19023*((1 - 2*x)^(1/2) - 1)^5)/(15625*(3^(1/2) - (5*x + 3)^(1/2))^5) + (19023*((1 - 2*x)^(1/2) - 1)^7)/(6250*(3^(1/2) - (5*x + 3)^(1/2))^7) - (7543*((1 - 2*x)^(1/2) - 1)^9)/(5000*(3^(1/2) - (5*x + 3)^(1/2))^9) + (363 *((1 - 2*x)^(1/2) - 1)^11)/(2000*(3^(1/2) - (5*x + 3)^(1/2))^11) + (1152*3 ^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(15625*(3^(1/2) - (5*x + 3)^(1/2))^2) - (1 1136*3^(1/2)*((1 - 2*x)^(1/2) - 1)^4)/(15625*(3^(1/2) - (5*x + 3)^(1/2))^4 ) + (15936*3^(1/2)*((1 - 2*x)^(1/2) - 1)^6)/(3125*(3^(1/2) - (5*x + 3)^(1/ 2))^6) - (2784*3^(1/2)*((1 - 2*x)^(1/2) - 1)^8)/(625*(3^(1/2) - (5*x + 3)^ (1/2))^8) + (72*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)