Integrand size = 24, antiderivative size = 94 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {99}{400} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {3}{40} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {1089 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{400 \sqrt {10}} \]
1089/4000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+3/40*(1-2*x)^(3/2)* (3+5*x)^(1/2)-1/10*(1-2*x)^(5/2)*(3+5*x)^(1/2)+99/400*(1-2*x)^(1/2)*(3+5*x )^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {10 \sqrt {1-2 x} \left (267+745 x+20 x^2-800 x^3\right )-1089 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{4000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(267 + 745*x + 20*x^2 - 800*x^3) - 1089*Sqrt[30 + 50*x]* ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(4000*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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)}{\sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {3}{4} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx-\frac {1}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{4} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {1}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{4} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {1}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {3}{4} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {1}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {3}{4} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {1}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\) |
-1/10*((1 - 2*x)^(5/2)*Sqrt[3 + 5*x]) + (3*(((1 - 2*x)^(3/2)*Sqrt[3 + 5*x] )/10 + (33*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/20))/4
3.24.63.3.1 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 = 1.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (-3200 x^{2} \sqrt {-10 x^{2}-x +3}+1089 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+2000 x \sqrt {-10 x^{2}-x +3}+1780 \sqrt {-10 x^{2}-x +3}\right )}{8000 \sqrt {-10 x^{2}-x +3}}\) | \(87\) |
risch | \(\frac {\left (160 x^{2}-100 x -89\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{400 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {1089 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{8000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(98\) |
1/8000*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(-3200*x^2*(-10*x^2-x+3)^(1/2)+1089*10^ (1/2)*arcsin(20/11*x+1/11)+2000*x*(-10*x^2-x+3)^(1/2)+1780*(-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 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=-\frac {1}{400} \, {\left (160 \, x^{2} - 100 \, x - 89\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {1089}{8000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/400*(160*x^2 - 100*x - 89)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 1089/8000*sqr t(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}} \cdot \left (3 x + 2\right )}{\sqrt {5 x + 3}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=-\frac {2}{5} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + \frac {1}{4} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {1089}{8000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {89}{400} \, \sqrt {-10 \, x^{2} - x + 3} \]
-2/5*sqrt(-10*x^2 - x + 3)*x^2 + 1/4*sqrt(-10*x^2 - x + 3)*x - 1089/8000*s qrt(10)*arcsin(-20/11*x - 1/11) + 89/400*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 = 0.35 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=-\frac {1}{20000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{2000} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \]
-1/20000*sqrt(5)*(2*(4*(40*x - 59)*(5*x + 3) + 1293)*sqrt(5*x + 3)*sqrt(-1 0*x + 5) + 4785*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) - 1/2000*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))) + 1/25*sqrt(5)*(11*sqrt(2)*arcsin(1/11*sqrt(22)* sqrt(5*x + 3)) + 2*sqrt(5*x + 3)*sqrt(-10*x + 5))
Timed out. \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )}{\sqrt {5\,x+3}} \,d x \]