Integrand size = 26, antiderivative size = 116 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {3+5 x}}+\frac {357 \sqrt {1-2 x} \sqrt {3+5 x}}{2000}+\frac {119 (1-2 x)^{3/2} \sqrt {3+5 x}}{2200}-\frac {3}{50} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {3927 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2000 \sqrt {10}} \]
3927/20000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/275*(1-2*x)^(5/2 )/(3+5*x)^(1/2)+119/2200*(1-2*x)^(3/2)*(3+5*x)^(1/2)-3/50*(1-2*x)^(5/2)*(3 +5*x)^(1/2)+357/2000*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.35 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {5 \sqrt {1-2 x} \left (1021+2575 x-180 x^2-2400 x^3\right )-3927 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{10000 \sqrt {3+5 x}} \]
(5*Sqrt[1 - 2*x]*(1021 + 2575*x - 180*x^2 - 2400*x^3) - 3927*Sqrt[30 + 50* x]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/(10000*Sqrt[3 + 5*x ])
Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {100, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^2}{(5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{275} \int \frac {5 (1-2 x)^{3/2} (99 x+71)}{2 \sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{55} \int \frac {(1-2 x)^{3/2} (99 x+71)}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{55} \left (\frac {119}{4} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx-\frac {33}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{55} \left (\frac {119}{4} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {33}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{55} \left (\frac {119}{4} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {33}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{55} \left (\frac {119}{4} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {33}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{55} \left (\frac {119}{4} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {33}{10} (1-2 x)^{5/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{5/2}}{275 \sqrt {5 x+3}}\) |
(-2*(1 - 2*x)^(5/2))/(275*Sqrt[3 + 5*x]) + ((-33*(1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/10 + (119*(((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)/55
3.24.72.3.1 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*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 = 1.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\left (-48000 x^{3} \sqrt {-10 x^{2}-x +3}+19635 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -3600 x^{2} \sqrt {-10 x^{2}-x +3}+11781 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+51500 x \sqrt {-10 x^{2}-x +3}+20420 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{40000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(116\) |
1/40000*(-48000*x^3*(-10*x^2-x+3)^(1/2)+19635*10^(1/2)*arcsin(20/11*x+1/11 )*x-3600*x^2*(-10*x^2-x+3)^(1/2)+11781*10^(1/2)*arcsin(20/11*x+1/11)+51500 *x*(-10*x^2-x+3)^(1/2)+20420*(-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 = 86, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {3927 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (2400 \, x^{3} + 180 \, x^{2} - 2575 \, x - 1021\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{40000 \, {\left (5 \, x + 3\right )}} \]
-1/40000*(3927*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*(2400*x^3 + 180*x^2 - 2575*x - 1021)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
\[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}} \left (3 x + 2\right )^{2}}{\left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.33 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {11979}{200000} i \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {23}{11}\right ) + \frac {957}{25000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {3}{125} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {99}{500} \, \sqrt {10 \, x^{2} + 23 \, x + \frac {51}{5}} x + \frac {2277}{10000} \, \sqrt {10 \, x^{2} + 23 \, x + \frac {51}{5}} + \frac {99}{1250} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{125 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{125 \, {\left (5 \, x + 3\right )}} - \frac {33 \, \sqrt {-10 \, x^{2} - x + 3}}{625 \, {\left (5 \, x + 3\right )}} \]
-11979/200000*I*sqrt(5)*sqrt(2)*arcsin(20/11*x + 23/11) + 957/25000*sqrt(5 )*sqrt(2)*arcsin(20/11*x + 1/11) + 3/125*(-10*x^2 - x + 3)^(3/2) + 99/500* sqrt(10*x^2 + 23*x + 51/5)*x + 2277/10000*sqrt(10*x^2 + 23*x + 51/5) + 99/ 1250*sqrt(-10*x^2 - x + 3) + 1/125*(-10*x^2 - x + 3)^(3/2)/(25*x^2 + 30*x + 9) + 3/125*(-10*x^2 - x + 3)^(3/2)/(5*x + 3) - 33/625*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Time = 0.49 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {1}{50000} \, {\left (12 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} - 69 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 199 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {3927}{20000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {11 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{6250 \, \sqrt {5 \, x + 3}} + \frac {22 \, \sqrt {10} \sqrt {5 \, x + 3}}{3125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
-1/50000*(12*(8*sqrt(5)*(5*x + 3) - 69*sqrt(5))*(5*x + 3) - 199*sqrt(5))*s qrt(5*x + 3)*sqrt(-10*x + 5) + 3927/20000*sqrt(10)*arcsin(1/11*sqrt(22)*sq rt(5*x + 3)) - 11/6250*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt( 5*x + 3) + 22/3125*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt( 22))
Timed out. \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}\,{\left (3\,x+2\right )}^2}{{\left (5\,x+3\right )}^{3/2}} \,d x \]