Integrand size = 26, antiderivative size = 121 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=\frac {9933 \sqrt {1-2 x} \sqrt {3+5 x}}{32000}+\frac {301 (1-2 x)^{3/2} \sqrt {3+5 x}}{3200}-\frac {119}{800} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {3}{40} (1-2 x)^{5/2} (2+3 x) \sqrt {3+5 x}+\frac {109263 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{32000 \sqrt {10}} \]
109263/320000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+301/3200*(1-2*x )^(3/2)*(3+5*x)^(1/2)-119/800*(1-2*x)^(5/2)*(3+5*x)^(1/2)-3/40*(1-2*x)^(5/ 2)*(2+3*x)*(3+5*x)^(1/2)+9933/32000*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=\frac {10 \sqrt {1-2 x} \left (10149+91975 x+96780 x^2-133600 x^3-144000 x^4\right )-109263 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{320000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(10149 + 91975*x + 96780*x^2 - 133600*x^3 - 144000*x^4) - 109263*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(320000*Sq rt[3 + 5*x])
Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {101, 27, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^2}{\sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 101 |
\(\displaystyle -\frac {1}{40} \int -\frac {7 (1-2 x)^{3/2} (51 x+32)}{2 \sqrt {5 x+3}}dx-\frac {3}{40} (3 x+2) \sqrt {5 x+3} (1-2 x)^{5/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{80} \int \frac {(1-2 x)^{3/2} (51 x+32)}{\sqrt {5 x+3}}dx-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {7}{80} \left (\frac {43}{4} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx-\frac {17}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7}{80} \left (\frac {43}{4} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {17}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {7}{80} \left (\frac {43}{4} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {17}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7}{80} \left (\frac {43}{4} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {17}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7}{80} \left (\frac {43}{4} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {17}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {3}{40} (1-2 x)^{5/2} (3 x+2) \sqrt {5 x+3}\) |
(-3*(1 - 2*x)^(5/2)*(2 + 3*x)*Sqrt[3 + 5*x])/40 + (7*((-17*(1 - 2*x)^(5/2) *Sqrt[3 + 5*x])/10 + (43*(((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))/80
3.24.62.3.1 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 = 1.13 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {\left (28800 x^{3}+9440 x^{2}-25020 x -3383\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{32000 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {109263 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{640000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) | \(103\) |
default | \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (-576000 x^{3} \sqrt {-10 x^{2}-x +3}-188800 x^{2} \sqrt {-10 x^{2}-x +3}+109263 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+500400 x \sqrt {-10 x^{2}-x +3}+67660 \sqrt {-10 x^{2}-x +3}\right )}{640000 \sqrt {-10 x^{2}-x +3}}\) | \(104\) |
1/32000*(28800*x^3+9440*x^2-25020*x-3383)*(-1+2*x)*(3+5*x)^(1/2)/(-(-1+2*x )*(3+5*x))^(1/2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+109263/640000*10^(1 /2)*arcsin(20/11*x+1/11)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^(1/ 2)
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.60 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=-\frac {1}{32000} \, {\left (28800 \, x^{3} + 9440 \, x^{2} - 25020 \, x - 3383\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {109263}{640000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]
-1/32000*(28800*x^3 + 9440*x^2 - 25020*x - 3383)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 109263/640000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3) *sqrt(-2*x + 1)/(10*x^2 + x - 3))
\[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}} \left (3 x + 2\right )^{2}}{\sqrt {5 x + 3}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=-\frac {9}{10} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {59}{200} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + \frac {1251}{1600} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {109263}{640000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {3383}{32000} \, \sqrt {-10 \, x^{2} - x + 3} \]
-9/10*sqrt(-10*x^2 - x + 3)*x^3 - 59/200*sqrt(-10*x^2 - x + 3)*x^2 + 1251/ 1600*sqrt(-10*x^2 - x + 3)*x - 109263/640000*sqrt(10)*arcsin(-20/11*x - 1/ 11) + 3383/32000*sqrt(-10*x^2 - x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (88) = 176\).
Time = 0.40 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.68 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=-\frac {3}{1600000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (60 \, x - 119\right )} {\left (5 \, x + 3\right )} + 6163\right )} {\left (5 \, x + 3\right )} - 66189\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 184305 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{8000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{500} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {2}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \]
-3/1600000*sqrt(5)*(2*(4*(8*(60*x - 119)*(5*x + 3) + 6163)*(5*x + 3) - 661 89)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 184305*sqrt(2)*arcsin(1/11*sqrt(22)*sq rt(5*x + 3))) - 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))) + 1/500*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))) + 2/25*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. \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{\sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}\,{\left (3\,x+2\right )}^2}{\sqrt {5\,x+3}} \,d x \]