Integrand size = 26, antiderivative size = 94 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {3+5 x}}+\frac {317 \sqrt {1-2 x} \sqrt {3+5 x}}{2200}-\frac {9}{100} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {317 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{200 \sqrt {10}} \] Output:
-2/275*(1-2*x)^(3/2)/(3+5*x)^(1/2)+317/2200*(1-2*x)^(1/2)*(3+5*x)^(1/2)-9/ 100*(1-2*x)^(3/2)*(3+5*x)^(1/2)+317/2000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2 ))*10^(1/2)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {10 \sqrt {1-2 x} \left (31+165 x+180 x^2\right )-317 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2000 \sqrt {3+5 x}} \] Input:
Integrate[(Sqrt[1 - 2*x]*(2 + 3*x)^2)/(3 + 5*x)^(3/2),x]
Output:
(10*Sqrt[1 - 2*x]*(31 + 165*x + 180*x^2) - 317*Sqrt[30 + 50*x]*ArcTan[Sqrt [5/2 - 5*x]/Sqrt[3 + 5*x]])/(2000*Sqrt[3 + 5*x])
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {100, 27, 90, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^2}{(5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{275} \int \frac {\sqrt {1-2 x} (495 x+359)}{2 \sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{275} \int \frac {\sqrt {1-2 x} (495 x+359)}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{275} \left (\frac {1585}{8} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {99}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{275} \left (\frac {1585}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {99}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{275} \left (\frac {1585}{8} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {99}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{275} \left (\frac {1585}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {99}{4} (1-2 x)^{3/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{275 \sqrt {5 x+3}}\) |
Input:
Int[(Sqrt[1 - 2*x]*(2 + 3*x)^2)/(3 + 5*x)^(3/2),x]
Output:
(-2*(1 - 2*x)^(3/2))/(275*Sqrt[3 + 5*x]) + ((-99*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/4 + (1585*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*S qrt[3 + 5*x]])/(5*Sqrt[10])))/8)/275
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 = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {\left (1585 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +3600 x^{2} \sqrt {-10 x^{2}-x +3}+951 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+3300 x \sqrt {-10 x^{2}-x +3}+620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{4000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(99\) |
Input:
int((1-2*x)^(1/2)*(2+3*x)^2/(3+5*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4000*(1585*10^(1/2)*arcsin(20/11*x+1/11)*x+3600*x^2*(-10*x^2-x+3)^(1/2)+ 951*10^(1/2)*arcsin(20/11*x+1/11)+3300*x*(-10*x^2-x+3)^(1/2)+620*(-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 = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {317 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (180 \, x^{2} + 165 \, x + 31\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4000 \, {\left (5 \, x + 3\right )}} \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)^2/(3+5*x)^(3/2),x, algorithm="fricas")
Output:
-1/4000*(317*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*(180*x^2 + 165*x + 31)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
\[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {\sqrt {1 - 2 x} \left (3 x + 2\right )^{2}}{\left (5 x + 3\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((1-2*x)**(1/2)*(2+3*x)**2/(3+5*x)**(3/2),x)
Output:
Integral(sqrt(1 - 2*x)*(3*x + 2)**2/(5*x + 3)**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {317}{4000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {9}{50} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {57}{1000} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{125 \, {\left (5 \, x + 3\right )}} \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)^2/(3+5*x)^(3/2),x, algorithm="maxima")
Output:
317/4000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 9/50*sqrt(-10*x^2 - x + 3)*x + 57/1000*sqrt(-10*x^2 - x + 3) - 2/125*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Time = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {3}{5000} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} - 17 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {317}{2000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{1250 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)^2/(3+5*x)^(3/2),x, algorithm="giac")
Output:
3/5000*(12*sqrt(5)*(5*x + 3) - 17*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 317/2000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/1250*sqrt(10)*( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/625*sqrt(10)*sqrt(5* x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^2}{{\left (5\,x+3\right )}^{3/2}} \,d x \] Input:
int(((1 - 2*x)^(1/2)*(3*x + 2)^2)/(5*x + 3)^(3/2),x)
Output:
int(((1 - 2*x)^(1/2)*(3*x + 2)^2)/(5*x + 3)^(3/2), x)
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {-317 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+1800 \sqrt {-2 x +1}\, x^{2}+1650 \sqrt {-2 x +1}\, x +310 \sqrt {-2 x +1}}{2000 \sqrt {5 x +3}} \] Input:
int((1-2*x)^(1/2)*(2+3*x)^2/(3+5*x)^(3/2),x)
Output:
( - 317*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 1800*sqrt( - 2*x + 1)*x**2 + 1650*sqrt( - 2*x + 1)*x + 310*sqrt( - 2*x + 1))/(2000*sqrt(5*x + 3))