Integrand size = 26, antiderivative size = 72 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {2 \sqrt {1-2 x}}{275 \sqrt {3+5 x}}-\frac {9}{50} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {123 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}} \]
123/500*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/275*(1-2*x)^(1/2)/( 3+5*x)^(1/2)-9/50*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\frac {-10 \sqrt {1-2 x} (301+495 x)-1353 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{5500 \sqrt {3+5 x}} \]
(-10*Sqrt[1 - 2*x]*(301 + 495*x) - 1353*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(5500*Sqrt[3 + 5*x])
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 27, 90, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{275} \int \frac {33 (15 x+11)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{25} \int \frac {15 x+11}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {3}{25} \left (\frac {41}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {3}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 \sqrt {1-2 x}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {3}{25} \left (\frac {41}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {3}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 \sqrt {1-2 x}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {3}{25} \left (\frac {41 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {3}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 \sqrt {1-2 x}}{275 \sqrt {5 x+3}}\) |
(-2*Sqrt[1 - 2*x])/(275*Sqrt[3 + 5*x]) + (3*((-3*Sqrt[1 - 2*x]*Sqrt[3 + 5* x])/2 + (41*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/25
3.25.100.3.1 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[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*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 1.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\left (6765 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +4059 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-9900 x \sqrt {-10 x^{2}-x +3}-6020 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{11000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(82\) |
1/11000*(6765*10^(1/2)*arcsin(20/11*x+1/11)*x+4059*10^(1/2)*arcsin(20/11*x +1/11)-9900*x*(-10*x^2-x+3)^(1/2)-6020*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/ (-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {1353 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (495 \, x + 301\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{11000 \, {\left (5 \, x + 3\right )}} \]
-1/11000*(1353*sqrt(10)*(5*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(495*x + 301)*sqrt(5*x + 3)*sq rt(-2*x + 1))/(5*x + 3)
\[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{2}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\frac {123}{1000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {9}{50} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{275 \, {\left (5 \, x + 3\right )}} \]
123/1000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 9/50*sqrt(-10*x^2 - x + 3) - 2/275*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {9}{250} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {123}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{2750 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{1375 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
-9/250*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 123/500*sqrt(10)*arcsin(1/1 1*sqrt(22)*sqrt(5*x + 3)) - 1/2750*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqr t(22))/sqrt(5*x + 3) + 2/1375*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^2}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{3/2}} \,d x \]